Answer the question [ 4 mark question ]

Question 14 Marks

If $A=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$ and $A^2-4 A=k I_3$ then find the value of $k$. Here $I_3$ has a order of $3 \times 3$ matrix.

Answer

If $A =\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$ and $A ^2-4 A=k I _3$
Here $I_3$ has a order of $3 \times 3$ matrix$
\begin{aligned}
A^2=A \cdot A & =\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right] \\
& =\left[\begin{array}{lll}
1+4+4 & 2+2+4 & 2+4+2 \\
2+2+4 & 4+1+4 & 4+2+2 \\
2+4+2 & 4+2+2 & 4+4+1
\end{array}\right] \\
\text { now } A^2-4 A & =\left[\begin{array}{lll}
9 & 8 & 8 \\
8 & 9 & 8 \\
8 & 8 & 9
\end{array}\right] \\
& =\left[\begin{array}{lll}
9 & 8 & 8 \\
8 & 9 & 8 \\
8 & 8 & 9
\end{array}\right]-4\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right] \\
A^2-4 A & \left.=5 \begin{array}{lll}
9 & 8 \\
8 & 9 & 8 \\
8 & 8 & 9
\end{array}\right]-\left[\begin{array}{lll}
4 & 8 & 8 \\
8 & 4 & 8 \\
8 & 8 & 4
\end{array}\right] \\
& =\left[\begin{array}{lll}
5 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & 5
\end{array}\right]=5\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]
\end{aligned}
$
It is prove that $k$ will be 5 .

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Question 24 Marks

If $A =\left[\begin{array}{ccc}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right]$,then find the value of $A ^{ 2 }- 3 A + 2 I$.

Answer

Given that$
A=\left[\begin{array}{ccc}
2 & 0 & 1 \\
2 & 1 & 3 \\
1 & -1 & 0
\end{array}\right]
$
$
\text { So, } \begin{aligned}
A^2 & =A \cdot A=\left[\begin{array}{ccc}
2 & 0 & 1 \\
2 & 1 & 3 \\
1 & -1 & 0
\end{array}\right]\left[\begin{array}{ccc}
2 & 0 & 1 \\
2 & 1 & 3 \\
1 & -1 & 0
\end{array}\right] \\
& =\left[\begin{array}{lll}
4+0+1 & 0+0-1 & 2+0+0 \\
4+2+3 & 0+1-3 & 2+3+0 \\
2-2+0 & 0-1+0 & 1-3+0
\end{array}\right] \\
& =\left[\begin{array}{lll}
5 & -1 & 2 \\
9 & -2 & 5 \\
0 & -1 & -2
\end{array}\right] \\
\therefore A^2-3 A & +2 I \\
& =\left[\begin{array}{ccc}
5 & -1 & 2 \\
9 & -2 & 5 \\
0 & -1 & -2
\end{array}\right]-3\left[\begin{array}{ccc}
2 & 0 & 1 \\
2 & 1 & 3 \\
1 & -1 & 0
\end{array}\right]+2\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]
\end{aligned}
$
$
\begin{array}{l}
=\left[\begin{array}{ccc}
5 & -1 & 2 \\
9 & -2 & 5 \\
0 & -1 & -2
\end{array}\right]-\left[\begin{array}{ccc}
6 & 0 & 3 \\
6 & 3 & 9 \\
3 & -3 & 0
\end{array}\right]+\left[\begin{array}{lll}
2 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 2
\end{array}\right] \\
=\left[\begin{array}{ccc}
1 & -1 & -1 \\
3 & -3 & -4 \\
-3 & 2 & 0
\end{array}\right]
\end{array}
$

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Mathematics Answer the question [ 4 mark question ]

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