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 exchange 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]AB^{T}[/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].
GroupWork 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].
GroupWork 2: Compute det [latex]A^3[/latex].
[latex]A = \begin{bmatrix}1 & 3 & 0\\1 & 2 & -2\\0 & 0 & 1\end{bmatrix}[/latex]
GroupWork 3: Show det [latex](A^{-1}) = \frac{1}{\text{det}A}[/latex] when[latex]A[/latex] is invertible.
GroupWork 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].