· ........................................ 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


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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


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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

∣∣∣∣


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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


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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.


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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 #


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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 #


= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #


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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 #


= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #


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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 #


= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #


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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 #


= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #


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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 #


= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #


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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 #


= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #


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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 #


= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #


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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 #


= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #


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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 #


= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #


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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 #


= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #


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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 #


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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 #


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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 #


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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 #


= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #


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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 #


= (−1)(−8) det(A)

= (−1)(−8)(−5) = −40 #


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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 #


= (−1)(−8) det(A)= (−1)(−8)(−5) = −40 #


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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 .


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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)


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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


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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 )


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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 #


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Solution


1. det(AB)

= det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #





= (36)(−10)2 = 3600 #


8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #


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Solution


1. det(AB) = det(A) det(B)

= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #





= (36)(−10)2 = 3600 #


8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #


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Solution


1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #





= (36)(−10)2 = 3600 #


8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #


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Solution


1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC)

= det(B) det(C)= (−10)(−9) = 90 #





= (36)(−10)2 = 3600 #


8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #


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Solution


1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)

= (−10)(−9) = 90 #





= (36)(−10)2 = 3600 #


8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #


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Solution


1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #





= (36)(−10)2 = 3600 #


8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #


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Solution


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 #




= (36)(−10)2 = 3600 #


8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #


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Solution


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 #




= (36)(−10)2 = 3600 #


8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #


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Solution


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 #




= (36)(−10)2 = 3600 #


8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #


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Solution


1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #





= (36)(−10)2 = 3600 #


8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #


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Solution


1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #


4. det(2AB)

= 22 det(AB)= 4 det(A) det(B)= 4(−1)(−10) = 40 #



= (36)(−10)2 = 3600 #


8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #


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Solution


1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #


4. det(2AB) = 22 det(AB)

= 4 det(A) det(B)= 4(−1)(−10) = 40 #



= (36)(−10)2 = 3600 #


8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #


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Solution


1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #


4. det(2AB) = 22 det(AB)= 4 det(A) det(B)

= 4(−1)(−10) = 40 #



= (36)(−10)2 = 3600 #


8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #


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Solution


1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #





= (36)(−10)2 = 3600 #


8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #


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Solution


1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #



5. det(−3ABTC)

= (−3)2 det(A) det(B) det(C)= 9(−1)(−10)(−9) = −810 #


= (36)(−10)2 = 3600 #


8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #


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Solution


1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #



5. det(−3ABTC) = (−3)2 det(A) det(B) det(C)

= 9(−1)(−10)(−9) = −810 #


= (36)(−10)2 = 3600 #


8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #


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Solution


1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #





= (36)(−10)2 = 3600 #


8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #


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Solution


1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #




6. det(2C(AB2)T )

= 22 det(C) det(A) det(B2)= 4(−9)(−1)[det(B)]2

= (36)(−10)2 = 3600 #


8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #


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Solution


1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #




6. det(2C(AB2)T ) = 22 det(C) det(A) det(B2)

= 4(−9)(−1)[det(B)]2

= (36)(−10)2 = 3600 #


8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #


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Solution


1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #





= (36)(−10)2 = 3600 #


8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #


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Solution


1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #





= (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 #


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Solution


1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #





= (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 #


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Solution


1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #





= (36)(−10)2 = 3600 #


8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #


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Solution


1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #





= (36)(−10)2 = 3600 #


8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]

det(A+AT ) = 2(10)− 5(5) = −5 #


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Solution


1. det(AB) = det(A) det(B)= (−1)(−10) = 10 #

2. det(BC) = det(B) det(C)= (−10)(−9) = 90 #





= (36)(−10)2 = 3600 #


8. det(A+AT )

A+AT =

[1 23 5

]+

[1 32 5

]=

[2 55 10

]det(A+AT ) = 2(10)− 5(5) = −5 #


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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 )


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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 #


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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

#


= 9(−2)2 det(A)

= 94(−6)

= − 38

#


= 4(−3)2 det(A) det(A−1) det(B)

= 4(9)(−6) 1−6

(−12) = −432 #


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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

#


= 9(−2)2 det(A)

= 94(−6)

= − 38

#


= 4(−3)2 det(A) det(A−1) det(B)

= 4(9)(−6) 1−6

(−12) = −432 #


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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

#


= 9(−2)2 det(A)

= 94(−6)

= − 38

#


= 4(−3)2 det(A) det(A−1) det(B)

= 4(9)(−6) 1−6

(−12) = −432 #


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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

#


= 9(−2)2 det(A)

= 94(−6)

= − 38

#


= 4(−3)2 det(A) det(A−1) det(B)

= 4(9)(−6) 1−6

(−12) = −432 #


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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

#


= 9(−2)2 det(A)

= 94(−6)

= − 38

#


= 4(−3)2 det(A) det(A−1) det(B)

= 4(9)(−6) 1−6

(−12) = −432 #


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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(−3)2 det(A) det(A−1) det(B)

= 4(9)(−6) 1−6

(−12) = −432 #


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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(−3)2 det(A) det(A−1) det(B)

= 4(9)(−6) 1−6

(−12) = −432 #


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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

#


= 9(−2)2 det(A)

= 94(−6)

= − 38

#


= 4(−3)2 det(A) det(A−1) det(B)

= 4(9)(−6) 1−6

(−12) = −432 #


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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

#


= 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 #


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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

#


= 9(−2)2 det(A)

= 94(−6)

= − 38

#


= 4(−3)2 det(A) det(A−1) det(B)

= 4(9)(−6) 1−6

(−12) = −432 #


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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.


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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


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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


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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.


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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


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Example (3.4.2)


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




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Example (3.4.2)


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




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Example (3.4.2)


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,





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Example (3.4.2)


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 Column

det(A) = 1 · C11 + 2 · C21 + 1 · C31= 1(−2) + 2(2) + 1(1) = 3



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Example (3.4.2)


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 Column det(A) = 1 · C11 + 2 · C21 + 1 · C31

= 1(−2) + 2(2) + 1(1) = 3



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Example (3.4.2)


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,





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Example (3.4.2)


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,



Third Row

det(A) = 1 · C31 + 2 · C32 + 0 · C33= 1(1) + 2(1) + 0(−3) = 3


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Example (3.4.2)


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,



Third Row det(A) = 1 · C31 + 2 · C32 + 0 · C33

= 1(1) + 2(1) + 0(−3) = 3


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Example (3.4.2)


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,





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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


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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 #


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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 #


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Solution


∣∣∣∣∣∣ = 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 #


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Solution


∣∣∣∣∣∣ = 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 #


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Solution


∣∣∣∣∣∣∣∣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 #


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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 #


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Solution


∣∣∣∣∣∣∣∣ = 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 #


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Solution


∣∣∣∣∣∣∣∣ = 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 #


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Solution


∣∣∣∣∣∣∣∣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 #


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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 #


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Solution


∣∣∣∣∣∣∣∣ = 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 #


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Solution


∣∣∣∣∣∣∣∣ = 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 #


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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 #


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Take a BreakFor 10 Minutes


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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) .


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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

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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

#


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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

#


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Example (3.4.6)


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

#


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Example (3.4.6)


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

#


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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


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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


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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


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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.


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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.


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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)


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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



xi =b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni

det(A)


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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



xi =

b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni

det(A)


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Sketch Proof for Cramer’s Rule (Continue)


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)


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det(A)

Set up,

Ai =



So, det(Ai) = b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni.

Therefore,xi =

det(Ai)

det(A)


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det(A)

Set up,

Ai =



So, det(Ai) = b1C1i + b2C2i + · · ·+ biCji + · · ·+ bnCni.Therefore,

xi =det(Ai)

det(A)


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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


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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


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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


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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


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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


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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


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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


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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


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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


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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


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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


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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


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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


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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


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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


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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


· ........................................ determinant prperties of thr determinant solution...

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