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

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MATHS Answer the question [ 4 mark question ]

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