Question
Show that $\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right]$ is a solution of the matrix equation $X^2-2 X-3 I=0$, Where $I$ is the unit matrix of order 2
✓
Answer
Given
$
x^2-2 x-31=0
$
Solution $=\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right]$
or
$
\begin{aligned}
& X=\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right] \\
& \therefore X^2=\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right] \\
& =\left[\begin{array}{ll}
1+4 & 2+2 \\
2+2 & 4+1
\end{array}\right] \\
& =\left[\begin{array}{ll}
5 & 4 \\
4 & 5
\end{array}\right]
\end{aligned}
$
Now $X^2-2 X-31$
$
\begin{aligned}
& =\left[\begin{array}{ll}
5 & 4 \\
4 & 5
\end{array}\right]-2\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]-3\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
& =\left[\begin{array}{ll}
5 & 4 \\
4 & 5
\end{array}\right]-\left[\begin{array}{ll}
2 & 4 \\
4 & 2
\end{array}\right]-\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right] \\
& =\left[\begin{array}{ll}
5-2-3 & 4-4+0 \\
4-4-0 & 5-2-3
\end{array}\right] \\
& =\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right] \\
& \therefore X ^2=2 X -31=0 \\
&
\end{aligned}
$
Hence proved.
$
x^2-2 x-31=0
$
Solution $=\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right]$
or
$
\begin{aligned}
& X=\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right] \\
& \therefore X^2=\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right] \\
& =\left[\begin{array}{ll}
1+4 & 2+2 \\
2+2 & 4+1
\end{array}\right] \\
& =\left[\begin{array}{ll}
5 & 4 \\
4 & 5
\end{array}\right]
\end{aligned}
$
Now $X^2-2 X-31$
$
\begin{aligned}
& =\left[\begin{array}{ll}
5 & 4 \\
4 & 5
\end{array}\right]-2\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]-3\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
& =\left[\begin{array}{ll}
5 & 4 \\
4 & 5
\end{array}\right]-\left[\begin{array}{ll}
2 & 4 \\
4 & 2
\end{array}\right]-\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right] \\
& =\left[\begin{array}{ll}
5-2-3 & 4-4+0 \\
4-4-0 & 5-2-3
\end{array}\right] \\
& =\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right] \\
& \therefore X ^2=2 X -31=0 \\
&
\end{aligned}
$
Hence proved.
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