Question
Find $k$, if $A=\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right]$ and $A^2=k A-2 l$.

Answer

$
\begin{aligned}
A^2= A \cdot A & =\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right] \\
& =\left[\begin{array}{rr}
9-8 & -6+4 \\
12-8 & -8+4
\end{array}\right]=\left[\begin{array}{ll}
1 & -2 \\
4 & -4
\end{array}\right] \\
k A -2 I & =k\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]-2\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
& =\left[\begin{array}{cc}
3 k & -2 k \\
4 k & -2 k
\end{array}\right]-\left[\begin{array}{ll}
2 & 0 \\
0 & 2
\end{array}\right] \\
& =\left[\begin{array}{cc}
3 k-2 & -2 k \\
4 k & -2 k-2
\end{array}\right]
\end{aligned}
$
But, $A ^2=k A -2 I$
$
\therefore\left[\begin{array}{rr}
1 & -2 \\
4 & -4
\end{array}\right]=\left[\begin{array}{rr}
3 k-2 & -2 k \\
4 k & -2 k-2
\end{array}\right]
$
By equality of matrices,
$
\begin{aligned}
& 1=3 k-2 \ldots \\
& -2=-2 k \ldots \ldots \\
& 4=4 k \ldots \ldots . \\
& -4=-2 k-2
\end{aligned}
$
From (2), $k =1$. $k =1$ also satisfies equation (1), (3) and (4). Hence, $k=1$.

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