Problem: Find out the inverse of the following elements.
-
$\begin{pmatrix}
1 & 0 \\
2 & 2 \\
\end{pmatrix}$ in $\left( \mathbb{Z} _3, \times \right) $.
-
$\begin{pmatrix}
2 & 3 \\
1 & 2 \\
\end{pmatrix}$ in $ \left( \mathbb{Z} _5, \times \right) $.
-
$\begin{pmatrix}
1 & 0 \\
2 & 2 \\
\end{pmatrix}$ in $\left( \mathbb{Z} _7,\times \right) $.
Solution: We will find the inverse of each matrix in $\mathbb{R} $ and then will convert them to the corresponding field.
-
The inverse of
\[
\begin{pmatrix}
1 & 0 \\
2 & 2 \\
\end{pmatrix} ^{-1} =
\frac{1}{2}
\begin{pmatrix}
2 & 0 \\
-2 & 1 \\
\end{pmatrix}
=
2^{-1}
\begin{pmatrix}
2 & 0 \\
-2 & 1 \\
\end{pmatrix} .
\]
Note that
\[
2^{-1} \equiv 2 \text{ mod } 3 \text{ and } -2 \equiv 1 \text{ mod } 3.
\]
Therefore,
\begin{align*}
\begin{pmatrix}
1 & 0 \\
2 & 2 \\
\end{pmatrix} ^{-1} & = 2^{-1}
\begin{pmatrix}
2 & 0 \\
-2 & 1 \\
\end{pmatrix}
\\[1ex]
& \equiv 2 \begin{pmatrix}
1 & 0 \\
1 & 1 \\
\end{pmatrix} \text{ mod } 3
\\[1ex]
& \equiv \begin{pmatrix}
1 & 0 \\
2 & 2 \\
\end{pmatrix} \text{ mod } 3.
\end{align*}
Hence,
\[
\begin{pmatrix}
1 & 0 \\
2 & 2 \\
\end{pmatrix} ^{-1} = \begin{pmatrix}
1 & 0 \\
2 & 2 \\
\end{pmatrix}.
\]
-
The inverse of
\[
\begin{pmatrix}
2 & 3 \\
1 & 2 \\
\end{pmatrix} ^{-1} =
\frac{1}{1}
\begin{pmatrix}
2 & -3 \\
-1 & 2 \\
\end{pmatrix}
=
\begin{pmatrix}
2 & -3 \\
-1 & 2 \\
\end{pmatrix} .
\]
Note that
\[
-3 \equiv 2\text{ mod } 5 \text{ and } -1 \equiv 4\text{ mod } 5.
\]
Therefore,
\begin{align*}
\begin{pmatrix}
2 & 3 \\
1 & 2 \\
\end{pmatrix} ^{-1} & =
\begin{pmatrix}
2 & -3 \\
-2 & 1 \\
\end{pmatrix}
\\[1ex]
& \equiv
\begin{pmatrix}
2 & 2 \\
4 & 2 \\
\end{pmatrix} \text{ mod } 5.
\end{align*}
Hence,
\[
\begin{pmatrix}
2 & 3 \\
1 & 2 \\
\end{pmatrix} ^{-1} = \begin{pmatrix}
2 & 2 \\
4 & 2 \\
\end{pmatrix}.
\]
-
The inverse of
\[
\begin{pmatrix}
3 & 6 \\
4 & 5
\end{pmatrix} ^{-1} =
\frac{1}{-9}
\begin{pmatrix}
5 & -6 \\
-4 & 3
\end{pmatrix}
=
(-9)^{-1} \begin{pmatrix}
5 & -6 \\
-4 & 3
\end{pmatrix}.
\]
Note that
\begin{align*}
(-9)^{-1} & \equiv 5^{-1} \text{ mod } 7 \equiv 3 \text{ mod } 7, \\
-4 & \equiv 3 \text{ mod } 7, \\
-6 & \equiv 1 \text{ mod } 7.
\end{align*}
Therefore,
\begin{align*}
\begin{pmatrix}
3 & 6 \\
4 & 5
\end{pmatrix} ^{-1} & =
(-9)^{-1} \begin{pmatrix}
5 & -6 \\
-4 & 3
\end{pmatrix}
\\[1ex]
& \equiv
3\begin{pmatrix}
5 & 1 \\
3 & 3
\end{pmatrix} \text{ mod } 7
\\[1ex]
& \equiv \begin{pmatrix}
1 & 3 \\
2 & 2 \\
\end{pmatrix} \text{ mod } 7.
\end{align*}
Hence,
\[
\begin{pmatrix}
3 & 6 \\
4 & 5
\end{pmatrix} ^{-1} = \begin{pmatrix}
1 & 3 \\
2 & 2 \\
\end{pmatrix}.
\]