chapter 3felix_kwok/math2207/ch3.pdf · › for every squarematrix a, one can associate to it a...

DeterminantsChapter 3

MATH 2207 – Linear Algebra

§3.1 Introduction to Determinants› For every square matrix A, one can associate to it a scalar number, called the determinant of A.

› Historically, determinants have been discovered before matrices.› Unlike formulas involving determinants, matrix techniques are now the preferred tool for solving linear systems and other problems in linear algebra.

› Determinants remain important for some purposes, e.g.• Determining whether certain matrices are invertible,• Eigenvalue problems,• Areas and volumes, c.f. Multivariate calculus.

CHAPTER 3: DETERMINANTS 2

Determinants as volumes:› Consider a parallelogram defined by two vectors in R2.

𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 = 𝑑𝑑 𝐴𝐴 − ℎ = 𝑑𝑑 𝐴𝐴 −𝑏𝑏𝑏𝑏𝑑𝑑

= 𝐴𝐴𝑑𝑑 − 𝑏𝑏𝑏𝑏


› For a 2x2 matrix 𝐴𝐴 = 𝐴𝐴 𝑏𝑏𝑏𝑏 𝑑𝑑 , we define the determinant of A to be

det 𝐴𝐴 = 𝐴𝐴𝑑𝑑 − 𝑏𝑏𝑏𝑏.

› This formula gives the signed area of the parallelogram: if we switch the order of the vectors, we get

det 𝑏𝑏 𝑑𝑑𝐴𝐴 𝑏𝑏 = 𝑏𝑏𝑏𝑏 − 𝐴𝐴𝑑𝑑 = −det 𝐴𝐴 𝑏𝑏

𝑏𝑏 𝑑𝑑 .


› The determinant is also defined in 𝑛𝑛 ≥ 3 dimensions. The formula is given recursively by

det𝐴𝐴 = 𝐴𝐴11det 𝐴𝐴11 − 𝐴𝐴12det 𝐴𝐴12 + −⋯+ −1 𝑛𝑛+1𝐴𝐴1𝑛𝑛det 𝐴𝐴1𝑛𝑛

where A1j is the submatrix of A with row 1 and column j deleted.› In R3, this represents the signed area of a parallelepiped formed by the three rows of the matrix.

› Example: calculate det1 2 34 5 67 8 10

.

det1 2 34 5 67 8 10

= 1 ⋅ det 5 68 10 − 2 ⋅ det 4 6

7 10 + 3 ⋅ det 4 57 8

= 1 ⋅ 2 − 2 ⋅ −2 + 3 ⋅ −3 = −3.


› Fact: one can calculate the determinant by expanding along any row or any column.

› For any choice of i* and j*, we have

det𝐴𝐴 = �𝑗𝑗=1

𝑛𝑛

−1 𝑖𝑖∗+𝑗𝑗𝐴𝐴𝑖𝑖∗𝑗𝑗 det𝐴𝐴𝑖𝑖∗𝑗𝑗 = �𝑖𝑖=1

𝑛𝑛

−1 𝑖𝑖+𝑗𝑗∗𝐴𝐴𝑖𝑖𝑗𝑗∗ det𝐴𝐴𝑖𝑖𝑗𝑗∗

› The formula on the previous slide corresponds to expansion along the first row, i.e., i* = 1.

› Warning: beware of the alternating sign pattern!

+ − + −− + − ++ − + −− + − +


› Example: calculate det1 2 34 5 67 8 10

by expanding along the 2nd

column.

det1 2 34 5 67 8 10

= −2 ⋅ det 4 67 10 + 5 ⋅ det 1 3

7 10 − 8 ⋅ det 1 34 6

= −2 ⋅ −2 + 5 ⋅ −11 − 8 ⋅ −6 = −3.

› Note: An alternate notation for det(A) is |A|, i.e., det 𝐴𝐴 𝑏𝑏𝑏𝑏 𝑑𝑑 = 𝐴𝐴 𝑏𝑏

𝑏𝑏 𝑑𝑑 .

› Don’t confuse this with the absolute value!


› Expansion along other rows/columns is useful when the matrix contains many zeros.

› Examples:

1. det

1 2 3 40 0 0 05 6 7 89 10 11 12

= 0.

2. det

1 2 3 40 5 6 70 0 8 90 0 0 10

= 1 ⋅ det5 6 70 8 90 0 10

= 5 ⋅ det 8 90 10 = 400.

3. det 𝐼𝐼𝑛𝑛 = 1.

› Warning: determinants only exist for n x n square matrices!


› Definition: A matrix is said to be • upper triangular if it has no non-zero entries below the diagonal,• lower triangular if it has no non-zero entries above the diagonal,• triangular if it is either upper triangular or lower triangular.

› Theorem: The determinant of a triangular matrix is the product of its diagonal entries.


§3.2 Properties of Determinants› For larger matrices, determinant calculations become cumbersome unless there are many zeros.

› One way to create zeros is to use row operations!› Theorem 3:

Suppose A is a square matrix, and B is obtained from A by performing one elementary row operation. Then the determinants of A and B are related as follows:• Linear combination (𝑅𝑅𝑖𝑖 ← 𝑅𝑅𝑖𝑖 + 𝑏𝑏 ⋅ 𝑅𝑅𝑗𝑗): det(B) = det(A)• Interchange (𝑅𝑅𝑖𝑖 ↔ 𝑅𝑅𝑗𝑗): det(B) = -det(A)• Scaling (𝑅𝑅𝑖𝑖 ← 𝑏𝑏 ⋅ 𝑅𝑅𝑖𝑖): det(B) = c det(A)


› This can be remembered as follows:1 0𝑘𝑘 1 = 1 0

0 1 , 0 11 0 = − 1 0

0 1 , 𝑘𝑘 00 1 = 𝑘𝑘 1 0

0 1 .

› Examples:

1. det 𝑘𝑘 ⋅ 𝐼𝐼3 =𝑘𝑘 0 00 𝑘𝑘 00 0 𝑘𝑘

= 𝑘𝑘 ⋅1 0 00 𝑘𝑘 00 0 𝑘𝑘

= ⋯ = 𝑘𝑘3.

2.1 −4 2−2 8 −9−1 7 0

=1 −4 20 0 −50 3 2

= −1 4 20 3 20 0 −5

= 15.


› Another way of phrasing Theorem 3 is in terms of elementary matrices:

› Theorem: If E is an elementary matrix, then det(EA) = det(E) det(A).

› Suppose we reduce a square matrix A to its RREF U by some elementary row operations. Then

𝐸𝐸𝑝𝑝⋯𝐸𝐸2𝐸𝐸1𝐴𝐴 = 𝑈𝑈det 𝐸𝐸𝑝𝑝 ⋯det 𝐸𝐸1

≠0

det 𝐴𝐴 = det 𝑈𝑈 .

› If A is invertible, then U has n pivots on the diagonal, so det 𝑈𝑈 ≠ 0.

› If A is not invertible, then U has at least one zero row, so det(U) = 0.


› Theorem 4:A square matrix A is invertible if and only if det(A) ≠ 0.

› Geometrically, we have a set of linearly dependent vectors, so the parallelepiped generated by them must have zero volume!

› Examples:

1. Let 𝐴𝐴 =1 2 12 4 13 6 1

. Then the columns of A are linearly

dependent (column 2 is a multiple of column 1), so A is not invertible. Thus, det(A) = 0.

2. If two rows (or columns) in a matrix are identical, or scalar multiples of each other, then A is not invertible, so det(A) = 0.


› Example: determine the values of s for which the matrix

𝐴𝐴 =𝑠𝑠 0 −2−1 𝑠𝑠 −10 −1 𝑠𝑠 + 2

is invertible.

› The determinant of A is det 𝐴𝐴 = 𝑠𝑠 𝑠𝑠 𝑠𝑠 + 2 − 1 + 1 ⋅ −2 = 𝑠𝑠3 + 2𝑠𝑠2 − 𝑠𝑠 − 2

= (𝑠𝑠2 − 1)(𝑠𝑠 + 2)

Thus, A is invertible for all values of s except -2, -1 and 1.

› Note: this example is related to eigenvalue problems, cf. Chapter 5.



Add: det(A) ≠ 0

› Theorem 5: For any square matrix A, det(AT) = det(A).› Proof: Expand along columns instead of rows.

› Theorem 6:If A and B are n x n matrices, then det(AB) = det(A) det(B).

› Proof:• If det(A) = 0 or det(B) = 0, then either A or B is non-invertible, so AB cannot be invertible, which means det(AB) = 0.

• If both A and B are invertible, then both matrices are row-equivalent to the identity matrix, so they can be written as products of elementary matrices. By Theorem 3, we have

det(𝐴𝐴) det 𝐵𝐵 = det 𝐸𝐸1 ⋯𝐸𝐸𝑝𝑝 det �𝐸𝐸1 ⋯ �𝐸𝐸𝑞𝑞= det(𝐸𝐸1 ⋯𝐸𝐸𝑝𝑝 �𝐸𝐸1 ⋯ �𝐸𝐸𝑞𝑞) = det(𝐴𝐴𝐵𝐵).


› Examples:

1. det 𝐴𝐴 𝑏𝑏𝑏𝑏 𝑑𝑑 = det 𝐴𝐴 𝑏𝑏

𝑏𝑏 𝑑𝑑 .

2. Let 𝐴𝐴 = 1 23 4 , 𝐵𝐵 = 2 1

1 1 , 𝐶𝐶 = 𝐴𝐴𝐵𝐵 = 4 310 7 . Then

det 𝐴𝐴 = −2, det 𝐵𝐵 = 1,det 𝐶𝐶 = 28 − 30 = −2 = −2 ⋅ 1.

3. If det(A) = 3, then det(A3) = det(AAA) = (det(A))3 = 27.

4. If A is n x n, then det 2𝐴𝐴 = det 2𝐼𝐼 det 𝐴𝐴 = 2𝑛𝑛 det 𝐴𝐴 .

› Warning: it is not true that det(A+B) = det(A) + det(B)!


§3.3 Applications of Determinants› Areas and volumes:

• We have seen (but not proved) that the area of a parallelogram generated by the columns of A is given by |det(A)|.

• Suppose b1 and b2 are vectors that generate a parallelogram S in R2. Then if T:R2 → R2 is a linear transformation, then T(S) is a parallelogram generated by T(b1) and T(b2), so its area is

det 𝑇𝑇𝒃𝒃1 𝑇𝑇𝒃𝒃2 = | det(𝑇𝑇 𝒃𝒃1,𝒃𝒃2 )| = |det(T)| ⋅ |det( 𝒃𝒃1,𝒃𝒃2 )|.

• Thus, |det(T)| is the amplification factor of the linear transformation T. The same is true in higher dimensions.


› In general, if we have a linear coordinate transformation T, the area/volume of any bounded region is amplified by |det(T)|. This fact will be used in Multivariate Calculus.


› Cramer’s rule:Let A be an invertible n x n matrix, and b be a vector in Rn. We denote by Ai(b) the matrix obtained by replacing the i-th column of A by the vector b.

𝐴𝐴𝑖𝑖 𝒃𝒃 = 𝒂𝒂1 ⋯ 𝒃𝒃 ⋯ 𝒂𝒂𝑛𝑛

Theorem 7 (Cramer’s rule): If A is invertible, then for any vector b, the solution x of the linear system Ax = b has entries given by

𝑥𝑥𝑖𝑖 =det𝐴𝐴𝑖𝑖 𝒃𝒃

det 𝐴𝐴› One can also obtain a formula for A-1 using Cramer’s rule n times, see Theorem 8 in the textbook.


› Proof: First note the equality𝐴𝐴 𝒆𝒆1 ⋯ 𝒙𝒙 ⋯ 𝒆𝒆𝑛𝑛 = 𝒂𝒂1 ⋯ 𝒃𝒃 ⋯ 𝒂𝒂𝑛𝑛 .

By taking determinants on both sides, we getdet 𝐴𝐴 det 𝒆𝒆1 ⋯ 𝒙𝒙 ⋯ 𝒆𝒆𝑛𝑛

=𝑥𝑥𝑖𝑖

= det 𝐴𝐴𝑖𝑖 𝒃𝒃

Thus, if det(A) is non-zero, we get

𝑥𝑥𝑖𝑖 =det𝐴𝐴𝑖𝑖 𝒃𝒃

det 𝐴𝐴.

› In general, it is much more efficient to use Gaussian elimination to solve linear systems than to use Cramer’s rule.

› Cramer’s rule is mostly used as a theoretical tool to deduce the form of the solution without actually calculating it.


› Example: Suppose we want to solve3𝑠𝑠 −2−6 𝑠𝑠

𝑥𝑥1𝑥𝑥2 = 4

1 ,

where s is a parameter. Since all the determinants that appear in Cramer’s rule are polynomials in s, we conclude that the solution x1 and x2 are rational functions of s.

› A detailed calculation shows that

𝑥𝑥1 =4𝑠𝑠 + 2

3𝑠𝑠2 − 12, 𝑥𝑥2 =

𝑠𝑠 + 8𝑠𝑠2 − 4

,

which are defined whenever 3𝑠𝑠2 − 12 ≠ 0, i.e., when the matrix is non-singular.


chapter 3felix_kwok/math2207/ch3.pdf · › for every squarematrix a, one can associate to it a...

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