1. ## Matrix Inverse

Using Row Operations, find the inverse of $\begin{pmatrix}2&3&4\\0&4&1\\1&0&3 \end{pmatrix}$ in $M_3 (\mathbb{F}_7)$.

2. Hello, funnyinga!

Using Row Operations, find the inverse of: . $A \;=\; \begin{bmatrix}2&3&4\\0&4&1\\1&0&3 \end{bmatrix}$

We have: . $\left[ \begin{array}{ccc|ccc}2 &3&4&1&0&0 \\ 0&4&1&0&1&0\\ 1&0&3&0&0&1\end{array}\right]$

. . $\begin{array}{c}\text{Switch} \\ R_1 \& R_3 \\ \end{array} \left[\begin{array}{ccc|ccc}1&0&3&0&0&1 \\ 0&4&1&0&1&0\\2&3&4&1&0&0 \end{array}\right]$

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

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

$\begin{array}{c}\\ \\ R_3-3R_2 \end{array} \left[\begin{array}{ccc|ccc} 1&0&3&0&0&1 \\ 0&1&3&\text{-}1&1&2 \\ 0&0&\text{-}11 &4&\text{-}3 & \text{-}8 \end{array}\right]$

. . $\begin{array}{c} \\ \\ \text{-}\frac{1}{11}R_3 \end{array} \left[\begin{array}{ccc|ccc} 1&0&3&0&0&1 \\0&1&3&\text{-}1&1&2 \\ 0&0&1&\text{-}\frac{4}{11} & \frac{3}{11} & \frac{8}{11} \end{array}\right]$

$\begin{array}{c}R_2-3R_3 \\ R_2-3R_3 \\ \end{array} \left[\begin{array}{ccc|ccc}1&0&0 & \frac{12}{11} & \text{-}\frac{9}{11} & \text{-}\frac{13}{11} \\ \\[-4mm] 0&1&0 & \frac{1}{11} & \frac{2}{11} & \text{-}\frac{2}{11} \\ \\[-4mm] 0&0&1& \text{-}\frac{4}{11} & \frac{3}{11} & \frac{8}{11}\end{array}\right]$

Therefore: . $A^{-1} \;=\;\frac{1}{11}\begin{bmatrix}12 &\text{-}9 &\text{-}13 \\ 1&2&\text{-}2 \\ \text{-}4 &3&8 \end{bmatrix}$

3. Originally Posted by Soroban
Hello, funnyinga!

We have: . $\left[ \begin{array}{ccc|ccc}2 &3&4&1&0&0 \\ 0&4&1&0&1&0\\ 1&0&3&0&0&1\end{array}\right]$

. . $\begin{array}{c}\text{Switch} \\ R_1 \& R_3 \\ \end{array} \left[\begin{array}{ccc|ccc}1&0&3&0&0&1 \\ 0&4&1&0&1&0\\2&3&4&1&0&0 \end{array}\right]$

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

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

$\begin{array}{c}\\ \\ R_3-3R_2 \end{array} \left[\begin{array}{ccc|ccc} 1&0&3&0&0&1 \\ 0&1&3&\text{-}1&1&2 \\ 0&0&\text{-}11 &4&\text{-}3 & \text{-}8 \end{array}\right]$

. . $\begin{array}{c} \\ \\ \text{-}\frac{1}{11}R_3 \end{array} \left[\begin{array}{ccc|ccc} 1&0&3&0&0&1 \\0&1&3&\text{-}1&1&2 \\ 0&0&1&\text{-}\frac{4}{11} & \frac{3}{11} & \frac{8}{11} \end{array}\right]$

$\begin{array}{c}R_2-3R_3 \\ R_2-3R_3 \\ \end{array} \left[\begin{array}{ccc|ccc}1&0&0 & \frac{12}{11} & \text{-}\frac{9}{11} & \text{-}\frac{13}{11} \\ \\[-4mm] 0&1&0 & \frac{1}{11} & \frac{2}{11} & \text{-}\frac{2}{11} \\ \\[-4mm] 0&0&1& \text{-}\frac{4}{11} & \frac{3}{11} & \frac{8}{11}\end{array}\right]$

Therefore: . $A^{-1} \;=\;\frac{1}{11}\begin{bmatrix}12 &\text{-}9 &\text{-}13 \\ 1&2&\text{-}2 \\ \text{-}4 &3&8 \end{bmatrix}$
i hate to add anything to Soroban's beautiful work, but the last step needs a little care! since everything here is in $\mathbb{F}_7,$ we have $\frac{1}{11}=11^{-1}=2$ and so: $A^{-1}=\begin{pmatrix}3 & 3 & 2 \\ 2 & 4 & 3 \\ 6 & 6 & 2 \end{pmatrix}.$

4. Hello, NonCommAlg!

I was so involved with the inverse, I forgot the modulo theme.

Thanks for catching it!