Show that the matrix
Question:

Show that the matrix $A=\left[\begin{array}{rr}5 & 3 \\ 12 & 7\end{array}\right]$ is root of the equation $A^{2}-12 A-1=0$

Solution:

Given : $A=\left[\begin{array}{cc}5 & 3 \\ 12 & 7\end{array}\right]$

Now,

$A^{2}=A A$

$\Rightarrow A^{2}=\left[\begin{array}{cc}5 & 3 \\ 12 & 7\end{array}\right]\left[\begin{array}{cc}5 & 3 \\ 12 & 7\end{array}\right]$

$\Rightarrow A^{2}=\left[\begin{array}{ll}25+36 & 15+21 \\ 60+84 & 36+49\end{array}\right]$

$\Rightarrow A^{2}=\left[\begin{array}{cc}61 & 36 \\ 144 & 85\end{array}\right]$

$A^{2}-12 A-I$

$\Rightarrow A^{2}-12 A-I=\left[\begin{array}{cc}61 & 36 \\ 144 & 85\end{array}\right]-12\left[\begin{array}{cc}5 & 3 \\ 12 & 7\end{array}\right]-\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

$\Rightarrow A^{2}-12 A-I=\left[\begin{array}{cc}61 & 36 \\ 144 & 85\end{array}\right]-\left[\begin{array}{cc}60 & 36 \\ 144 & 84\end{array}\right]-\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

$\Rightarrow A^{2}-12 A-I=\left[\begin{array}{cc}61-60-1 & 36-36+0 \\ 144-144+0 & 85-84-1\end{array}\right]$

$\Rightarrow A^{2}-12 A-I=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$

Since $A$ is satisfying the equation $A^{2}-12 A-I, A$ is the root of the equation $A^{2}-12 A-I$.

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