Section 3.2 Determinants and Matrix Inverses

Theorem: Let [latex]A[/latex] be a square matrix:

(a) If a multiple of one row of [latex]A[/latex] is added to another row to produce a matrix [latex]B[/latex], then det[latex]B =[/latex]det[latex]A[/latex].

(b) If two rows of [latex]A[/latex] are interchanged to produce [latex]B[/latex], then det[latex]B = -[/latex]det[latex]A[/latex]. [latex]([/latex]det[latex]A = -[/latex]det[latex]B)[/latex].

(c) If one row of [latex]A[/latex] is multiplied by [latex]c[/latex] to produce [latex]B[/latex], then det[latex]B = c[/latex]det[latex]A([/latex]det[latex]A = (1/c)[/latex]det[latex]B)[/latex].

Fact: The above theorem could be rewrite as:

(a) If [latex]E[/latex] is obtained by a multiple of one row of [latex]I_{n}[/latex] is added to another row, then det[latex]EA =[/latex]det[latex]E[/latex]det[latex]A =[/latex]det[latex]A[/latex].

(b) If [latex]E[/latex] is obtained by two rows of [latex]I_{n}[/latex] are interchanged, then det[latex]EA =[/latex]det[latex]E[/latex]det[latex]A = -[/latex]det[latex]A[/latex].

(c) If [latex]E[/latex] is obtained by one row of [latex]I_{n}[/latex] is multiplied by [latex]c[/latex], then det[latex]EA =[/latex]det[latex]E[/latex]det[latex]A = c[/latex]det[latex]A([/latex]det[latex]A = (1/c)[/latex]det[latex]EA)[/latex]

Moreover,

det[latex]E = \begin{cases}1 & \text{ if } E \text{ is the row placement of } I_{n}\\-1 & \text{ if } E \text{ is the row exchnage of } I_{n}\\c & \text{ if } E \text{ is a row scale of } I_{n}\end{cases}[/latex]

Sketch of the Proof: 1. Show the case of [latex]n = 2[/latex] is true. 2. Use mathematical induction on [latex]n[/latex]. We use the fact that [latex]n - 1[/latex] is true to show the case [latex]n[/latex] is true. Because most of you did not know mathematical induction, we will not prove it here.

Example 1: Compute the determinant of [latex]A = \begin{bmatrix}1 & 0 & 2 & 3\\-1 & 2 & 4 & 5\\0 & 1 & -1 & 2\\2 & 3 & 0 & -1\end{bmatrix}[/latex].

Exercise 1: Compute the determinant of [latex]A = \begin{bmatrix}1 & 0 & 0 & 3\\1 & 4 & 2 & 0\\-2 & 3 & -1 & 2\\0 & 1 & -2 & -1\end{bmatrix}[/latex].

Theorem: A square matrix [latex]A[/latex] is invertible if and only if det[latex]A \neq 0[/latex].

Fact: 1. An [latex]n \times n[/latex] matrix [latex]A[/latex] is invertible if and only if [latex]A^T[/latex] is invertible. When [latex]A[/latex] is not invertible then [latex]A^T[/latex] is not invertible, then [latex]A^T[/latex] has less than [latex]n[/latex] pivot positions, less than [latex]n[/latex] pivot columns. Hence [latex]A^{T}\vec{x} = \vec{0}[/latex] has nontrivial solution.

2. If we do the row inter-exchange and row replacement on an [latex]n \times n[/latex] matrix [latex]A[/latex] to obtain the Echelon form of [latex]A[/latex], [latex]U[/latex], then det[latex]A = (-1)^{r}[/latex]det[latex]U[/latex] where [latex]r[/latex] is the number of row exchanges. Notice that [latex]U[/latex] is a triangular matrix, hence det[latex]U[/latex] is the product of the diagonal entries. When [latex]U[/latex] does not have [latex]n[/latex] pivot positions, then the det[latex]U = 0[/latex], i.e det[latex]A = 0[/latex] and [latex]A[/latex] is not invertible.

Example 2: Find the determinant of [latex]A = \begin{bmatrix}0 & -1 & 3 & 1\\2 & 3 & 1 & 2\\3 & 5 & -2 & -2\\1 & 2 & 4 & 6\end{bmatrix}[/latex].

Exercise 2: Find the determinant of [latex]A = \begin{bmatrix}1 & 0 & 3 & -1\\0 & 2 & 4 & 3\\1 & 2 & 7 & 2\\3 & 4 & 5 & -2\end{bmatrix}[/latex].

Theorem: If [latex]A[/latex] is an [latex]n \times n[/latex] matrix then det[latex]A =[/latex]det[latex]A^T[/latex].

Theorem: If [latex]A[/latex] and [latex]B[/latex] are [latex]n \times n[/latex] matrices then det[latex]AB =[/latex]det[latex]A[/latex]det[latex]B[/latex].

Example 3: Compute det[latex]AB[/latex] without finding [latex]AB[/latex], where [latex]A = \begin{bmatrix}1 & 2 \\ -1 & 3\end{bmatrix}[/latex], [latex]B = \begin{bmatrix}2 & 3 \\ -2 & 1\end{bmatrix}[/latex].

Exercise 3: Compute det[latex]AB[/latex] without finding [latex]AB[/latex], where [latex]A = \begin{bmatrix}0 & 1 \\ 1 & 4\end{bmatrix}[/latex], [latex]B = \begin{bmatrix}1 & 4 \\ -2 & 1\end{bmatrix}[/latex].

Example 4: Compute det[latex]A^{T}B[/latex] without finding [latex]A^{T}B[/latex], where [latex]A = \begin{bmatrix}1 & 2 \\ -2 & 4\end{bmatrix}[/latex], [latex]B = \begin{bmatrix}2 & -3 \\ 3 & -1\end{bmatrix}[/latex].

Exercise 4: Compute det[latex]A^{T}B[/latex] without finding [latex]A^{T}B[/latex], where [latex]A = \begin{bmatrix}2 & 1 \\ 0 & 4\end{bmatrix}[/latex], [latex]B = \begin{bmatrix}-1 & 3 \\ 1 & -1\end{bmatrix}[/latex].

Group Work 1: Mark each statement True or False. Justify each answer. All matrices are [latex]n \times n[/latex] matrices.

a. A row replacement does not affect the determinant of a matrix.

b. The determinant of [latex]A[/latex] is the product of the diagonal in any echelon form [latex]U[/latex] of [latex]A[/latex], multiplied by [latex](-1)^r[/latex], where [latex]r[/latex] is the number of row interchange made during the row operation.

c. det[latex](A + B) =[/latex]det[latex]A +[/latex]det[latex]B[/latex]

d. If two row exchange are made in succession, then the new determinant
equals the old determinant.

e. The determinant of [latex]A[/latex] is the product of the diagonal entries.

f. If det[latex]A[/latex] is zero, then two rows or two columns are the same, or a row or a column is zero.

g. det[latex]A^T = (-1)[/latex]det[latex]A[/latex].

Group Work 2: Compute det[latex]A^3[/latex].

[latex]A = \begin{bmatrix}1 & 3 & 0\\1 & 2 & -2\\0 & 0 & 1\end{bmatrix}[/latex]

Group Work 3: Show det[latex](A^{-1}) = \frac{1}{\text{det}A}[/latex] when [latex]A[/latex] is invertible.

Group Work 4: In each case either prove the statement or give an example
showing that it is false. All matrices are [latex]n \times n[/latex] matrices.

a. det[latex]AB =[/latex]det[latex]B^{T}A[/latex].

b. If det[latex]A \neq 0[/latex] and [latex]AB = AC[/latex], then [latex]B = C[/latex].

c. If [latex]AB[/latex] is invertible, then [latex]A[/latex] and [latex]B[/latex] are invertible.

d. det[latex](I + A)=1+[/latex]det[latex]A[/latex].

e. [latex]A[/latex] and [latex]P[/latex] are square matrices and [latex]P[/latex] is invertible then det[latex]PAP^{-1} =[/latex]det[latex]A[/latex].

f. If [latex]A^T = -A[/latex], then det[latex]A = -1[/latex].