· ........................................ determinant prperties of thr determinant solution...
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MAT2305 LINEAR AL :Week4
Thanatyod Jampawai, Ph.D.
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 1 / 40
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Review Week 3
Diagonal MatrixSymmetric Matrix(Upper and Lower) Triangular MatrixLU DecompositionPermutation and InversionEven and OddElementary ProductSigned Elementary ProductDeterminat Function
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 2 / 40
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Outline for Week 3
Chapter 3 Determinant3.3 Properties of the Determinant3.4 Adjoint of a Matrix3.5 Cramer’s Rule
Assignment 4
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 3 / 40
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Determinant Prperties of thr Determinant
3.3 Properties of the Determinant
Theorem (3.3.1 Additional Determinant)
Let A,B and C be n× n matrices that differ only in a single row, say the rth,and assume that the rth row of C can be obtained by adding correspondingentries in the rth rows of A and B. Then
det(C) = det(A) + det(B)
∣∣∣∣a bc d
∣∣∣∣+ ∣∣∣∣a bx y
∣∣∣∣ = (ad− bc) + (ay − bx) = a(d+ y)− b(c+ x) =
∣∣∣∣ a bc+ x d+ y
∣∣∣∣
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 4 / 40
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Determinant Prperties of thr Determinant
3.3 Properties of the Determinant
Theorem (3.3.1 Additional Determinant)
Let A,B and C be n× n matrices that differ only in a single row, say the rth,and assume that the rth row of C can be obtained by adding correspondingentries in the rth rows of A and B. Then
det(C) = det(A) + det(B)
∣∣∣∣a bc d
∣∣∣∣+ ∣∣∣∣a bx y
∣∣∣∣ = (ad− bc) + (ay − bx) = a(d+ y)− b(c+ x) =
∣∣∣∣ a bc+ x d+ y
∣∣∣∣
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 4 / 40
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Determinant Prperties of thr Determinant
Example (3.3.1)
By evaluating the determinant, check
det
1 7 52 0 3
1 + 0 4 + 1 7 + (−1)
= det
1 7 52 0 31 4 7
+ det
1 7 52 0 30 1 −1
Solution
∣∣∣∣∣∣1 7 52 0 3
1 + 0 4 + 1 7 + (−1)
∣∣∣∣∣∣ =∣∣∣∣∣∣1 7 52 0 31 5 6
∣∣∣∣∣∣ = −28
∣∣∣∣∣∣1 7 52 0 31 4 7
∣∣∣∣∣∣+∣∣∣∣∣∣1 7 52 0 30 1 −1
∣∣∣∣∣∣ = −49 + 21 = −28
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 5 / 40
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Determinant Prperties of thr Determinant
Example (3.3.1)
By evaluating the determinant, check
det
1 7 52 0 3
1 + 0 4 + 1 7 + (−1)
= det
1 7 52 0 31 4 7
+ det
1 7 52 0 30 1 −1
Solution
∣∣∣∣∣∣1 7 52 0 3
1 + 0 4 + 1 7 + (−1)
∣∣∣∣∣∣ =∣∣∣∣∣∣1 7 52 0 31 5 6
∣∣∣∣∣∣ = −28
∣∣∣∣∣∣1 7 52 0 31 4 7
∣∣∣∣∣∣+∣∣∣∣∣∣1 7 52 0 30 1 −1
∣∣∣∣∣∣ = −49 + 21 = −28
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 5 / 40
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Determinant Prperties of thr Determinant
Scalar Multiplication to Determinant
Theorem (3.3.2)
Let A be an n× n matrix and k be any scalar. Then
det(kA) = kn det(A)
Example (3.3.2)
Evaluate the determinant of the matrix
det(2A), det(−A), det(3AT ) and det(−(−2A)T )
if det(A) = −5 for 3× 3 matrix A.
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 6 / 40
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Determinant Prperties of thr Determinant
Scalar Multiplication to Determinant
Theorem (3.3.2)
Let A be an n× n matrix and k be any scalar. Then
det(kA) = kn det(A)
Example (3.3.2)
Evaluate the determinant of the matrix
det(2A), det(−A), det(3AT ) and det(−(−2A)T )
if det(A) = −5 for 3× 3 matrix A.
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 6 / 40
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Determinant Prperties of thr Determinant
Solution
det(2A)
= 23 det(A) = 8(−5) = −40 #
det(−A) = det((−1)A) = (−1)3 det(A)= −1(−5) = 5 #
det(3AT ) = 33 det(AT )= 27 det(A)= 27(−5) = −135 #
det(−(−2A)T ) = (−1)3 det((−2A)T )= (−1) det(−2AT )= (−1)(−2)3 det(AT )
= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40
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Determinant Prperties of thr Determinant
Solution
det(2A) = 23 det(A)
= 8(−5) = −40 #
det(−A) = det((−1)A) = (−1)3 det(A)= −1(−5) = 5 #
det(3AT ) = 33 det(AT )= 27 det(A)= 27(−5) = −135 #
det(−(−2A)T ) = (−1)3 det((−2A)T )= (−1) det(−2AT )= (−1)(−2)3 det(AT )
= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40
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Determinant Prperties of thr Determinant
Solution
det(2A) = 23 det(A) = 8(−5) = −40 #
det(−A) = det((−1)A) = (−1)3 det(A)= −1(−5) = 5 #
det(3AT ) = 33 det(AT )= 27 det(A)= 27(−5) = −135 #
det(−(−2A)T ) = (−1)3 det((−2A)T )= (−1) det(−2AT )= (−1)(−2)3 det(AT )
= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40
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Determinant Prperties of thr Determinant
Solution
det(2A) = 23 det(A) = 8(−5) = −40 #
det(−A)
= det((−1)A) = (−1)3 det(A)= −1(−5) = 5 #
det(3AT ) = 33 det(AT )= 27 det(A)= 27(−5) = −135 #
det(−(−2A)T ) = (−1)3 det((−2A)T )= (−1) det(−2AT )= (−1)(−2)3 det(AT )
= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40
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Determinant Prperties of thr Determinant
Solution
det(2A) = 23 det(A) = 8(−5) = −40 #
det(−A) = det((−1)A)
= (−1)3 det(A)= −1(−5) = 5 #
det(3AT ) = 33 det(AT )= 27 det(A)= 27(−5) = −135 #
det(−(−2A)T ) = (−1)3 det((−2A)T )= (−1) det(−2AT )= (−1)(−2)3 det(AT )
= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40
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Determinant Prperties of thr Determinant
Solution
det(2A) = 23 det(A) = 8(−5) = −40 #
det(−A) = det((−1)A) = (−1)3 det(A)
= −1(−5) = 5 #
det(3AT ) = 33 det(AT )= 27 det(A)= 27(−5) = −135 #
det(−(−2A)T ) = (−1)3 det((−2A)T )= (−1) det(−2AT )= (−1)(−2)3 det(AT )
= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40
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Determinant Prperties of thr Determinant
Solution
det(2A) = 23 det(A) = 8(−5) = −40 #
det(−A) = det((−1)A) = (−1)3 det(A)= −1(−5) = 5 #
det(3AT ) = 33 det(AT )= 27 det(A)= 27(−5) = −135 #
det(−(−2A)T ) = (−1)3 det((−2A)T )= (−1) det(−2AT )= (−1)(−2)3 det(AT )
= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40
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Determinant Prperties of thr Determinant
Solution
det(2A) = 23 det(A) = 8(−5) = −40 #
det(−A) = det((−1)A) = (−1)3 det(A)= −1(−5) = 5 #
det(3AT )
= 33 det(AT )= 27 det(A)= 27(−5) = −135 #
det(−(−2A)T ) = (−1)3 det((−2A)T )= (−1) det(−2AT )= (−1)(−2)3 det(AT )
= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40
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Determinant Prperties of thr Determinant
Solution
det(2A) = 23 det(A) = 8(−5) = −40 #
det(−A) = det((−1)A) = (−1)3 det(A)= −1(−5) = 5 #
det(3AT ) = 33 det(AT )
= 27 det(A)= 27(−5) = −135 #
det(−(−2A)T ) = (−1)3 det((−2A)T )= (−1) det(−2AT )= (−1)(−2)3 det(AT )
= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40
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Determinant Prperties of thr Determinant
Solution
det(2A) = 23 det(A) = 8(−5) = −40 #
det(−A) = det((−1)A) = (−1)3 det(A)= −1(−5) = 5 #
det(3AT ) = 33 det(AT )= 27 det(A)
= 27(−5) = −135 #
det(−(−2A)T ) = (−1)3 det((−2A)T )= (−1) det(−2AT )= (−1)(−2)3 det(AT )
= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40
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Determinant Prperties of thr Determinant
Solution
det(2A) = 23 det(A) = 8(−5) = −40 #
det(−A) = det((−1)A) = (−1)3 det(A)= −1(−5) = 5 #
det(3AT ) = 33 det(AT )= 27 det(A)= 27(−5) = −135 #
det(−(−2A)T ) = (−1)3 det((−2A)T )= (−1) det(−2AT )= (−1)(−2)3 det(AT )
= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40
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Determinant Prperties of thr Determinant
Solution
det(2A) = 23 det(A) = 8(−5) = −40 #
det(−A) = det((−1)A) = (−1)3 det(A)= −1(−5) = 5 #
det(3AT ) = 33 det(AT )= 27 det(A)= 27(−5) = −135 #
det(−(−2A)T )
= (−1)3 det((−2A)T )= (−1) det(−2AT )= (−1)(−2)3 det(AT )
= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40
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Determinant Prperties of thr Determinant
Solution
det(2A) = 23 det(A) = 8(−5) = −40 #
det(−A) = det((−1)A) = (−1)3 det(A)= −1(−5) = 5 #
det(3AT ) = 33 det(AT )= 27 det(A)= 27(−5) = −135 #
det(−(−2A)T ) = (−1)3 det((−2A)T )
= (−1) det(−2AT )= (−1)(−2)3 det(AT )
= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40
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Determinant Prperties of thr Determinant
Solution
det(2A) = 23 det(A) = 8(−5) = −40 #
det(−A) = det((−1)A) = (−1)3 det(A)= −1(−5) = 5 #
det(3AT ) = 33 det(AT )= 27 det(A)= 27(−5) = −135 #
det(−(−2A)T ) = (−1)3 det((−2A)T )= (−1) det(−2AT )
= (−1)(−2)3 det(AT )
= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40
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Determinant Prperties of thr Determinant
Solution
det(2A) = 23 det(A) = 8(−5) = −40 #
det(−A) = det((−1)A) = (−1)3 det(A)= −1(−5) = 5 #
det(3AT ) = 33 det(AT )= 27 det(A)= 27(−5) = −135 #
det(−(−2A)T ) = (−1)3 det((−2A)T )= (−1) det(−2AT )= (−1)(−2)3 det(AT )
= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40
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Determinant Prperties of thr Determinant
Solution
det(2A) = 23 det(A) = 8(−5) = −40 #
det(−A) = det((−1)A) = (−1)3 det(A)= −1(−5) = 5 #
det(3AT ) = 33 det(AT )= 27 det(A)= 27(−5) = −135 #
det(−(−2A)T ) = (−1)3 det((−2A)T )= (−1) det(−2AT )= (−1)(−2)3 det(AT )
= (−1)(−8) det(A)
= (−1)(−8)(−5) = −40 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40
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Determinant Prperties of thr Determinant
Solution
det(2A) = 23 det(A) = 8(−5) = −40 #
det(−A) = det((−1)A) = (−1)3 det(A)= −1(−5) = 5 #
det(3AT ) = 33 det(AT )= 27 det(A)= 27(−5) = −135 #
det(−(−2A)T ) = (−1)3 det((−2A)T )= (−1) det(−2AT )= (−1)(−2)3 det(AT )
= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 7 / 40
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Determinant Prperties of thr Determinant
Invertible Test
Theorem (3.3.3)
If B is an n× n matrix and E is n× n elementary matrix, then
det(EB) = det(E) det(B)
Moreover, if E1, E2, ..., Er are n× n elementary matrices, then
det(E1E2 · · ·ErB) = det(E1) det(E2) · · · det(Er) det(B).
Theorem (3.3.4 Invertible Test)
A square matrix A is invertible if and only if det(A) ̸= 0 .
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 8 / 40
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Determinant Prperties of thr Determinant
Invertible Test
Theorem (3.3.3)
If B is an n× n matrix and E is n× n elementary matrix, then
det(EB) = det(E) det(B)
Moreover, if E1, E2, ..., Er are n× n elementary matrices, then
det(E1E2 · · ·ErB) = det(E1) det(E2) · · · det(Er) det(B).
Theorem (3.3.4 Invertible Test)
A square matrix A is invertible if and only if det(A) ̸= 0 .
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 8 / 40
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Determinant Prperties of thr Determinant
Example (3.3.3)
Use theorem 3.3.4 (Invertible Test) to determine which of the followingmatrices are invertible.
1. A =
1 0 10 2 40 0 4
det(A) = 1(2)(4) = 8 ̸= 0 (Invertible)
2. A =
1 2 30 −5 72 4 6
det(A) = 0 (NOT invertible)
3. A =
0 1 11 0 00 1 0
det(A) = 1 ̸= 0 (Invertible)
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 9 / 40
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Determinant Prperties of thr Determinant
Example (3.3.3)
Use theorem 3.3.4 (Invertible Test) to determine which of the followingmatrices are invertible.
1. A =
1 0 10 2 40 0 4
det(A) = 1(2)(4) = 8 ̸= 0 (Invertible)
2. A =
1 2 30 −5 72 4 6
det(A) = 0 (NOT invertible)
3. A =
0 1 11 0 00 1 0
det(A) = 1 ̸= 0 (Invertible)
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 9 / 40
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Determinant Prperties of thr Determinant
Example (3.3.3)
Use theorem 3.3.4 (Invertible Test) to determine which of the followingmatrices are invertible.
1. A =
1 0 10 2 40 0 4
det(A) = 1(2)(4) = 8 ̸= 0 (Invertible)
2. A =
1 2 30 −5 72 4 6
det(A) = 0 (NOT invertible)
3. A =
0 1 11 0 00 1 0
det(A) = 1 ̸= 0 (Invertible)
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 9 / 40
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Determinant Prperties of thr Determinant
Example (3.3.3)
Use theorem 3.3.4 (Invertible Test) to determine which of the followingmatrices are invertible.
1. A =
1 0 10 2 40 0 4
det(A) = 1(2)(4) = 8 ̸= 0 (Invertible)
2. A =
1 2 30 −5 72 4 6
det(A) = 0 (NOT invertible)
3. A =
0 1 11 0 00 1 0
det(A) = 1 ̸= 0 (Invertible)
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 9 / 40
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Determinant Prperties of thr Determinant
Example (3.3.3)
Use theorem 3.3.4 (Invertible Test) to determine which of the followingmatrices are invertible.
1. A =
1 0 10 2 40 0 4
det(A) = 1(2)(4) = 8 ̸= 0 (Invertible)
2. A =
1 2 30 −5 72 4 6
det(A) = 0 (NOT invertible)
3. A =
0 1 11 0 00 1 0
det(A) = 1 ̸= 0 (Invertible)
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 9 / 40
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Determinant Prperties of thr Determinant
Example (3.3.3)
Use theorem 3.3.4 (Invertible Test) to determine which of the followingmatrices are invertible.
1. A =
1 0 10 2 40 0 4
det(A) = 1(2)(4) = 8 ̸= 0 (Invertible)
2. A =
1 2 30 −5 72 4 6
det(A) = 0 (NOT invertible)
3. A =
0 1 11 0 00 1 0
det(A) = 1 ̸= 0 (Invertible)
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 9 / 40
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Determinant Prperties of thr Determinant
Product Determinant
Theorem (3.3.5)
If A and B are square matrix of the same size, then
det(AB) = det(A) det(B)
If m is a positive integer,
det(Am) = [det(A)]m
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 10 / 40
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Determinant Prperties of thr Determinant
Example (3.3.4)
Let A =
[1 23 5
], B =
[−2 26 −1
]and C =
[11 45 1
]. Find
1. det(AB)
2. det(BC)
3. det(ATC)
4. det(2AB)
5. det(−3ABTC)
6. det(2C(AB2)T )
7. det(A2B3C2)
8. det(A+AT )
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 11 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −1, det(B) = −10 and det(C) = −9
1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #
2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #
3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #
4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #
5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #
6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2
= (36)(−10)2 = 3600 #
7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #
8. det(A+AT )
A+AT =
[1 23 5
]+
[1 32 5
]=
[2 55 10
]det(A+AT ) = 2(10)− 5(5) = −5 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −1, det(B) = −10 and det(C) = −9
1. det(AB)
= det(A) det(B)= (−1)(−10) = 10 #
2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #
3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #
4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #
5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #
6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2
= (36)(−10)2 = 3600 #
7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #
8. det(A+AT )
A+AT =
[1 23 5
]+
[1 32 5
]=
[2 55 10
]det(A+AT ) = 2(10)− 5(5) = −5 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −1, det(B) = −10 and det(C) = −9
1. det(AB) = det(A) det(B)
= (−1)(−10) = 10 #
2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #
3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #
4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #
5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #
6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2
= (36)(−10)2 = 3600 #
7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #
8. det(A+AT )
A+AT =
[1 23 5
]+
[1 32 5
]=
[2 55 10
]det(A+AT ) = 2(10)− 5(5) = −5 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −1, det(B) = −10 and det(C) = −9
1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #
2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #
3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #
4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #
5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #
6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2
= (36)(−10)2 = 3600 #
7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #
8. det(A+AT )
A+AT =
[1 23 5
]+
[1 32 5
]=
[2 55 10
]det(A+AT ) = 2(10)− 5(5) = −5 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −1, det(B) = −10 and det(C) = −9
1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #
2. det(BC)
= det(B) det(C)= (−10)(−9) = 90 #
3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #
4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #
5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #
6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2
= (36)(−10)2 = 3600 #
7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #
8. det(A+AT )
A+AT =
[1 23 5
]+
[1 32 5
]=
[2 55 10
]det(A+AT ) = 2(10)− 5(5) = −5 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −1, det(B) = −10 and det(C) = −9
1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #
2. det(BC) = det(B) det(C)
= (−10)(−9) = 90 #
3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #
4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #
5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #
6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2
= (36)(−10)2 = 3600 #
7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #
8. det(A+AT )
A+AT =
[1 23 5
]+
[1 32 5
]=
[2 55 10
]det(A+AT ) = 2(10)− 5(5) = −5 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −1, det(B) = −10 and det(C) = −9
1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #
2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #
3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #
4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #
5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #
6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2
= (36)(−10)2 = 3600 #
7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #
8. det(A+AT )
A+AT =
[1 23 5
]+
[1 32 5
]=
[2 55 10
]det(A+AT ) = 2(10)− 5(5) = −5 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −1, det(B) = −10 and det(C) = −9
1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #
2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #
3. det(ATC)
= det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #
4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #
5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #
6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2
= (36)(−10)2 = 3600 #
7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #
8. det(A+AT )
A+AT =
[1 23 5
]+
[1 32 5
]=
[2 55 10
]det(A+AT ) = 2(10)− 5(5) = −5 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −1, det(B) = −10 and det(C) = −9
1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #
2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #
3. det(ATC) = det(AT ) det(C)
= det(A) det(C)= (−1)(−9) = 18 #
4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #
5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #
6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2
= (36)(−10)2 = 3600 #
7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #
8. det(A+AT )
A+AT =
[1 23 5
]+
[1 32 5
]=
[2 55 10
]det(A+AT ) = 2(10)− 5(5) = −5 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −1, det(B) = −10 and det(C) = −9
1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #
2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #
3. det(ATC) = det(AT ) det(C)= det(A) det(C)
= (−1)(−9) = 18 #
4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #
5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #
6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2
= (36)(−10)2 = 3600 #
7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #
8. det(A+AT )
A+AT =
[1 23 5
]+
[1 32 5
]=
[2 55 10
]det(A+AT ) = 2(10)− 5(5) = −5 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −1, det(B) = −10 and det(C) = −9
1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #
2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #
3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #
4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #
5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #
6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2
= (36)(−10)2 = 3600 #
7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #
8. det(A+AT )
A+AT =
[1 23 5
]+
[1 32 5
]=
[2 55 10
]det(A+AT ) = 2(10)− 5(5) = −5 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −1, det(B) = −10 and det(C) = −9
1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #
2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #
3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #
4. det(2AB)
= 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #
5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #
6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2
= (36)(−10)2 = 3600 #
7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #
8. det(A+AT )
A+AT =
[1 23 5
]+
[1 32 5
]=
[2 55 10
]det(A+AT ) = 2(10)− 5(5) = −5 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −1, det(B) = −10 and det(C) = −9
1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #
2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #
3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #
4. det(2AB) = 22 det(AB)
= 4 det(A) det(B)= 4(−1)(−10) = 40 #
5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #
6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2
= (36)(−10)2 = 3600 #
7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #
8. det(A+AT )
A+AT =
[1 23 5
]+
[1 32 5
]=
[2 55 10
]det(A+AT ) = 2(10)− 5(5) = −5 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −1, det(B) = −10 and det(C) = −9
1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #
2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #
3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #
4. det(2AB) = 22 det(AB)= 4 det(A) det(B)
= 4(−1)(−10) = 40 #
5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #
6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2
= (36)(−10)2 = 3600 #
7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #
8. det(A+AT )
A+AT =
[1 23 5
]+
[1 32 5
]=
[2 55 10
]det(A+AT ) = 2(10)− 5(5) = −5 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −1, det(B) = −10 and det(C) = −9
1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #
2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #
3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #
4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #
5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #
6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2
= (36)(−10)2 = 3600 #
7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #
8. det(A+AT )
A+AT =
[1 23 5
]+
[1 32 5
]=
[2 55 10
]det(A+AT ) = 2(10)− 5(5) = −5 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −1, det(B) = −10 and det(C) = −9
1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #
2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #
3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #
4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #
5. det(−3ABTC)
= (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #
6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2
= (36)(−10)2 = 3600 #
7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #
8. det(A+AT )
A+AT =
[1 23 5
]+
[1 32 5
]=
[2 55 10
]det(A+AT ) = 2(10)− 5(5) = −5 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −1, det(B) = −10 and det(C) = −9
1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #
2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #
3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #
4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #
5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)
= 9(−1)(−10)(−9) = −810 #
6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2
= (36)(−10)2 = 3600 #
7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #
8. det(A+AT )
A+AT =
[1 23 5
]+
[1 32 5
]=
[2 55 10
]det(A+AT ) = 2(10)− 5(5) = −5 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −1, det(B) = −10 and det(C) = −9
1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #
2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #
3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #
4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #
5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #
6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2
= (36)(−10)2 = 3600 #
7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #
8. det(A+AT )
A+AT =
[1 23 5
]+
[1 32 5
]=
[2 55 10
]det(A+AT ) = 2(10)− 5(5) = −5 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −1, det(B) = −10 and det(C) = −9
1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #
2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #
3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #
4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #
5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #
6. det(2C(AB2)T )
= 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2
= (36)(−10)2 = 3600 #
7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #
8. det(A+AT )
A+AT =
[1 23 5
]+
[1 32 5
]=
[2 55 10
]det(A+AT ) = 2(10)− 5(5) = −5 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −1, det(B) = −10 and det(C) = −9
1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #
2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #
3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #
4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #
5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #
6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)
= 4(−9)(−1)[det(B)]2
= (36)(−10)2 = 3600 #
7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #
8. det(A+AT )
A+AT =
[1 23 5
]+
[1 32 5
]=
[2 55 10
]det(A+AT ) = 2(10)− 5(5) = −5 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −1, det(B) = −10 and det(C) = −9
1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #
2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #
3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #
4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #
5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #
6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2
= (36)(−10)2 = 3600 #
7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #
8. det(A+AT )
A+AT =
[1 23 5
]+
[1 32 5
]=
[2 55 10
]det(A+AT ) = 2(10)− 5(5) = −5 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −1, det(B) = −10 and det(C) = −9
1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #
2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #
3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #
4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #
5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #
6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2
= (36)(−10)2 = 3600 #
7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #
8. det(A+AT )
A+AT =
[1 23 5
]+
[1 32 5
]=
[2 55 10
]det(A+AT ) = 2(10)− 5(5) = −5 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −1, det(B) = −10 and det(C) = −9
1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #
2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #
3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #
4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #
5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #
6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2
= (36)(−10)2 = 3600 #
7. det(A2B3C2)
= [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #
8. det(A+AT )
A+AT =
[1 23 5
]+
[1 32 5
]=
[2 55 10
]det(A+AT ) = 2(10)− 5(5) = −5 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −1, det(B) = −10 and det(C) = −9
1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #
2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #
3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #
4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #
5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #
6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2
= (36)(−10)2 = 3600 #
7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2
= (−1)2(−10)3(−9)2 = −81000 #
8. det(A+AT )
A+AT =
[1 23 5
]+
[1 32 5
]=
[2 55 10
]det(A+AT ) = 2(10)− 5(5) = −5 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −1, det(B) = −10 and det(C) = −9
1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #
2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #
3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #
4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #
5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #
6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2
= (36)(−10)2 = 3600 #
7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #
8. det(A+AT )
A+AT =
[1 23 5
]+
[1 32 5
]=
[2 55 10
]det(A+AT ) = 2(10)− 5(5) = −5 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −1, det(B) = −10 and det(C) = −9
1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #
2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #
3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #
4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #
5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #
6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2
= (36)(−10)2 = 3600 #
7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #
8. det(A+AT )
A+AT =
[1 23 5
]+
[1 32 5
]=
[2 55 10
]det(A+AT ) = 2(10)− 5(5) = −5 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −1, det(B) = −10 and det(C) = −9
1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #
2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #
3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #
4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #
5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #
6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2
= (36)(−10)2 = 3600 #
7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #
8. det(A+AT )
A+AT =
[1 23 5
]+
[1 32 5
]=
[2 55 10
]
det(A+AT ) = 2(10)− 5(5) = −5 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −1, det(B) = −10 and det(C) = −9
1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #
2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #
3. det(ATC) = det(AT ) det(C)= det(A) det(C)= (−1)(−9) = 18 #
4. det(2AB) = 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #
5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #
6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2
= (36)(−10)2 = 3600 #
7. det(A2B3C2) = [det(A)]2[det(B)]3[det(C)]2= (−1)2(−10)3(−9)2 = −81000 #
8. det(A+AT )
A+AT =
[1 23 5
]+
[1 32 5
]=
[2 55 10
]det(A+AT ) = 2(10)− 5(5) = −5 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 12 / 40
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Determinant Prperties of thr Determinant
Determinant of Inverse
Theorem (3.3.6)
If A is invertible, then
det(A−1) =1
det(A)
Example (3.3.6)
Let A =
[6 −4−3 1
]and B =
[8 45 1
]. Find
1. det(A−1)
2. det(AB−1)
3. det(3(−2AT )−1)
4. det(2(−3A)TA−1BT )
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 13 / 40
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Determinant Prperties of thr Determinant
Determinant of Inverse
Theorem (3.3.6)
If A is invertible, then
det(A−1) =1
det(A)
Example (3.3.6)
Let A =
[6 −4−3 1
]and B =
[8 45 1
]. Find
1. det(A−1)
2. det(AB−1)
3. det(3(−2AT )−1)
4. det(2(−3A)TA−1BT )
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 13 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −6 and det(B) = −12
1. det(A−1) = 1det(A)
= − 16
#
2. det(AB−1) = det(A) det(B−1)= (−6) 1−12
= 12
#
3. det(3(−2AT )−1) = 32 det((−2AT )−1)= 9 1det(−2AT )
= 9(−2)2 det(A)
= 94(−6)
= − 38
#
4. det(2(−3A)TA−1BT )= 22 det(−3AT ) det(A−1) det(BT )
= 4(−3)2 det(A) det(A−1) det(B)
= 4(9)(−6) 1−6
(−12) = −432 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 14 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −6 and det(B) = −12
1. det(A−1)
= 1det(A)
= − 16
#
2. det(AB−1) = det(A) det(B−1)= (−6) 1−12
= 12
#
3. det(3(−2AT )−1) = 32 det((−2AT )−1)= 9 1det(−2AT )
= 9(−2)2 det(A)
= 94(−6)
= − 38
#
4. det(2(−3A)TA−1BT )= 22 det(−3AT ) det(A−1) det(BT )
= 4(−3)2 det(A) det(A−1) det(B)
= 4(9)(−6) 1−6
(−12) = −432 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 14 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −6 and det(B) = −12
1. det(A−1) = 1det(A)
= − 16
#
2. det(AB−1) = det(A) det(B−1)= (−6) 1−12
= 12
#
3. det(3(−2AT )−1) = 32 det((−2AT )−1)= 9 1det(−2AT )
= 9(−2)2 det(A)
= 94(−6)
= − 38
#
4. det(2(−3A)TA−1BT )= 22 det(−3AT ) det(A−1) det(BT )
= 4(−3)2 det(A) det(A−1) det(B)
= 4(9)(−6) 1−6
(−12) = −432 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 14 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −6 and det(B) = −12
1. det(A−1) = 1det(A)
= − 16
#
2. det(AB−1)
= det(A) det(B−1)= (−6) 1−12
= 12
#
3. det(3(−2AT )−1) = 32 det((−2AT )−1)= 9 1det(−2AT )
= 9(−2)2 det(A)
= 94(−6)
= − 38
#
4. det(2(−3A)TA−1BT )= 22 det(−3AT ) det(A−1) det(BT )
= 4(−3)2 det(A) det(A−1) det(B)
= 4(9)(−6) 1−6
(−12) = −432 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 14 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −6 and det(B) = −12
1. det(A−1) = 1det(A)
= − 16
#
2. det(AB−1) = det(A) det(B−1)
= (−6) 1−12
= 12
#
3. det(3(−2AT )−1) = 32 det((−2AT )−1)= 9 1det(−2AT )
= 9(−2)2 det(A)
= 94(−6)
= − 38
#
4. det(2(−3A)TA−1BT )= 22 det(−3AT ) det(A−1) det(BT )
= 4(−3)2 det(A) det(A−1) det(B)
= 4(9)(−6) 1−6
(−12) = −432 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 14 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −6 and det(B) = −12
1. det(A−1) = 1det(A)
= − 16
#
2. det(AB−1) = det(A) det(B−1)= (−6) 1−12
= 12
#
3. det(3(−2AT )−1) = 32 det((−2AT )−1)= 9 1det(−2AT )
= 9(−2)2 det(A)
= 94(−6)
= − 38
#
4. det(2(−3A)TA−1BT )= 22 det(−3AT ) det(A−1) det(BT )
= 4(−3)2 det(A) det(A−1) det(B)
= 4(9)(−6) 1−6
(−12) = −432 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 14 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −6 and det(B) = −12
1. det(A−1) = 1det(A)
= − 16
#
2. det(AB−1) = det(A) det(B−1)= (−6) 1−12
= 12
#
3. det(3(−2AT )−1)
= 32 det((−2AT )−1)= 9 1det(−2AT )
= 9(−2)2 det(A)
= 94(−6)
= − 38
#
4. det(2(−3A)TA−1BT )= 22 det(−3AT ) det(A−1) det(BT )
= 4(−3)2 det(A) det(A−1) det(B)
= 4(9)(−6) 1−6
(−12) = −432 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 14 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −6 and det(B) = −12
1. det(A−1) = 1det(A)
= − 16
#
2. det(AB−1) = det(A) det(B−1)= (−6) 1−12
= 12
#
3. det(3(−2AT )−1) = 32 det((−2AT )−1)
= 9 1det(−2AT )
= 9(−2)2 det(A)
= 94(−6)
= − 38
#
4. det(2(−3A)TA−1BT )= 22 det(−3AT ) det(A−1) det(BT )
= 4(−3)2 det(A) det(A−1) det(B)
= 4(9)(−6) 1−6
(−12) = −432 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 14 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −6 and det(B) = −12
1. det(A−1) = 1det(A)
= − 16
#
2. det(AB−1) = det(A) det(B−1)= (−6) 1−12
= 12
#
3. det(3(−2AT )−1) = 32 det((−2AT )−1)= 9 1det(−2AT )
= 9(−2)2 det(A)
= 94(−6)
= − 38
#
4. det(2(−3A)TA−1BT )= 22 det(−3AT ) det(A−1) det(BT )
= 4(−3)2 det(A) det(A−1) det(B)
= 4(9)(−6) 1−6
(−12) = −432 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 14 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −6 and det(B) = −12
1. det(A−1) = 1det(A)
= − 16
#
2. det(AB−1) = det(A) det(B−1)= (−6) 1−12
= 12
#
3. det(3(−2AT )−1) = 32 det((−2AT )−1)= 9 1det(−2AT )
= 9(−2)2 det(A)
= 94(−6)
= − 38
#
4. det(2(−3A)TA−1BT )= 22 det(−3AT ) det(A−1) det(BT )
= 4(−3)2 det(A) det(A−1) det(B)
= 4(9)(−6) 1−6
(−12) = −432 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 14 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −6 and det(B) = −12
1. det(A−1) = 1det(A)
= − 16
#
2. det(AB−1) = det(A) det(B−1)= (−6) 1−12
= 12
#
3. det(3(−2AT )−1) = 32 det((−2AT )−1)= 9 1det(−2AT )
= 9(−2)2 det(A)
= 94(−6)
= − 38
#
4. det(2(−3A)TA−1BT )= 22 det(−3AT ) det(A−1) det(BT )
= 4(−3)2 det(A) det(A−1) det(B)
= 4(9)(−6) 1−6
(−12) = −432 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 14 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −6 and det(B) = −12
1. det(A−1) = 1det(A)
= − 16
#
2. det(AB−1) = det(A) det(B−1)= (−6) 1−12
= 12
#
3. det(3(−2AT )−1) = 32 det((−2AT )−1)= 9 1det(−2AT )
= 9(−2)2 det(A)
= 94(−6)
= − 38
#
4. det(2(−3A)TA−1BT )
= 22 det(−3AT ) det(A−1) det(BT )
= 4(−3)2 det(A) det(A−1) det(B)
= 4(9)(−6) 1−6
(−12) = −432 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 14 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −6 and det(B) = −12
1. det(A−1) = 1det(A)
= − 16
#
2. det(AB−1) = det(A) det(B−1)= (−6) 1−12
= 12
#
3. det(3(−2AT )−1) = 32 det((−2AT )−1)= 9 1det(−2AT )
= 9(−2)2 det(A)
= 94(−6)
= − 38
#
4. det(2(−3A)TA−1BT )= 22 det(−3AT ) det(A−1) det(BT )
= 4(−3)2 det(A) det(A−1) det(B)
= 4(9)(−6) 1−6
(−12) = −432 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 14 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −6 and det(B) = −12
1. det(A−1) = 1det(A)
= − 16
#
2. det(AB−1) = det(A) det(B−1)= (−6) 1−12
= 12
#
3. det(3(−2AT )−1) = 32 det((−2AT )−1)= 9 1det(−2AT )
= 9(−2)2 det(A)
= 94(−6)
= − 38
#
4. det(2(−3A)TA−1BT )= 22 det(−3AT ) det(A−1) det(BT )
= 4(−3)2 det(A) det(A−1) det(B)
= 4(9)(−6) 1−6
(−12) = −432 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 14 / 40
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Determinant Prperties of thr Determinant
Solution
det(A) = −6 and det(B) = −12
1. det(A−1) = 1det(A)
= − 16
#
2. det(AB−1) = det(A) det(B−1)= (−6) 1−12
= 12
#
3. det(3(−2AT )−1) = 32 det((−2AT )−1)= 9 1det(−2AT )
= 9(−2)2 det(A)
= 94(−6)
= − 38
#
4. det(2(−3A)TA−1BT )= 22 det(−3AT ) det(A−1) det(BT )
= 4(−3)2 det(A) det(A−1) det(B)
= 4(9)(−6) 1−6
(−12) = −432 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 14 / 40
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Determinant Prperties of thr Determinant
Equivalent Statements
Theorem (3.3.7)
Let A be an n× n matrix. Then TFAE.
1 A is invertible.
2 Ax = 0 has only trivial solution.
3 The RREF of A is In.
4 A is expressible as a product of elementary matrices.
5 Ax = b is consistent for every n× 1 matrix b.
6 Ax = b has exactly one solution for every n× 1 matrix b.
7 det(A) ̸= 0.
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 15 / 40
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Determinant Adjoint of a Matrix
3.4 Adjoint of a Matrix
Definition (3.4.1 Minor and Cofactor)
Let A be a square matrix.
The minor of entry aij is denoted by Mij and is defined to be thedeterminant of the submatrix that remains after the ith row and jthcolumn are deleted from A.
The number (−1)i+jMij is denoted by Cij and is called the cofactor ofentry aij,
Cij = (−1)i+jMij
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 16 / 40
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Determinant Adjoint of a Matrix
Example (3.4.1)
Find all minors and all cofactors of matrix
A =
1 2 12 1 11 2 0
Solution
M11 =
∣∣∣∣1 12 0
∣∣∣∣ = −2
M12 =
∣∣∣∣2 11 0
∣∣∣∣ = −1
M13 =
∣∣∣∣2 11 2
∣∣∣∣ = 3
M21 =
∣∣∣∣2 12 0
∣∣∣∣ = −2
M22 =
∣∣∣∣1 11 0
∣∣∣∣ = −1
M23 =
∣∣∣∣1 21 2
∣∣∣∣ = 0
M31 =
∣∣∣∣2 11 1
∣∣∣∣ = 1
M32 =
∣∣∣∣1 12 1
∣∣∣∣ = −1
M33 =
∣∣∣∣1 22 1
∣∣∣∣ = −3
Thus,
C11 = (−1)1+1M11 = −2
C12 = (−1)1+2M12 = 1
C13 = (−1)1+3M13 = 3
C21 = (−1)2+1M21 = 2
C22 = (−1)2+2M22 = −1
C23 = (−1)2+3M23 = 0
C31 = (−1)3+1M31 = 1
C32 = (−1)3+2M32 = 1
C33 = (−1)3+3M33 = −3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 17 / 40
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Determinant Adjoint of a Matrix
Example (3.4.1)
Find all minors and all cofactors of matrix
A =
1 2 12 1 11 2 0
Solution
M11 =
∣∣∣∣1 12 0
∣∣∣∣ = −2
M12 =
∣∣∣∣2 11 0
∣∣∣∣ = −1
M13 =
∣∣∣∣2 11 2
∣∣∣∣ = 3
M21 =
∣∣∣∣2 12 0
∣∣∣∣ = −2
M22 =
∣∣∣∣1 11 0
∣∣∣∣ = −1
M23 =
∣∣∣∣1 21 2
∣∣∣∣ = 0
M31 =
∣∣∣∣2 11 1
∣∣∣∣ = 1
M32 =
∣∣∣∣1 12 1
∣∣∣∣ = −1
M33 =
∣∣∣∣1 22 1
∣∣∣∣ = −3
Thus,
C11 = (−1)1+1M11 = −2
C12 = (−1)1+2M12 = 1
C13 = (−1)1+3M13 = 3
C21 = (−1)2+1M21 = 2
C22 = (−1)2+2M22 = −1
C23 = (−1)2+3M23 = 0
C31 = (−1)3+1M31 = 1
C32 = (−1)3+2M32 = 1
C33 = (−1)3+3M33 = −3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 17 / 40
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Determinant Adjoint of a Matrix
Example (3.4.1)
Find all minors and all cofactors of matrix
A =
1 2 12 1 11 2 0
Solution
M11 =
∣∣∣∣1 12 0
∣∣∣∣ = −2
M12 =
∣∣∣∣2 11 0
∣∣∣∣ = −1
M13 =
∣∣∣∣2 11 2
∣∣∣∣ = 3
M21 =
∣∣∣∣2 12 0
∣∣∣∣ = −2
M22 =
∣∣∣∣1 11 0
∣∣∣∣ = −1
M23 =
∣∣∣∣1 21 2
∣∣∣∣ = 0
M31 =
∣∣∣∣2 11 1
∣∣∣∣ = 1
M32 =
∣∣∣∣1 12 1
∣∣∣∣ = −1
M33 =
∣∣∣∣1 22 1
∣∣∣∣ = −3
Thus,
C11 = (−1)1+1M11 = −2
C12 = (−1)1+2M12 = 1
C13 = (−1)1+3M13 = 3
C21 = (−1)2+1M21 = 2
C22 = (−1)2+2M22 = −1
C23 = (−1)2+3M23 = 0
C31 = (−1)3+1M31 = 1
C32 = (−1)3+2M32 = 1
C33 = (−1)3+3M33 = −3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 17 / 40
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Determinant Adjoint of a Matrix
Example (3.4.1)
Find all minors and all cofactors of matrix
A =
1 2 12 1 11 2 0
Solution
M11 =
∣∣∣∣1 12 0
∣∣∣∣ = −2
M12 =
∣∣∣∣2 11 0
∣∣∣∣ = −1
M13 =
∣∣∣∣2 11 2
∣∣∣∣ = 3
M21 =
∣∣∣∣2 12 0
∣∣∣∣ = −2
M22 =
∣∣∣∣1 11 0
∣∣∣∣ = −1
M23 =
∣∣∣∣1 21 2
∣∣∣∣ = 0
M31 =
∣∣∣∣2 11 1
∣∣∣∣ = 1
M32 =
∣∣∣∣1 12 1
∣∣∣∣ = −1
M33 =
∣∣∣∣1 22 1
∣∣∣∣ = −3
Thus,
C11 = (−1)1+1M11 = −2
C12 = (−1)1+2M12 = 1
C13 = (−1)1+3M13 = 3
C21 = (−1)2+1M21 = 2
C22 = (−1)2+2M22 = −1
C23 = (−1)2+3M23 = 0
C31 = (−1)3+1M31 = 1
C32 = (−1)3+2M32 = 1
C33 = (−1)3+3M33 = −3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 17 / 40
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Determinant Adjoint of a Matrix
Example (3.4.1)
Find all minors and all cofactors of matrix
A =
1 2 12 1 11 2 0
Solution
M11 =
∣∣∣∣1 12 0
∣∣∣∣ = −2
M12 =
∣∣∣∣2 11 0
∣∣∣∣ = −1
M13 =
∣∣∣∣2 11 2
∣∣∣∣ = 3
M21 =
∣∣∣∣2 12 0
∣∣∣∣ = −2
M22 =
∣∣∣∣1 11 0
∣∣∣∣ = −1
M23 =
∣∣∣∣1 21 2
∣∣∣∣ = 0
M31 =
∣∣∣∣2 11 1
∣∣∣∣ = 1
M32 =
∣∣∣∣1 12 1
∣∣∣∣ = −1
M33 =
∣∣∣∣1 22 1
∣∣∣∣ = −3
Thus,
C11 = (−1)1+1M11 = −2
C12 = (−1)1+2M12 = 1
C13 = (−1)1+3M13 = 3
C21 = (−1)2+1M21 = 2
C22 = (−1)2+2M22 = −1
C23 = (−1)2+3M23 = 0
C31 = (−1)3+1M31 = 1
C32 = (−1)3+2M32 = 1
C33 = (−1)3+3M33 = −3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 17 / 40
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Determinant Adjoint of a Matrix
Example (3.4.1)
Find all minors and all cofactors of matrix
A =
1 2 12 1 11 2 0
Solution
M11 =
∣∣∣∣1 12 0
∣∣∣∣ = −2
M12 =
∣∣∣∣2 11 0
∣∣∣∣ = −1
M13 =
∣∣∣∣2 11 2
∣∣∣∣ = 3
M21 =
∣∣∣∣2 12 0
∣∣∣∣ = −2
M22 =
∣∣∣∣1 11 0
∣∣∣∣ = −1
M23 =
∣∣∣∣1 21 2
∣∣∣∣ = 0
M31 =
∣∣∣∣2 11 1
∣∣∣∣ = 1
M32 =
∣∣∣∣1 12 1
∣∣∣∣ = −1
M33 =
∣∣∣∣1 22 1
∣∣∣∣ = −3
Thus,
C11 = (−1)1+1M11 = −2
C12 = (−1)1+2M12 = 1
C13 = (−1)1+3M13 = 3
C21 = (−1)2+1M21 = 2
C22 = (−1)2+2M22 = −1
C23 = (−1)2+3M23 = 0
C31 = (−1)3+1M31 = 1
C32 = (−1)3+2M32 = 1
C33 = (−1)3+3M33 = −3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 17 / 40
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Determinant Adjoint of a Matrix
Example (3.4.1)
Find all minors and all cofactors of matrix
A =
1 2 12 1 11 2 0
Solution
M11 =
∣∣∣∣1 12 0
∣∣∣∣ = −2
M12 =
∣∣∣∣2 11 0
∣∣∣∣ = −1
M13 =
∣∣∣∣2 11 2
∣∣∣∣ = 3
M21 =
∣∣∣∣2 12 0
∣∣∣∣ = −2
M22 =
∣∣∣∣1 11 0
∣∣∣∣ = −1
M23 =
∣∣∣∣1 21 2
∣∣∣∣ = 0
M31 =
∣∣∣∣2 11 1
∣∣∣∣ = 1
M32 =
∣∣∣∣1 12 1
∣∣∣∣ = −1
M33 =
∣∣∣∣1 22 1
∣∣∣∣ = −3
Thus,
C11 = (−1)1+1M11 = −2
C12 = (−1)1+2M12 = 1
C13 = (−1)1+3M13 = 3
C21 = (−1)2+1M21 = 2
C22 = (−1)2+2M22 = −1
C23 = (−1)2+3M23 = 0
C31 = (−1)3+1M31 = 1
C32 = (−1)3+2M32 = 1
C33 = (−1)3+3M33 = −3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 17 / 40
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Determinant Adjoint of a Matrix
Example (3.4.1)
Find all minors and all cofactors of matrix
A =
1 2 12 1 11 2 0
Solution
M11 =
∣∣∣∣1 12 0
∣∣∣∣ = −2
M12 =
∣∣∣∣2 11 0
∣∣∣∣ = −1
M13 =
∣∣∣∣2 11 2
∣∣∣∣ = 3
M21 =
∣∣∣∣2 12 0
∣∣∣∣ = −2
M22 =
∣∣∣∣1 11 0
∣∣∣∣ = −1
M23 =
∣∣∣∣1 21 2
∣∣∣∣ = 0
M31 =
∣∣∣∣2 11 1
∣∣∣∣ = 1
M32 =
∣∣∣∣1 12 1
∣∣∣∣ = −1
M33 =
∣∣∣∣1 22 1
∣∣∣∣ = −3
Thus,
C11 = (−1)1+1M11 = −2
C12 = (−1)1+2M12 = 1
C13 = (−1)1+3M13 = 3
C21 = (−1)2+1M21 = 2
C22 = (−1)2+2M22 = −1
C23 = (−1)2+3M23 = 0
C31 = (−1)3+1M31 = 1
C32 = (−1)3+2M32 = 1
C33 = (−1)3+3M33 = −3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 17 / 40
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Determinant Adjoint of a Matrix
Example (3.4.1)
Find all minors and all cofactors of matrix
A =
1 2 12 1 11 2 0
Solution
M11 =
∣∣∣∣1 12 0
∣∣∣∣ = −2
M12 =
∣∣∣∣2 11 0
∣∣∣∣ = −1
M13 =
∣∣∣∣2 11 2
∣∣∣∣ = 3
M21 =
∣∣∣∣2 12 0
∣∣∣∣ = −2
M22 =
∣∣∣∣1 11 0
∣∣∣∣ = −1
M23 =
∣∣∣∣1 21 2
∣∣∣∣ = 0
M31 =
∣∣∣∣2 11 1
∣∣∣∣ = 1
M32 =
∣∣∣∣1 12 1
∣∣∣∣ = −1
M33 =
∣∣∣∣1 22 1
∣∣∣∣ = −3
Thus,
C11 = (−1)1+1M11 = −2
C12 = (−1)1+2M12 = 1
C13 = (−1)1+3M13 = 3
C21 = (−1)2+1M21 = 2
C22 = (−1)2+2M22 = −1
C23 = (−1)2+3M23 = 0
C31 = (−1)3+1M31 = 1
C32 = (−1)3+2M32 = 1
C33 = (−1)3+3M33 = −3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 17 / 40
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Determinant Adjoint of a Matrix
Example (3.4.1)
Find all minors and all cofactors of matrix
A =
1 2 12 1 11 2 0
Solution
M11 =
∣∣∣∣1 12 0
∣∣∣∣ = −2
M12 =
∣∣∣∣2 11 0
∣∣∣∣ = −1
M13 =
∣∣∣∣2 11 2
∣∣∣∣ = 3
M21 =
∣∣∣∣2 12 0
∣∣∣∣ = −2
M22 =
∣∣∣∣1 11 0
∣∣∣∣ = −1
M23 =
∣∣∣∣1 21 2
∣∣∣∣ = 0
M31 =
∣∣∣∣2 11 1
∣∣∣∣ = 1
M32 =
∣∣∣∣1 12 1
∣∣∣∣ = −1
M33 =
∣∣∣∣1 22 1
∣∣∣∣ = −3
Thus,
C11 = (−1)1+1M11 = −2
C12 = (−1)1+2M12 = 1
C13 = (−1)1+3M13 = 3
C21 = (−1)2+1M21 = 2
C22 = (−1)2+2M22 = −1
C23 = (−1)2+3M23 = 0
C31 = (−1)3+1M31 = 1
C32 = (−1)3+2M32 = 1
C33 = (−1)3+3M33 = −3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 17 / 40
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Determinant Adjoint of a Matrix
Example (3.4.1)
Find all minors and all cofactors of matrix
A =
1 2 12 1 11 2 0
Solution
M11 =
∣∣∣∣1 12 0
∣∣∣∣ = −2
M12 =
∣∣∣∣2 11 0
∣∣∣∣ = −1
M13 =
∣∣∣∣2 11 2
∣∣∣∣ = 3
M21 =
∣∣∣∣2 12 0
∣∣∣∣ = −2
M22 =
∣∣∣∣1 11 0
∣∣∣∣ = −1
M23 =
∣∣∣∣1 21 2
∣∣∣∣ = 0
M31 =
∣∣∣∣2 11 1
∣∣∣∣ = 1
M32 =
∣∣∣∣1 12 1
∣∣∣∣ = −1
M33 =
∣∣∣∣1 22 1
∣∣∣∣ = −3
Thus,
C11 = (−1)1+1M11 = −2
C12 = (−1)1+2M12 = 1
C13 = (−1)1+3M13 = 3
C21 = (−1)2+1M21 = 2
C22 = (−1)2+2M22 = −1
C23 = (−1)2+3M23 = 0
C31 = (−1)3+1M31 = 1
C32 = (−1)3+2M32 = 1
C33 = (−1)3+3M33 = −3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 17 / 40
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Determinant Adjoint of a Matrix
Example (3.4.1)
Find all minors and all cofactors of matrix
A =
1 2 12 1 11 2 0
Solution
M11 =
∣∣∣∣1 12 0
∣∣∣∣ = −2
M12 =
∣∣∣∣2 11 0
∣∣∣∣ = −1
M13 =
∣∣∣∣2 11 2
∣∣∣∣ = 3
M21 =
∣∣∣∣2 12 0
∣∣∣∣ = −2
M22 =
∣∣∣∣1 11 0
∣∣∣∣ = −1
M23 =
∣∣∣∣1 21 2
∣∣∣∣ = 0
M31 =
∣∣∣∣2 11 1
∣∣∣∣ = 1
M32 =
∣∣∣∣1 12 1
∣∣∣∣ = −1
M33 =
∣∣∣∣1 22 1
∣∣∣∣ = −3
Thus,
C11 = (−1)1+1M11 = −2
C12 = (−1)1+2M12 = 1
C13 = (−1)1+3M13 = 3
C21 = (−1)2+1M21 = 2
C22 = (−1)2+2M22 = −1
C23 = (−1)2+3M23 = 0
C31 = (−1)3+1M31 = 1
C32 = (−1)3+2M32 = 1
C33 = (−1)3+3M33 = −3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 17 / 40
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Determinant Adjoint of a Matrix
Example (3.4.1)
Find all minors and all cofactors of matrix
A =
1 2 12 1 11 2 0
Solution
M11 =
∣∣∣∣1 12 0
∣∣∣∣ = −2
M12 =
∣∣∣∣2 11 0
∣∣∣∣ = −1
M13 =
∣∣∣∣2 11 2
∣∣∣∣ = 3
M21 =
∣∣∣∣2 12 0
∣∣∣∣ = −2
M22 =
∣∣∣∣1 11 0
∣∣∣∣ = −1
M23 =
∣∣∣∣1 21 2
∣∣∣∣ = 0
M31 =
∣∣∣∣2 11 1
∣∣∣∣ = 1
M32 =
∣∣∣∣1 12 1
∣∣∣∣ = −1
M33 =
∣∣∣∣1 22 1
∣∣∣∣ = −3
Thus,
C11 = (−1)1+1M11 = −2
C12 = (−1)1+2M12 = 1
C13 = (−1)1+3M13 = 3
C21 = (−1)2+1M21 = 2
C22 = (−1)2+2M22 = −1
C23 = (−1)2+3M23 = 0
C31 = (−1)3+1M31 = 1
C32 = (−1)3+2M32 = 1
C33 = (−1)3+3M33 = −3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 17 / 40
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Determinant Adjoint of a Matrix
Example (3.4.1)
Find all minors and all cofactors of matrix
A =
1 2 12 1 11 2 0
Solution
M11 =
∣∣∣∣1 12 0
∣∣∣∣ = −2
M12 =
∣∣∣∣2 11 0
∣∣∣∣ = −1
M13 =
∣∣∣∣2 11 2
∣∣∣∣ = 3
M21 =
∣∣∣∣2 12 0
∣∣∣∣ = −2
M22 =
∣∣∣∣1 11 0
∣∣∣∣ = −1
M23 =
∣∣∣∣1 21 2
∣∣∣∣ = 0
M31 =
∣∣∣∣2 11 1
∣∣∣∣ = 1
M32 =
∣∣∣∣1 12 1
∣∣∣∣ = −1
M33 =
∣∣∣∣1 22 1
∣∣∣∣ = −3
Thus,
C11 = (−1)1+1M11 = −2
C12 = (−1)1+2M12 = 1
C13 = (−1)1+3M13 = 3
C21 = (−1)2+1M21 = 2
C22 = (−1)2+2M22 = −1
C23 = (−1)2+3M23 = 0
C31 = (−1)3+1M31 = 1
C32 = (−1)3+2M32 = 1
C33 = (−1)3+3M33 = −3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 17 / 40
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Determinant Adjoint of a Matrix
Expansions by Cofactors
Theorem (3.4.1)
The determinant of an n× n matrix A = [aij ] can be computed by
cofactor expansion along the jth column:
det(A) = a1jC1j + a2jC2j + · · ·+ anjCnj
cofactor expansion along the ith row:
det(A) = ai1Ci1 + ai2Ci2 + · · ·+ ainCin
for each 1 ≤ i ≤ n and 1 ≤ j ≤ n.
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 18 / 40
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Determinant Adjoint of a Matrix
Example (3.4.2)
Find det(A) in example 3.4.1 by cofactor expansions.
A =
1 2 12 1 11 2 0
Solution
C11 = −2
C12 = 1
C13 = 3
C21 = 2
C22 = −1
C23 = 0
C31 = 1
C32 = 1
C33 = −3
Thus,
First Row det(A) = 1 · C11 + 2 · C12 + 1 · C13= 1(−2) + 2(1) + 1(3) = 3
First Column det(A) = 1 · C11 + 2 · C21 + 1 · C31= 1(−2) + 2(2) + 1(1) = 3
Third Row det(A) = 1 · C31 + 2 · C32 + 0 · C33= 1(1) + 2(1) + 0(−3) = 3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 19 / 40
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Determinant Adjoint of a Matrix
Example (3.4.2)
Find det(A) in example 3.4.1 by cofactor expansions.
A =
1 2 12 1 11 2 0
Solution
C11 = −2
C12 = 1
C13 = 3
C21 = 2
C22 = −1
C23 = 0
C31 = 1
C32 = 1
C33 = −3
Thus,
First Row det(A) = 1 · C11 + 2 · C12 + 1 · C13= 1(−2) + 2(1) + 1(3) = 3
First Column det(A) = 1 · C11 + 2 · C21 + 1 · C31= 1(−2) + 2(2) + 1(1) = 3
Third Row det(A) = 1 · C31 + 2 · C32 + 0 · C33= 1(1) + 2(1) + 0(−3) = 3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 19 / 40
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Determinant Adjoint of a Matrix
Example (3.4.2)
Find det(A) in example 3.4.1 by cofactor expansions.
A =
1 2 12 1 11 2 0
Solution
C11 = −2
C12 = 1
C13 = 3
C21 = 2
C22 = −1
C23 = 0
C31 = 1
C32 = 1
C33 = −3
Thus,
First Row
det(A) = 1 · C11 + 2 · C12 + 1 · C13= 1(−2) + 2(1) + 1(3) = 3
First Column det(A) = 1 · C11 + 2 · C21 + 1 · C31= 1(−2) + 2(2) + 1(1) = 3
Third Row det(A) = 1 · C31 + 2 · C32 + 0 · C33= 1(1) + 2(1) + 0(−3) = 3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 19 / 40
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Determinant Adjoint of a Matrix
Example (3.4.2)
Find det(A) in example 3.4.1 by cofactor expansions.
A =
1 2 12 1 11 2 0
Solution
C11 = −2
C12 = 1
C13 = 3
C21 = 2
C22 = −1
C23 = 0
C31 = 1
C32 = 1
C33 = −3
Thus,
First Row det(A) = 1 · C11 + 2 · C12 + 1 · C13
= 1(−2) + 2(1) + 1(3) = 3
First Column det(A) = 1 · C11 + 2 · C21 + 1 · C31= 1(−2) + 2(2) + 1(1) = 3
Third Row det(A) = 1 · C31 + 2 · C32 + 0 · C33= 1(1) + 2(1) + 0(−3) = 3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 19 / 40
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Determinant Adjoint of a Matrix
Example (3.4.2)
Find det(A) in example 3.4.1 by cofactor expansions.
A =
1 2 12 1 11 2 0
Solution
C11 = −2
C12 = 1
C13 = 3
C21 = 2
C22 = −1
C23 = 0
C31 = 1
C32 = 1
C33 = −3
Thus,
First Row det(A) = 1 · C11 + 2 · C12 + 1 · C13= 1(−2) + 2(1) + 1(3) = 3
First Column det(A) = 1 · C11 + 2 · C21 + 1 · C31= 1(−2) + 2(2) + 1(1) = 3
Third Row det(A) = 1 · C31 + 2 · C32 + 0 · C33= 1(1) + 2(1) + 0(−3) = 3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 19 / 40
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Determinant Adjoint of a Matrix
Example (3.4.2)
Find det(A) in example 3.4.1 by cofactor expansions.
A =
1 2 12 1 11 2 0
Solution
C11 = −2
C12 = 1
C13 = 3
C21 = 2
C22 = −1
C23 = 0
C31 = 1
C32 = 1
C33 = −3
Thus,
First Row det(A) = 1 · C11 + 2 · C12 + 1 · C13= 1(−2) + 2(1) + 1(3) = 3
First Column
det(A) = 1 · C11 + 2 · C21 + 1 · C31= 1(−2) + 2(2) + 1(1) = 3
Third Row det(A) = 1 · C31 + 2 · C32 + 0 · C33= 1(1) + 2(1) + 0(−3) = 3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 19 / 40
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Determinant Adjoint of a Matrix
Example (3.4.2)
Find det(A) in example 3.4.1 by cofactor expansions.
A =
1 2 12 1 11 2 0
Solution
C11 = −2
C12 = 1
C13 = 3
C21 = 2
C22 = −1
C23 = 0
C31 = 1
C32 = 1
C33 = −3
Thus,
First Row det(A) = 1 · C11 + 2 · C12 + 1 · C13= 1(−2) + 2(1) + 1(3) = 3
First Column det(A) = 1 · C11 + 2 · C21 + 1 · C31
= 1(−2) + 2(2) + 1(1) = 3
Third Row det(A) = 1 · C31 + 2 · C32 + 0 · C33= 1(1) + 2(1) + 0(−3) = 3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 19 / 40
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Determinant Adjoint of a Matrix
Example (3.4.2)
Find det(A) in example 3.4.1 by cofactor expansions.
A =
1 2 12 1 11 2 0
Solution
C11 = −2
C12 = 1
C13 = 3
C21 = 2
C22 = −1
C23 = 0
C31 = 1
C32 = 1
C33 = −3
Thus,
First Row det(A) = 1 · C11 + 2 · C12 + 1 · C13= 1(−2) + 2(1) + 1(3) = 3
First Column det(A) = 1 · C11 + 2 · C21 + 1 · C31= 1(−2) + 2(2) + 1(1) = 3
Third Row det(A) = 1 · C31 + 2 · C32 + 0 · C33= 1(1) + 2(1) + 0(−3) = 3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 19 / 40
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Determinant Adjoint of a Matrix
Example (3.4.2)
Find det(A) in example 3.4.1 by cofactor expansions.
A =
1 2 12 1 11 2 0
Solution
C11 = −2
C12 = 1
C13 = 3
C21 = 2
C22 = −1
C23 = 0
C31 = 1
C32 = 1
C33 = −3
Thus,
First Row det(A) = 1 · C11 + 2 · C12 + 1 · C13= 1(−2) + 2(1) + 1(3) = 3
First Column det(A) = 1 · C11 + 2 · C21 + 1 · C31= 1(−2) + 2(2) + 1(1) = 3
Third Row
det(A) = 1 · C31 + 2 · C32 + 0 · C33= 1(1) + 2(1) + 0(−3) = 3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 19 / 40
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Determinant Adjoint of a Matrix
Example (3.4.2)
Find det(A) in example 3.4.1 by cofactor expansions.
A =
1 2 12 1 11 2 0
Solution
C11 = −2
C12 = 1
C13 = 3
C21 = 2
C22 = −1
C23 = 0
C31 = 1
C32 = 1
C33 = −3
Thus,
First Row det(A) = 1 · C11 + 2 · C12 + 1 · C13= 1(−2) + 2(1) + 1(3) = 3
First Column det(A) = 1 · C11 + 2 · C21 + 1 · C31= 1(−2) + 2(2) + 1(1) = 3
Third Row det(A) = 1 · C31 + 2 · C32 + 0 · C33
= 1(1) + 2(1) + 0(−3) = 3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 19 / 40
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Determinant Adjoint of a Matrix
Example (3.4.2)
Find det(A) in example 3.4.1 by cofactor expansions.
A =
1 2 12 1 11 2 0
Solution
C11 = −2
C12 = 1
C13 = 3
C21 = 2
C22 = −1
C23 = 0
C31 = 1
C32 = 1
C33 = −3
Thus,
First Row det(A) = 1 · C11 + 2 · C12 + 1 · C13= 1(−2) + 2(1) + 1(3) = 3
First Column det(A) = 1 · C11 + 2 · C21 + 1 · C31= 1(−2) + 2(2) + 1(1) = 3
Third Row det(A) = 1 · C31 + 2 · C32 + 0 · C33= 1(1) + 2(1) + 0(−3) = 3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 19 / 40
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Determinant Adjoint of a Matrix
Example (3.4.3)
Find determinant of the following matrices by row operations or cofactorexpansions or both.
1.
1 3 50 2 20 3 −1
2.
3 5 1 50 0 −1 12 4 0 5−1 0 9 0
3.
3 0 0 01 2 −1 12 4 0 53 4 5 3
4.
3 5 −2 61 2 −1 12 4 1 53 7 5 3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 20 / 40
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Determinant Adjoint of a Matrix
Solution
1. Cofactor expansion
∣∣∣∣∣∣1 3 50 2 20 3 −1
∣∣∣∣∣∣ = 1 · C11 + 0 · C21 + 0 · C31 = 1 · (−1)1+1
∣∣∣∣2 23 −1
∣∣∣∣ = −8 #
Row Operation∣∣∣∣∣∣1 3 50 2 20 3 −1
∣∣∣∣∣∣ = 2
∣∣∣∣∣∣1 3 50 1 10 3 −1
∣∣∣∣∣∣ = 2
∣∣∣∣∣∣1 3 50 1 10 0 −4
∣∣∣∣∣∣ = 2(1)(1)(−4) = −8 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 21 / 40
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Determinant Adjoint of a Matrix
Solution
1. Cofactor expansion∣∣∣∣∣∣1 3 50 2 20 3 −1
∣∣∣∣∣∣ = 1 · C11 + 0 · C21 + 0 · C31 = 1 · (−1)1+1
∣∣∣∣2 23 −1
∣∣∣∣ = −8 #
Row Operation∣∣∣∣∣∣1 3 50 2 20 3 −1
∣∣∣∣∣∣ = 2
∣∣∣∣∣∣1 3 50 1 10 3 −1
∣∣∣∣∣∣ = 2
∣∣∣∣∣∣1 3 50 1 10 0 −4
∣∣∣∣∣∣ = 2(1)(1)(−4) = −8 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 21 / 40
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Determinant Adjoint of a Matrix
Solution
1. Cofactor expansion∣∣∣∣∣∣1 3 50 2 20 3 −1
∣∣∣∣∣∣ = 1 · C11 + 0 · C21 + 0 · C31 = 1 · (−1)1+1
∣∣∣∣2 23 −1
∣∣∣∣ = −8 #
Row Operation
∣∣∣∣∣∣1 3 50 2 20 3 −1
∣∣∣∣∣∣ = 2
∣∣∣∣∣∣1 3 50 1 10 3 −1
∣∣∣∣∣∣ = 2
∣∣∣∣∣∣1 3 50 1 10 0 −4
∣∣∣∣∣∣ = 2(1)(1)(−4) = −8 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 21 / 40
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Determinant Adjoint of a Matrix
Solution
1. Cofactor expansion∣∣∣∣∣∣1 3 50 2 20 3 −1
∣∣∣∣∣∣ = 1 · C11 + 0 · C21 + 0 · C31 = 1 · (−1)1+1
∣∣∣∣2 23 −1
∣∣∣∣ = −8 #
Row Operation∣∣∣∣∣∣1 3 50 2 20 3 −1
∣∣∣∣∣∣ = 2
∣∣∣∣∣∣1 3 50 1 10 3 −1
∣∣∣∣∣∣ = 2
∣∣∣∣∣∣1 3 50 1 10 0 −4
∣∣∣∣∣∣ = 2(1)(1)(−4) = −8 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 21 / 40
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Determinant Adjoint of a Matrix
Solution
2. Cofactor expansion
∣∣∣∣∣∣∣∣3 5 1 50 0 −1 12 4 0 5−1 0 9 0
∣∣∣∣∣∣∣∣ = 0 · C21 + 0 · C22 − 1 · C23 + 1 · C24
= −1 · (−1)2+3
∣∣∣∣∣∣3 5 52 4 5−1 0 0
∣∣∣∣∣∣+ 1 · (−1)2+4
∣∣∣∣∣∣3 5 12 4 0−1 0 9
∣∣∣∣∣∣= −5 + 22 = 17 #
Row Operation∣∣∣∣∣∣∣∣3 5 1 50 0 −1 12 4 0 5−1 0 9 0
∣∣∣∣∣∣∣∣ = −
∣∣∣∣∣∣∣∣−1 0 9 00 0 −1 12 4 0 53 5 1 5
∣∣∣∣∣∣∣∣ = −
∣∣∣∣∣∣∣∣−1 0 9 00 0 −1 10 4 18 50 5 28 5
∣∣∣∣∣∣∣∣= −(−1) · C11 = −1(−1)1+1
∣∣∣∣∣∣0 −1 14 18 55 28 5
∣∣∣∣∣∣= 17 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 22 / 40
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Determinant Adjoint of a Matrix
Solution
2. Cofactor expansion∣∣∣∣∣∣∣∣3 5 1 50 0 −1 12 4 0 5−1 0 9 0
∣∣∣∣∣∣∣∣ = 0 · C21 + 0 · C22 − 1 · C23 + 1 · C24
= −1 · (−1)2+3
∣∣∣∣∣∣3 5 52 4 5−1 0 0
∣∣∣∣∣∣+ 1 · (−1)2+4
∣∣∣∣∣∣3 5 12 4 0−1 0 9
∣∣∣∣∣∣= −5 + 22 = 17 #
Row Operation∣∣∣∣∣∣∣∣3 5 1 50 0 −1 12 4 0 5−1 0 9 0
∣∣∣∣∣∣∣∣ = −
∣∣∣∣∣∣∣∣−1 0 9 00 0 −1 12 4 0 53 5 1 5
∣∣∣∣∣∣∣∣ = −
∣∣∣∣∣∣∣∣−1 0 9 00 0 −1 10 4 18 50 5 28 5
∣∣∣∣∣∣∣∣= −(−1) · C11 = −1(−1)1+1
∣∣∣∣∣∣0 −1 14 18 55 28 5
∣∣∣∣∣∣= 17 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 22 / 40
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Determinant Adjoint of a Matrix
Solution
2. Cofactor expansion∣∣∣∣∣∣∣∣3 5 1 50 0 −1 12 4 0 5−1 0 9 0
∣∣∣∣∣∣∣∣ = 0 · C21 + 0 · C22 − 1 · C23 + 1 · C24
= −1 · (−1)2+3
∣∣∣∣∣∣3 5 52 4 5−1 0 0
∣∣∣∣∣∣+ 1 · (−1)2+4
∣∣∣∣∣∣3 5 12 4 0−1 0 9
∣∣∣∣∣∣= −5 + 22 = 17 #
Row Operation
∣∣∣∣∣∣∣∣3 5 1 50 0 −1 12 4 0 5−1 0 9 0
∣∣∣∣∣∣∣∣ = −
∣∣∣∣∣∣∣∣−1 0 9 00 0 −1 12 4 0 53 5 1 5
∣∣∣∣∣∣∣∣ = −
∣∣∣∣∣∣∣∣−1 0 9 00 0 −1 10 4 18 50 5 28 5
∣∣∣∣∣∣∣∣= −(−1) · C11 = −1(−1)1+1
∣∣∣∣∣∣0 −1 14 18 55 28 5
∣∣∣∣∣∣= 17 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 22 / 40
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Determinant Adjoint of a Matrix
Solution
2. Cofactor expansion∣∣∣∣∣∣∣∣3 5 1 50 0 −1 12 4 0 5−1 0 9 0
∣∣∣∣∣∣∣∣ = 0 · C21 + 0 · C22 − 1 · C23 + 1 · C24
= −1 · (−1)2+3
∣∣∣∣∣∣3 5 52 4 5−1 0 0
∣∣∣∣∣∣+ 1 · (−1)2+4
∣∣∣∣∣∣3 5 12 4 0−1 0 9
∣∣∣∣∣∣= −5 + 22 = 17 #
Row Operation∣∣∣∣∣∣∣∣3 5 1 50 0 −1 12 4 0 5−1 0 9 0
∣∣∣∣∣∣∣∣ = −
∣∣∣∣∣∣∣∣−1 0 9 00 0 −1 12 4 0 53 5 1 5
∣∣∣∣∣∣∣∣ = −
∣∣∣∣∣∣∣∣−1 0 9 00 0 −1 10 4 18 50 5 28 5
∣∣∣∣∣∣∣∣= −(−1) · C11 = −1(−1)1+1
∣∣∣∣∣∣0 −1 14 18 55 28 5
∣∣∣∣∣∣= 17 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 22 / 40
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Determinant Adjoint of a Matrix
Solution
3. Cofactor expansion
∣∣∣∣∣∣∣∣3 0 0 01 2 −1 12 4 0 53 4 5 3
∣∣∣∣∣∣∣∣ = 3C11 = 3
∣∣∣∣∣∣2 −1 14 0 54 5 3
∣∣∣∣∣∣ = 3(−38) = −114 #
Row Operation∣∣∣∣∣∣∣∣3 0 0 01 2 −1 12 4 0 53 4 5 3
∣∣∣∣∣∣∣∣ = −
∣∣∣∣∣∣∣∣1 2 −1 13 0 0 02 4 0 53 4 5 3
∣∣∣∣∣∣∣∣ = −
∣∣∣∣∣∣∣∣1 2 −1 10 −6 3 −30 0 2 30 −2 8 0
∣∣∣∣∣∣∣∣= −3
∣∣∣∣∣∣∣∣1 2 −1 10 −2 1 −10 0 2 30 −2 8 0
∣∣∣∣∣∣∣∣ = −3
∣∣∣∣∣∣∣∣1 2 −1 10 −2 1 −10 0 2 30 0 7 1
∣∣∣∣∣∣∣∣= −3
∣∣∣∣∣∣∣∣1 2 −1 10 −2 1 −10 0 2 30 0 0 −9.5
∣∣∣∣∣∣∣∣ = (−3)(1)(−2)(2)(−9.5) = 114 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 23 / 40
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Determinant Adjoint of a Matrix
Solution
3. Cofactor expansion∣∣∣∣∣∣∣∣3 0 0 01 2 −1 12 4 0 53 4 5 3
∣∣∣∣∣∣∣∣ = 3C11 = 3
∣∣∣∣∣∣2 −1 14 0 54 5 3
∣∣∣∣∣∣ = 3(−38) = −114 #
Row Operation∣∣∣∣∣∣∣∣3 0 0 01 2 −1 12 4 0 53 4 5 3
∣∣∣∣∣∣∣∣ = −
∣∣∣∣∣∣∣∣1 2 −1 13 0 0 02 4 0 53 4 5 3
∣∣∣∣∣∣∣∣ = −
∣∣∣∣∣∣∣∣1 2 −1 10 −6 3 −30 0 2 30 −2 8 0
∣∣∣∣∣∣∣∣= −3
∣∣∣∣∣∣∣∣1 2 −1 10 −2 1 −10 0 2 30 −2 8 0
∣∣∣∣∣∣∣∣ = −3
∣∣∣∣∣∣∣∣1 2 −1 10 −2 1 −10 0 2 30 0 7 1
∣∣∣∣∣∣∣∣= −3
∣∣∣∣∣∣∣∣1 2 −1 10 −2 1 −10 0 2 30 0 0 −9.5
∣∣∣∣∣∣∣∣ = (−3)(1)(−2)(2)(−9.5) = 114 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 23 / 40
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Determinant Adjoint of a Matrix
Solution
3. Cofactor expansion∣∣∣∣∣∣∣∣3 0 0 01 2 −1 12 4 0 53 4 5 3
∣∣∣∣∣∣∣∣ = 3C11 = 3
∣∣∣∣∣∣2 −1 14 0 54 5 3
∣∣∣∣∣∣ = 3(−38) = −114 #
Row Operation
∣∣∣∣∣∣∣∣3 0 0 01 2 −1 12 4 0 53 4 5 3
∣∣∣∣∣∣∣∣ = −
∣∣∣∣∣∣∣∣1 2 −1 13 0 0 02 4 0 53 4 5 3
∣∣∣∣∣∣∣∣ = −
∣∣∣∣∣∣∣∣1 2 −1 10 −6 3 −30 0 2 30 −2 8 0
∣∣∣∣∣∣∣∣= −3
∣∣∣∣∣∣∣∣1 2 −1 10 −2 1 −10 0 2 30 −2 8 0
∣∣∣∣∣∣∣∣ = −3
∣∣∣∣∣∣∣∣1 2 −1 10 −2 1 −10 0 2 30 0 7 1
∣∣∣∣∣∣∣∣= −3
∣∣∣∣∣∣∣∣1 2 −1 10 −2 1 −10 0 2 30 0 0 −9.5
∣∣∣∣∣∣∣∣ = (−3)(1)(−2)(2)(−9.5) = 114 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 23 / 40
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Determinant Adjoint of a Matrix
Solution
3. Cofactor expansion∣∣∣∣∣∣∣∣3 0 0 01 2 −1 12 4 0 53 4 5 3
∣∣∣∣∣∣∣∣ = 3C11 = 3
∣∣∣∣∣∣2 −1 14 0 54 5 3
∣∣∣∣∣∣ = 3(−38) = −114 #
Row Operation∣∣∣∣∣∣∣∣3 0 0 01 2 −1 12 4 0 53 4 5 3
∣∣∣∣∣∣∣∣ = −
∣∣∣∣∣∣∣∣1 2 −1 13 0 0 02 4 0 53 4 5 3
∣∣∣∣∣∣∣∣ = −
∣∣∣∣∣∣∣∣1 2 −1 10 −6 3 −30 0 2 30 −2 8 0
∣∣∣∣∣∣∣∣= −3
∣∣∣∣∣∣∣∣1 2 −1 10 −2 1 −10 0 2 30 −2 8 0
∣∣∣∣∣∣∣∣ = −3
∣∣∣∣∣∣∣∣1 2 −1 10 −2 1 −10 0 2 30 0 7 1
∣∣∣∣∣∣∣∣= −3
∣∣∣∣∣∣∣∣1 2 −1 10 −2 1 −10 0 2 30 0 0 −9.5
∣∣∣∣∣∣∣∣ = (−3)(1)(−2)(2)(−9.5) = 114 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 23 / 40
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Determinant Adjoint of a Matrix
Solution
4.
3 5 −2 61 2 −1 12 4 1 53 7 5 3
Row Operation∣∣∣∣∣∣∣∣
3 5 −2 61 2 −1 12 4 1 53 7 5 3
∣∣∣∣∣∣∣∣ = −
∣∣∣∣∣∣∣∣1 2 −1 13 5 −2 62 4 1 53 7 5 3
∣∣∣∣∣∣∣∣ = −
∣∣∣∣∣∣∣∣1 2 −1 10 −1 1 30 0 3 30 1 8 0
∣∣∣∣∣∣∣∣= −
∣∣∣∣∣∣∣∣1 2 −1 10 −1 1 30 0 3 30 0 9 3
∣∣∣∣∣∣∣∣ = −
∣∣∣∣∣∣∣∣1 2 −1 10 −1 1 30 0 3 30 0 0 −6
∣∣∣∣∣∣∣∣= (−1)(1)(−1)(3)(−6) = −18 #
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 24 / 40
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Determinant Adjoint of a Matrix
Take a BreakFor 10 Minutes
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 25 / 40
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Determinant Adjoint of a Matrix
Adjoint Matrix
Definition (3.4.2)
Let A be any n× n matrix and Cij be the cofactor of aij. The the matrixC11 C12 · · · C1n
C21 C22 · · · C2n
......
...Cn1 Cn2 · · · Cnn
is called the matrix of cofactor from A. The transpose of this matrix is
called the adjoint of A and denoted by adj(A) .
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 26 / 40
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Determinant Adjoint of a Matrix
Example (3.4.4)
Find the matrix of cofactor and adjoint of the matrix A in example 3.4.1
A =
1 2 12 1 11 2 0
Solution
C11 = −2
C12 = 1
C13 = 3
C21 = 2
C22 = −1
C23 = 0
C31 = 1
C32 = 1
C33 = −3
Co(A) =
−2 1 32 −1 01 1 −3
and adj(A) = [Co(A)]T =
−2 2 11 −1 13 0 −3
#
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 27 / 40
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Determinant Adjoint of a Matrix
Example (3.4.4)
Find the matrix of cofactor and adjoint of the matrix A in example 3.4.1
A =
1 2 12 1 11 2 0
Solution
C11 = −2
C12 = 1
C13 = 3
C21 = 2
C22 = −1
C23 = 0
C31 = 1
C32 = 1
C33 = −3
Co(A) =
−2 1 32 −1 01 1 −3
and adj(A) = [Co(A)]T =
−2 2 11 −1 13 0 −3
#
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 27 / 40
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Determinant Adjoint of a Matrix
Example (3.4.4)
Find the matrix of cofactor and adjoint of the matrix A in example 3.4.1
A =
1 2 12 1 11 2 0
Solution
C11 = −2
C12 = 1
C13 = 3
C21 = 2
C22 = −1
C23 = 0
C31 = 1
C32 = 1
C33 = −3
Co(A) =
−2 1 32 −1 01 1 −3
and adj(A) = [Co(A)]T =
−2 2 11 −1 13 0 −3
#
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 27 / 40
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Determinant Adjoint of a Matrix
Example (3.4.4)
Find the matrix of cofactor and adjoint of the matrix A in example 3.4.1
A =
1 2 12 1 11 2 0
Solution
C11 = −2
C12 = 1
C13 = 3
C21 = 2
C22 = −1
C23 = 0
C31 = 1
C32 = 1
C33 = −3
Co(A) =
−2 1 32 −1 01 1 −3
and adj(A) = [Co(A)]T =
−2 2 11 −1 13 0 −3
#
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 27 / 40
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Determinant Adjoint of a Matrix
Inverse of Matrix Using its Adjoint
Theorem (3.4.2)
If A is invertible matrix, then
A−1 =1
det(A)adj(A)
Example (3.4.5)
Use adjoint to find the inverse of the matrix A in example 3.4.1
A =
1 2 12 1 11 2 0
Solution
A−1 = 1det(A)
adj(A) = 13
−2 2 11 −1 13 0 −3
#
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 28 / 40
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Determinant Adjoint of a Matrix
Inverse of Matrix Using its Adjoint
Theorem (3.4.2)
If A is invertible matrix, then
A−1 =1
det(A)adj(A)
Example (3.4.5)
Use adjoint to find the inverse of the matrix A in example 3.4.1
A =
1 2 12 1 11 2 0
Solution
A−1 = 1det(A)
adj(A) = 13
−2 2 11 −1 13 0 −3
#
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 28 / 40
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Determinant Adjoint of a Matrix
Inverse of Matrix Using its Adjoint
Theorem (3.4.2)
If A is invertible matrix, then
A−1 =1
det(A)adj(A)
Example (3.4.5)
Use adjoint to find the inverse of the matrix A in example 3.4.1
A =
1 2 12 1 11 2 0
Solution
A−1 = 1det(A)
adj(A) = 13
−2 2 11 −1 13 0 −3
#
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 28 / 40
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Determinant Adjoint of a Matrix
Example (3.4.6)
Use adjoint to find the inverse of the matrix A where
A =
1 3 42 4 26 3 −1
Solution
Then det(A) = −40. Thus,
A−1
=1
det(A)adj(A) = −
1
40
C11 C21 C31C12 C22 C32C13 C23 C33
= −1
40
+
∣∣∣∣4 23 −1
∣∣∣∣ −∣∣∣∣3 43 −1
∣∣∣∣ +
∣∣∣∣3 44 2
∣∣∣∣−
∣∣∣∣2 26 −1
∣∣∣∣ +
∣∣∣∣1 46 −1
∣∣∣∣ −∣∣∣∣1 42 2
∣∣∣∣+
∣∣∣∣2 46 3
∣∣∣∣ −∣∣∣∣1 36 3
∣∣∣∣ +
∣∣∣∣1 32 4
∣∣∣∣
= −1
40
−10 15 −1014 −25 6−18 15 −2
#
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 29 / 40
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Determinant Adjoint of a Matrix
Example (3.4.6)
Use adjoint to find the inverse of the matrix A where
A =
1 3 42 4 26 3 −1
Solution
Then det(A) = −40.
Thus,
A−1
=1
det(A)adj(A) = −
1
40
C11 C21 C31C12 C22 C32C13 C23 C33
= −1
40
+
∣∣∣∣4 23 −1
∣∣∣∣ −∣∣∣∣3 43 −1
∣∣∣∣ +
∣∣∣∣3 44 2
∣∣∣∣−
∣∣∣∣2 26 −1
∣∣∣∣ +
∣∣∣∣1 46 −1
∣∣∣∣ −∣∣∣∣1 42 2
∣∣∣∣+
∣∣∣∣2 46 3
∣∣∣∣ −∣∣∣∣1 36 3
∣∣∣∣ +
∣∣∣∣1 32 4
∣∣∣∣
= −1
40
−10 15 −1014 −25 6−18 15 −2
#
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 29 / 40
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Determinant Adjoint of a Matrix
Example (3.4.6)
Use adjoint to find the inverse of the matrix A where
A =
1 3 42 4 26 3 −1
Solution
Then det(A) = −40. Thus,
A−1
=1
det(A)adj(A) = −
1
40
C11 C21 C31C12 C22 C32C13 C23 C33
= −1
40
+
∣∣∣∣4 23 −1
∣∣∣∣ −∣∣∣∣3 43 −1
∣∣∣∣ +
∣∣∣∣3 44 2
∣∣∣∣−
∣∣∣∣2 26 −1
∣∣∣∣ +
∣∣∣∣1 46 −1
∣∣∣∣ −∣∣∣∣1 42 2
∣∣∣∣+
∣∣∣∣2 46 3
∣∣∣∣ −∣∣∣∣1 36 3
∣∣∣∣ +
∣∣∣∣1 32 4
∣∣∣∣
= −1
40
−10 15 −1014 −25 6−18 15 −2
#
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 29 / 40
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Determinant Adjoint of a Matrix
Entry of Inverse Matrix
Theorem (3.4.3)
Let A = [aij ] be an n× n matrix and Cij be the cofactor of aij. If A isinvertible and A−1 = [a∗ij ], then
a∗ij =1
det(A)Cji
[a∗ij ] = A−1 =1
det(A)
C11 C21 · · · Cj1 · · · Cn1
C12 C22 · · · Cj2 · · · Cn2
...C1i C2i · · · Cji · · · Cni
...C1n C2n · · · Cjn · · · Cnn
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 30 / 40
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Determinant Adjoint of a Matrix
Entry of Inverse Matrix
Theorem (3.4.3)
Let A = [aij ] be an n× n matrix and Cij be the cofactor of aij. If A isinvertible and A−1 = [a∗ij ], then
a∗ij =1
det(A)Cji
[a∗ij ] = A−1 =1
det(A)
C11 C21 · · · Cj1 · · · Cn1
C12 C22 · · · Cj2 · · · Cn2
...C1i C2i · · · Cji · · · Cni
...C1n C2n · · · Cjn · · · Cnn
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 30 / 40
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Determinant Adjoint of a Matrix
Example (3.4.7)
Use theorem 3.4.3 to compute entry of the inverse of matrix A where
A =
1 3 4 12 4 2 00 2 1 01 3 −1 0
If A−1 = [a∗ij ], find a∗12 and a∗42
Solution
det(A) = 1 · C14 = −
∣∣∣∣∣∣2 4 20 2 11 3 −1
∣∣∣∣∣∣ =∣∣∣∣∣∣1 3 −10 2 12 4 2
∣∣∣∣∣∣R13 =
∣∣∣∣∣∣1 3 −10 2 10 −2 4
∣∣∣∣∣∣R3 − 2R1
=
∣∣∣∣∣∣1 3 −10 2 10 0 5
∣∣∣∣∣∣R3 +R2
= (1)(2)(5) = 10
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 31 / 40
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Determinant Adjoint of a Matrix
Example (3.4.7)
Use theorem 3.4.3 to compute entry of the inverse of matrix A where
A =
1 3 4 12 4 2 00 2 1 01 3 −1 0
If A−1 = [a∗ij ], find a∗12 and a∗42
Solution
det(A) = 1 · C14 = −
∣∣∣∣∣∣2 4 20 2 11 3 −1
∣∣∣∣∣∣ =∣∣∣∣∣∣1 3 −10 2 12 4 2
∣∣∣∣∣∣R13 =
∣∣∣∣∣∣1 3 −10 2 10 −2 4
∣∣∣∣∣∣R3 − 2R1
=
∣∣∣∣∣∣1 3 −10 2 10 0 5
∣∣∣∣∣∣R3 +R2
= (1)(2)(5) = 10
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 31 / 40
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Determinant Adjoint of a Matrix
Solution
a∗12 =1
det(A)C21 =
1
10(−1)2+1
∣∣∣∣∣∣3 4 12 1 03 −1 0
∣∣∣∣∣∣= −
1
10· 1 · (−1)1+3
∣∣∣∣2 13 −1
∣∣∣∣ = 5
10=
1
2#
a∗42 =1
det(A)C24 =
1
10(−1)2+4
∣∣∣∣∣∣1 3 40 2 11 3 −1
∣∣∣∣∣∣ = 1
10
∣∣∣∣∣∣1 3 40 2 10 0 −5
∣∣∣∣∣∣R3 −R1
=1
10(1)(2)(−5) = −1 #
Computed by www.wolframalpha.com A−1 = 110
0 5 −10 00 −1 4 20 2 2 −410 −10 −10 10
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 32 / 40
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Determinant Adjoint of a Matrix
Solution
a∗12 =1
det(A)C21 =
1
10(−1)2+1
∣∣∣∣∣∣3 4 12 1 03 −1 0
∣∣∣∣∣∣= −
1
10· 1 · (−1)1+3
∣∣∣∣2 13 −1
∣∣∣∣ = 5
10=
1
2#
a∗42 =1
det(A)C24 =
1
10(−1)2+4
∣∣∣∣∣∣1 3 40 2 11 3 −1
∣∣∣∣∣∣ = 1
10
∣∣∣∣∣∣1 3 40 2 10 0 −5
∣∣∣∣∣∣R3 −R1
=1
10(1)(2)(−5) = −1 #
Computed by www.wolframalpha.com A−1 = 110
0 5 −10 00 −1 4 20 2 2 −410 −10 −10 10
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 32 / 40
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Determinant Adjoint of a Matrix
Solution
a∗12 =1
det(A)C21 =
1
10(−1)2+1
∣∣∣∣∣∣3 4 12 1 03 −1 0
∣∣∣∣∣∣= −
1
10· 1 · (−1)1+3
∣∣∣∣2 13 −1
∣∣∣∣ = 5
10=
1
2#
a∗42 =1
det(A)C24 =
1
10(−1)2+4
∣∣∣∣∣∣1 3 40 2 11 3 −1
∣∣∣∣∣∣ = 1
10
∣∣∣∣∣∣1 3 40 2 10 0 −5
∣∣∣∣∣∣R3 −R1
=1
10(1)(2)(−5) = −1 #
Computed by www.wolframalpha.com A−1 = 110
0 5 −10 00 −1 4 20 2 2 −410 −10 −10 10
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 32 / 40
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Determinant Cramer’s Rule
3.5 Cramer’s Rule
Gabriel Cramer (31 July 1704 to 4 January 1752)
He was a Swiss mathematician, born in Geneva. He was the son of physician Jean Cramerand Anne Mallet Cramer.
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 33 / 40
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Determinant Cramer’s Rule
Theorem (3.5.1)
Let a system of n equations in n unknowns be
Ax = b
with det(A) ̸= 0. Then the system has a unique solution and this solution is
x1 = det(A1)det(A) , x2 = det(A2)
det(A) , ..., xn = det(An)det(A)
where Aj is the matrix obtained by replacing the entrices in the jth column ofA by the entries in the matrix b.
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 34 / 40
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Determinant Cramer’s Rule
Sketch Proof for Cramer’s Rule
Ax = b → x = A−1b
x = A−1b =1
det(A)adj(A)b =
1
det(A)
C11 C21 · · · Cj1 · · · Cn1
C12 C22 · · · Cj2 · · · Cn2
...C1i C2i · · · Cji · · · Cni
...C1n C2n · · · Cjn · · · Cnn
b1b2...bi...bn
x =
x1
x2
...xi
...xn
=
1
det(A)
b1C11 + b2C21 + · · ·+ biCj1 + · · ·+ bnCn1
b1C12 + b2C22 + · · ·+ biCj2 + · · ·+ bnCn2
...b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni
...b1C1n + b2C2n + · · · biCjn + · · ·+ bnCnn
xi =
b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni
det(A)
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 35 / 40
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Determinant Cramer’s Rule
Sketch Proof for Cramer’s Rule
Ax = b → x = A−1b
x = A−1b =1
det(A)adj(A)b =
1
det(A)
C11 C21 · · · Cj1 · · · Cn1
C12 C22 · · · Cj2 · · · Cn2
...C1i C2i · · · Cji · · · Cni
...C1n C2n · · · Cjn · · · Cnn
b1b2...bi...bn
x =
x1
x2
...xi
...xn
=
1
det(A)
b1C11 + b2C21 + · · ·+ biCj1 + · · ·+ bnCn1
b1C12 + b2C22 + · · ·+ biCj2 + · · ·+ bnCn2
...b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni
...b1C1n + b2C2n + · · · biCjn + · · ·+ bnCnn
xi =b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni
det(A)
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 35 / 40
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Determinant Cramer’s Rule
Sketch Proof for Cramer’s Rule
Ax = b → x = A−1b
x = A−1b =1
det(A)adj(A)b =
1
det(A)
C11 C21 · · · Cj1 · · · Cn1
C12 C22 · · · Cj2 · · · Cn2
...C1i C2i · · · Cji · · · Cni
...C1n C2n · · · Cjn · · · Cnn
b1b2...bi...bn
x =
x1
x2
...xi
...xn
=
1
det(A)
b1C11 + b2C21 + · · ·+ biCj1 + · · ·+ bnCn1
b1C12 + b2C22 + · · ·+ biCj2 + · · ·+ bnCn2
...b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni
...b1C1n + b2C2n + · · · biCjn + · · ·+ bnCnn
xi =
b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni
det(A)
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 35 / 40
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Determinant Cramer’s Rule
Sketch Proof for Cramer’s Rule (Continue)
xi =b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni
det(A)
Set up,
Ai =
a11 a12 · · · a1i−1 b1 a1i+1 · · · a1na21 a22 · · · a2i−1 b2 a2i+1 · · · a2n
...an1 an2 · · · ani−1 bn ani+1 · · · ann
So, det(Ai) = b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni.Therefore,
xi =det(Ai)
det(A)
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 36 / 40
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Determinant Cramer’s Rule
Sketch Proof for Cramer’s Rule (Continue)
xi =b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni
det(A)
Set up,
Ai =
a11 a12 · · · a1i−1 b1 a1i+1 · · · a1na21 a22 · · · a2i−1 b2 a2i+1 · · · a2n
...an1 an2 · · · ani−1 bn ani+1 · · · ann
So, det(Ai) = b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni.Therefore,
xi =det(Ai)
det(A)
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 36 / 40
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Determinant Cramer’s Rule
Sketch Proof for Cramer’s Rule (Continue)
xi =b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni
det(A)
Set up,
Ai =
a11 a12 · · · a1i−1 b1 a1i+1 · · · a1na21 a22 · · · a2i−1 b2 a2i+1 · · · a2n
...an1 an2 · · · ani−1 bn ani+1 · · · ann
So, det(Ai) = b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni.
Therefore,xi =
det(Ai)
det(A)
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 36 / 40
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Determinant Cramer’s Rule
Sketch Proof for Cramer’s Rule (Continue)
xi =b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni
det(A)
Set up,
Ai =
a11 a12 · · · a1i−1 b1 a1i+1 · · · a1na21 a22 · · · a2i−1 b2 a2i+1 · · · a2n
...an1 an2 · · · ani−1 bn ani+1 · · · ann
So, det(Ai) = b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni.Therefore,
xi =det(Ai)
det(A)
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 36 / 40
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Determinant Cramer’s Rule
Example (3.5.1)
Use Cramer’s rule to solve
1.
x1 + x3 = 4
x1 + x2 − x3 = −2
2x1 − x2 + x3 = 5
2.
x+ y + 2z = 1
2x− y − 3z = −8
x− y + z = 2
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 37 / 40
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Determinant Cramer’s Rule
Solution
1.
x1 + x3 = 4
x1 + x2 − x3 = −2
2x1 − x2 + x3 = 5
↔
1 0 11 1 −12 −1 1
x1
x2
x3
=
4−25
x1 =
∣∣∣∣∣∣∣4 0 1−2 1 −15 −1 1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1
∣∣∣∣∣∣∣=
−3
−3= 1
x2 =
∣∣∣∣∣∣∣1 4 11 −2 −12 5 1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1
∣∣∣∣∣∣∣=
0
−3= 0
x3 =
∣∣∣∣∣∣∣1 0 41 1 −22 −1 5
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1
∣∣∣∣∣∣∣=
−9
−3= 3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 38 / 40
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Determinant Cramer’s Rule
Solution
1.
x1 + x3 = 4
x1 + x2 − x3 = −2
2x1 − x2 + x3 = 5
↔
1 0 11 1 −12 −1 1
x1
x2
x3
=
4−25
x1 =
∣∣∣∣∣∣∣4 0 1−2 1 −15 −1 1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1
∣∣∣∣∣∣∣
=−3
−3= 1
x2 =
∣∣∣∣∣∣∣1 4 11 −2 −12 5 1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1
∣∣∣∣∣∣∣=
0
−3= 0
x3 =
∣∣∣∣∣∣∣1 0 41 1 −22 −1 5
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1
∣∣∣∣∣∣∣=
−9
−3= 3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 38 / 40
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Determinant Cramer’s Rule
Solution
1.
x1 + x3 = 4
x1 + x2 − x3 = −2
2x1 − x2 + x3 = 5
↔
1 0 11 1 −12 −1 1
x1
x2
x3
=
4−25
x1 =
∣∣∣∣∣∣∣4 0 1−2 1 −15 −1 1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1
∣∣∣∣∣∣∣=
−3
−3= 1
x2 =
∣∣∣∣∣∣∣1 4 11 −2 −12 5 1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1
∣∣∣∣∣∣∣=
0
−3= 0
x3 =
∣∣∣∣∣∣∣1 0 41 1 −22 −1 5
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1
∣∣∣∣∣∣∣=
−9
−3= 3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 38 / 40
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Determinant Cramer’s Rule
Solution
1.
x1 + x3 = 4
x1 + x2 − x3 = −2
2x1 − x2 + x3 = 5
↔
1 0 11 1 −12 −1 1
x1
x2
x3
=
4−25
x1 =
∣∣∣∣∣∣∣4 0 1−2 1 −15 −1 1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1
∣∣∣∣∣∣∣=
−3
−3= 1
x2 =
∣∣∣∣∣∣∣1 4 11 −2 −12 5 1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1
∣∣∣∣∣∣∣
=0
−3= 0
x3 =
∣∣∣∣∣∣∣1 0 41 1 −22 −1 5
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1
∣∣∣∣∣∣∣=
−9
−3= 3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 38 / 40
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Determinant Cramer’s Rule
Solution
1.
x1 + x3 = 4
x1 + x2 − x3 = −2
2x1 − x2 + x3 = 5
↔
1 0 11 1 −12 −1 1
x1
x2
x3
=
4−25
x1 =
∣∣∣∣∣∣∣4 0 1−2 1 −15 −1 1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1
∣∣∣∣∣∣∣=
−3
−3= 1
x2 =
∣∣∣∣∣∣∣1 4 11 −2 −12 5 1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1
∣∣∣∣∣∣∣=
0
−3= 0
x3 =
∣∣∣∣∣∣∣1 0 41 1 −22 −1 5
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1
∣∣∣∣∣∣∣=
−9
−3= 3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 38 / 40
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Determinant Cramer’s Rule
Solution
1.
x1 + x3 = 4
x1 + x2 − x3 = −2
2x1 − x2 + x3 = 5
↔
1 0 11 1 −12 −1 1
x1
x2
x3
=
4−25
x1 =
∣∣∣∣∣∣∣4 0 1−2 1 −15 −1 1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1
∣∣∣∣∣∣∣=
−3
−3= 1
x2 =
∣∣∣∣∣∣∣1 4 11 −2 −12 5 1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1
∣∣∣∣∣∣∣=
0
−3= 0
x3 =
∣∣∣∣∣∣∣1 0 41 1 −22 −1 5
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1
∣∣∣∣∣∣∣
=−9
−3= 3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 38 / 40
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Determinant Cramer’s Rule
Solution
1.
x1 + x3 = 4
x1 + x2 − x3 = −2
2x1 − x2 + x3 = 5
↔
1 0 11 1 −12 −1 1
x1
x2
x3
=
4−25
x1 =
∣∣∣∣∣∣∣4 0 1−2 1 −15 −1 1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1
∣∣∣∣∣∣∣=
−3
−3= 1
x2 =
∣∣∣∣∣∣∣1 4 11 −2 −12 5 1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1
∣∣∣∣∣∣∣=
0
−3= 0
x3 =
∣∣∣∣∣∣∣1 0 41 1 −22 −1 5
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 0 11 1 −12 −1 1
∣∣∣∣∣∣∣=
−9
−3= 3
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 38 / 40
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Determinant Cramer’s Rule
Solution
2.
x+ y + 2z = 1
2x− y − 3z = −8
x− y + z = 2
↔
1 1 22 −1 −31 −1 1
xyz
=
1−82
x1 =
∣∣∣∣∣∣∣1 1 2−8 −1 −32 −1 1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1
∣∣∣∣∣∣∣=
18
−11= −
18
11
x2 =
∣∣∣∣∣∣∣1 1 22 −8 −31 2 1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1
∣∣∣∣∣∣∣=
17
−11= −
17
11
x3 =
∣∣∣∣∣∣∣1 1 12 −1 −81 −1 2
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1
∣∣∣∣∣∣∣=
−23
−11=
23
11
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 39 / 40
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Determinant Cramer’s Rule
Solution
2.
x+ y + 2z = 1
2x− y − 3z = −8
x− y + z = 2
↔
1 1 22 −1 −31 −1 1
xyz
=
1−82
x1 =
∣∣∣∣∣∣∣1 1 2−8 −1 −32 −1 1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1
∣∣∣∣∣∣∣
=18
−11= −
18
11
x2 =
∣∣∣∣∣∣∣1 1 22 −8 −31 2 1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1
∣∣∣∣∣∣∣=
17
−11= −
17
11
x3 =
∣∣∣∣∣∣∣1 1 12 −1 −81 −1 2
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1
∣∣∣∣∣∣∣=
−23
−11=
23
11
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 39 / 40
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Determinant Cramer’s Rule
Solution
2.
x+ y + 2z = 1
2x− y − 3z = −8
x− y + z = 2
↔
1 1 22 −1 −31 −1 1
xyz
=
1−82
x1 =
∣∣∣∣∣∣∣1 1 2−8 −1 −32 −1 1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1
∣∣∣∣∣∣∣=
18
−11= −
18
11
x2 =
∣∣∣∣∣∣∣1 1 22 −8 −31 2 1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1
∣∣∣∣∣∣∣=
17
−11= −
17
11
x3 =
∣∣∣∣∣∣∣1 1 12 −1 −81 −1 2
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1
∣∣∣∣∣∣∣=
−23
−11=
23
11
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 39 / 40
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Determinant Cramer’s Rule
Solution
2.
x+ y + 2z = 1
2x− y − 3z = −8
x− y + z = 2
↔
1 1 22 −1 −31 −1 1
xyz
=
1−82
x1 =
∣∣∣∣∣∣∣1 1 2−8 −1 −32 −1 1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1
∣∣∣∣∣∣∣=
18
−11= −
18
11
x2 =
∣∣∣∣∣∣∣1 1 22 −8 −31 2 1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1
∣∣∣∣∣∣∣
=17
−11= −
17
11
x3 =
∣∣∣∣∣∣∣1 1 12 −1 −81 −1 2
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1
∣∣∣∣∣∣∣=
−23
−11=
23
11
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 39 / 40
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Determinant Cramer’s Rule
Solution
2.
x+ y + 2z = 1
2x− y − 3z = −8
x− y + z = 2
↔
1 1 22 −1 −31 −1 1
xyz
=
1−82
x1 =
∣∣∣∣∣∣∣1 1 2−8 −1 −32 −1 1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1
∣∣∣∣∣∣∣=
18
−11= −
18
11
x2 =
∣∣∣∣∣∣∣1 1 22 −8 −31 2 1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1
∣∣∣∣∣∣∣=
17
−11= −
17
11
x3 =
∣∣∣∣∣∣∣1 1 12 −1 −81 −1 2
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1
∣∣∣∣∣∣∣=
−23
−11=
23
11
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 39 / 40
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Determinant Cramer’s Rule
Solution
2.
x+ y + 2z = 1
2x− y − 3z = −8
x− y + z = 2
↔
1 1 22 −1 −31 −1 1
xyz
=
1−82
x1 =
∣∣∣∣∣∣∣1 1 2−8 −1 −32 −1 1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1
∣∣∣∣∣∣∣=
18
−11= −
18
11
x2 =
∣∣∣∣∣∣∣1 1 22 −8 −31 2 1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1
∣∣∣∣∣∣∣=
17
−11= −
17
11
x3 =
∣∣∣∣∣∣∣1 1 12 −1 −81 −1 2
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1
∣∣∣∣∣∣∣
=−23
−11=
23
11
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 39 / 40
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Determinant Cramer’s Rule
Solution
2.
x+ y + 2z = 1
2x− y − 3z = −8
x− y + z = 2
↔
1 1 22 −1 −31 −1 1
xyz
=
1−82
x1 =
∣∣∣∣∣∣∣1 1 2−8 −1 −32 −1 1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1
∣∣∣∣∣∣∣=
18
−11= −
18
11
x2 =
∣∣∣∣∣∣∣1 1 22 −8 −31 2 1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1
∣∣∣∣∣∣∣=
17
−11= −
17
11
x3 =
∣∣∣∣∣∣∣1 1 12 −1 −81 −1 2
∣∣∣∣∣∣∣∣∣∣∣∣∣∣1 1 22 −1 −31 −1 1
∣∣∣∣∣∣∣=
−23
−11=
23
11
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 39 / 40
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Determinant Cramer’s Rule
Assignment 4
1. Find determinant by EROs and Cofactor expansion.
(a) (EVEN)
2 2 2 −24 3 −4 1−6 −6 −16 01 1 3 5
(b) (ODD)
3 5 5 11 2 3 2−2 −5 −9 01 0 2 −1
2. (EVEN and ODD) Let A =
[1 3−2 −8
]2and B−1 =
[6 75 6
]. Use determinant properties
to evaluate their determinants.
(a) det(A)
(b) det(B)
(c) det(2A3B)
(d) det(3BTA−1)
(e) det((2A)−1BAT )
(f) det(AB +ABT )
3. Use Cramer’s rule to solve
(a) (EVEN)
x1 + x2 + x3 = 6
2x1 − x2 + x3 = 3
x1 + 3x2 − 2x3 = 1
(b) (ODD)
x1 − x2 + 3x3 = 10
x1 + x2 + 2x3 = 2
2x1 + 4x2 + x3 = −8
Thanatyod Jampawai, Ph.D. MAT2305 LINEAR AL :Week4 40 / 40
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