# Section 3.2 Determinants and Matrix Inverses

Theorem: Let $A$ be a square matrix:

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

(b) If two rows of $A$ are interchanged to produce $B$, then det$B = -$det$A$. $($det$A = -$det$B)$.

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

Fact: The above theorem could be rewrite as:

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

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

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

Moreover,

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

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

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

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

Theorem: A square matrix $A$ is invertible if and only if det$A \neq 0$.

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

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

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

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

Theorem: If $A$ is an $n \times n$ matrix then det$A =$det$A^T$.

Theorem: If $A$ and $B$ are $n \times n$ matrices then det$AB =$det$A$det$B$.

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

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

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

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

Group Work 1: Mark each statement True or False. Justify each answer. All matrices are $n \times n$ matrices.

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

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

c. det$(A + B) =$det$A +$det$B$

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

e. The determinant of $A$ is the product of the diagonal entries.

f. If det$A$ is zero, then two rows or two columns are the same, or a row or a column is zero.

g. det$A^T = (-1)$det$A$.

Group Work 2: Compute det$A^3$.

$A = \begin{bmatrix}1 & 3 & 0\\1 & 2 & -2\\0 & 0 & 1\end{bmatrix}$

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

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

a. det$AB =$det$B^{T}A$.

b. If det$A \neq 0$ and $AB = AC$, then $B = C$.

c. If $AB$ is invertible, then $A$ and $B$ are invertible.

d. det$(I + A)=1+$det$A$.

e. $A$ and $P$ are square matrices and $P$ is invertible then det$PAP^{-1} =$det$A$.

f. If $A^T = -A$, then det$A = -1$.