Question
If $A=\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right]$, Find $\mathrm{k}$, so that $A^2-\mathrm{kA}+2 \mathrm{I}$ $=\mathrm{O}$, where $\mathrm{I}$ is an identity matrix and $\mathrm{O}$ is null matrix of order 2 .
✓
Answer
Given $A^2-\mathrm{kA}+2 \mathrm{I}=\mathrm{O}$
$
\begin{aligned}
& \therefore \text { Here, } A^2=\mathrm{AA}=\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]\left[\begin{array}{rr}
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}{rr}
1 & -2 \\
4 & -4
\end{array}\right] \\
& \therefore A^2-\mathrm{kA}+2 \mathrm{I}=\mathrm{O} \\
& \therefore\left[\begin{array}{ll}
1 & -2 \\
4 & -4
\end{array}\right]-\mathrm{k}\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]+2\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\mathrm{O} \\
& \therefore\left[\begin{array}{rr}
1 & -2 \\
4 & -4
\end{array}\right]-\left[\begin{array}{ll}
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}{ll}
0 & 0 \\
0 & 0
\end{array}\right] \\
& \therefore\left[\begin{array}{rr}
1-3 k+2 & -2+2 k \\
4-4 k & -4+2 k+2
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]
\end{aligned}
$
$\therefore$ Using definition of equality of matrices, we have
$
\left.\begin{array}{lll}
1-3 \mathrm{k}+2=0 & \therefore & 3 \mathrm{k}=3 \\
-2+k=0 & \therefore & 2 \mathrm{k}=2 \\
4-4 k=0 & \therefore & 4 \mathrm{k}=4 \\
-4+2 \mathrm{k}+2=0 & \therefore & 2 \mathrm{k}=2
\end{array}\right\} \quad \mathrm{k}=1
$
$
\begin{aligned}
& \therefore \text { Here, } A^2=\mathrm{AA}=\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]\left[\begin{array}{rr}
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}{rr}
1 & -2 \\
4 & -4
\end{array}\right] \\
& \therefore A^2-\mathrm{kA}+2 \mathrm{I}=\mathrm{O} \\
& \therefore\left[\begin{array}{ll}
1 & -2 \\
4 & -4
\end{array}\right]-\mathrm{k}\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]+2\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\mathrm{O} \\
& \therefore\left[\begin{array}{rr}
1 & -2 \\
4 & -4
\end{array}\right]-\left[\begin{array}{ll}
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}{ll}
0 & 0 \\
0 & 0
\end{array}\right] \\
& \therefore\left[\begin{array}{rr}
1-3 k+2 & -2+2 k \\
4-4 k & -4+2 k+2
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]
\end{aligned}
$
$\therefore$ Using definition of equality of matrices, we have
$
\left.\begin{array}{lll}
1-3 \mathrm{k}+2=0 & \therefore & 3 \mathrm{k}=3 \\
-2+k=0 & \therefore & 2 \mathrm{k}=2 \\
4-4 k=0 & \therefore & 4 \mathrm{k}=4 \\
-4+2 \mathrm{k}+2=0 & \therefore & 2 \mathrm{k}=2
\end{array}\right\} \quad \mathrm{k}=1
$
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