# 32.5: Determinants

The determinant is a useful value that can be computed from the elements of a square matrix

Consider row reducing the standard 2x2 matrix. Suppose that $$a$$ is nonzero.

$\begin{pmatrix} a &b \\ c &d \end{pmatrix}$

$\frac{1}{a} R_1 \rightarrow R_1, \;\;\; R_2-cR_1 \rightarrow R_2$

$\begin{pmatrix} 1 &\frac{b}{a} \\ c &d \end{pmatrix}$

$\begin{pmatrix} 1 & \frac{b}{a} \\ 0 & d-\frac{cb}{a}\end{pmatrix}$

Now notice that we cannot make the lower right corner a 1 if

$d - \frac{cb}{a} = 0$

or

$ad - bc = 0.$

##### Definition: The Determinant

We call $$ad - bc$$ the determinant of the 2 by 2 matrix

$\begin{pmatrix} a &b \\ c &d \end{pmatrix}$

it tells us when it is possible to row reduce the matrix and find a solution to the linear system.

##### Example 32.5.1 :

The determinant of the matrix

$\begin{pmatrix} 3 & 1\\ 5 & 2 \end{pmatrix}$

is

$3(2) - 1(5) = 6 - 5 = 1.$

## Determinants of 3 x 3 Matrices

We define the determinant of a triangular matrix

$\begin{pmatrix} a &d &e \\ 0 &b &f \\ 0 &0 &c \end{pmatrix}$

by

$\text{det} = abc.$

Notice that if we multiply a row by a constant $$k$$ then the new determinant is $$k$$ times the old one. We list the effect of all three row operations below.

##### Theorem

The effect of the the three basic row operations on the determinant are as follows

1. Multiplication of a row by a constant multiplies the determinant by that constant.
2. Switching two rows changes the sign of the determinant.
3. Replacing one row by that row + a multiply of another row has no effect on the determinant.

To find the determinant of a matrix we use the operations to make the matrix triangular and then work backwards.

##### Example 32.5.2 :

Find the determinant of

$\begin{pmatrix} 2 & 6 &10 \\ 2 &4 &-3 \\ 0 &4 &2 \end{pmatrix}$

We use row operations until the matrix is triangular.

$\dfrac{1}{2}R_1 \rightarrow R_1 \text{(Multiplies the determinant by } \dfrac{1}{2})$

$\begin{pmatrix} 1 & 3 &5 \\ 2 &4 &-3 \\ 0 &4 &2 \end{pmatrix}$

$R_2 - 2R_1 \rightarrow R_2 \text{ (No effect on the determinant)}$

$\begin{pmatrix} 1 & 3 &5 \\ 0 &-2 &-13 \\ 0 &4 &2 \end{pmatrix}$

Note that we do not need to zero out the upper middle number. We only need to zero out the bottom left numbers.

$R_3 + 2R_2 \rightarrow R_3 \text{ (No effect on the determinant)}.$

$\begin{pmatrix} 1 & 3 &5 \\ 0 &-2 &-13 \\ 0 &0 &-24 \end{pmatrix}$

Note that we do not need to make the middle number a 1.

The determinant of this matrix is 48. Since this matrix has $$\frac{1}{2}$$ the determinant of the original matrix, the determinant of the original matrix has

$\text{determinant} = 48(2) = 96.$

## Inverses

We call the square matrix I with all 1's down the diagonal and zeros everywhere else the identity matrix. It has the unique property that if $$A$$ is a square matrix with the same dimensions then

$AI = IA = A.$

Definition

If $$A$$ is a square matrix then the inverse $$A^{-1}$$ of $$A$$ is the unique matrix such that

$AA^{-1}=A^{-1}A=I.$

##### Example 32.5.3 :

Let

$A=\begin{pmatrix} 2 &5 \\ 1 &3 \end{pmatrix}$

then

$A^{-1}= \begin{pmatrix} 3 &-5 \\ -1 &2 \end{pmatrix}$

Verify this!

Theorem: ExistEnce

The inverse of a matrix exists if and only if the determinant is nonzero.

To find the inverse of a matrix, we write a new extended matrix with the identity on the right. Then we completely row reduce, the resulting matrix on the right will be the inverse matrix.

##### Example 32.5.4 :

$\begin{pmatrix} 2 &-1 \\ 1 &-1 \end{pmatrix}$

First note that the determinant of this matrix is

$-2 + 1 = -1$

hence the inverse exists. Now we set the augmented matrix as

$\begin{pmatrix}\begin{array}{cc|cc}2&-1&1&0 \\1&-1&0&1\end{array}\end{pmatrix}$

$R_1 {\leftrightarrow} R_2$

$\begin{pmatrix}\begin{array}{cc|cc}1&-1&0&1 \\2&-1&1&0\end{array}\end{pmatrix}$

$R_2 - 2R_1 {\rightarrow} R_2$

$\begin{pmatrix}\begin{array}{cc|cc}1&-1&0&1 \\0&1&1&-2\end{array}\end{pmatrix}$

$R_1 + R_2 {\rightarrow} R_1$

$\begin{pmatrix}\begin{array}{cc|cc}1&0&1&-1 \\0&1&1&-2\end{array}\end{pmatrix}$

Notice that the left hand part is now the identity. The right hand side is the inverse. Hence

$A^{-1}= \begin{pmatrix} 1&-1 \\ 1&-2 \end{pmatrix}$

## Solving Equations Using Matrices

##### Example 32.5.5 :

Suppose we have the system

$2x - y = 3$

$x - y = 4$

Then we can write this in matrix form

$Ax = b$

where

$A=\begin{pmatrix} 2&-1 \\ 1&-1 \end{pmatrix}, \;\;\; x= \begin{pmatrix} x \\ y \end{pmatrix}, \;\;\; \text{and} \; b=\begin{pmatrix} 3\\4 \end{pmatrix}$

We can multiply both sides by $$A^{-1}$$:

$A^{-1}A x = A^{-1}b$

or

$x = A^{-1}b$

From before,

$A^{-1}=\begin{pmatrix} 1&-1 \\ 1&-2 \end{pmatrix}$

Hence our solution is

$\begin{pmatrix} -1&-5 \end{pmatrix}$

or

$x = -1 \text{ and } y = 5$