# Thread: [SOLVED] Inverse of Matrices

1. ## [SOLVED] Inverse of Matrices

How can I find the inverse of the following matrices, if possible?

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

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

2. Hello, jennifer1004!

I must assume you know the Augmented Matrix method.

$A=\begin{pmatrix}2 & 1 & 3\\ 0 & 1 & 2\\1 & 0 & 3\end{pmatrix}$
Immediately evaluate its determinant.
. . If the determinant is zero, there is no inverse.

For this matrix, the determinant is 5.
. . (We can expect the inverse to have denominators of 5.)

We have: . $\left(\begin{array}{ccc|ccc}
2 & 1 & 3 & 1 & 0 & 0 \\ 0 & 1 & 2 & 0 & 1 & 0 \\ 1& 0 & 3 & 0 & 0 & 1\end{array}\right)$

$\begin{array}{c}\text{Switch }R_1\\ \text{and }R_3 \\ \end{array}\left(\begin{array}{ccc|ccc}1&0&3&0&0&1 \\ 0&1&2&0&1&0\\ 2&1&3&1&0&0\end{array}\right)$

$\begin{array}{c}\\ \\ R_3-2R_1\end{array}\left(\begin{array}{ccc|ccc}1&0&3&0 &0&1 \\ 0&1&2&0&1&0 \\ 0&1&\text{-}3 & 1 & 0 & \text{-}2 \end{array}\right)$

$\begin{array}{c}\\ \\ R_3-R_2 \end{array}\left(\begin{array}{ccc|ccc}1&0&3&0&0&1 \\ 0&1&2&0&1&0 \\ 0&0&\text{-}5 & 1&\text{-}1&\text{-}2\end{array}\right)$

. . $\begin{array}{ccc}\\ \\ \text{-}\frac{1}{5}R_3 \end{array}\left(\begin{array}{ccc|ccc}1&0&3&0&0&1 \\ 0&1&2&0&1&0 \\ 0&0&1&\text{-}\frac{1}{5} & \frac{1}{5} & \frac{2}{5} \end{array}\right)$

$\begin{array}{c}R_1-3R_3 \\ R_2-2R_3 \\ \\ \end{array} \left(\begin{array}{ccc|ccc}1&0&0 & \frac{3}{5}&\text{-}\frac{3}{5} & \text{-}\frac{1}{5} \\ \\[-4mm]0&1&0& \frac{2}{5} & \frac{3}{5} & \text{-}\frac{4}{5} \\ \\[-4mm] 0&0&1& \text{-}\frac{1}{5} & \frac{1}{5} & \frac{2}{5} \end{array}\right)$

Therefore: . $A^{-1} \;=\;\begin{pmatrix}\dfrac{3}{5} & \text{-}\dfrac{3}{5} & \text{-}\dfrac{1}{5} \\ \\[-3mm] \dfrac{2}{5} & \dfrac{3}{5} & \text{-}\dfrac{4}{5} \\ \\[-3mm] \text{-}\dfrac{1}{5} & \dfrac{1}{5} & \dfrac{2}{5} \end{pmatrix}$