Question 4, Exercise 2.1
Solutions of Question 4 of Exercise 2.1 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.
Question 4
Let $A= \begin{bmatrix}1 & 4 & 4 \\ 4 & 1 & 4 \\ 4 & 4 & 1 \end{bmatrix}$. Show that $\dfrac{1}{3}A^2-2A-9I=0$.
Solution
Given: $A=\begin{bmatrix} 1 & 4 & 4 \\ 4 & 1 & 4 \\ 4 & 4 & 1 \end{bmatrix}$.
Now
\begin{align}\frac{1}{3}A^2&=\frac{1}{3}\left[ \begin{matrix}
1 & 4 & 4 \\
4 & 1 & 4 \\
4 & 4 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
1 & 4 & 4 \\
4 & 1 & 4 \\
4 & 4 & 1 \\
\end{matrix} \right]\\& =\frac{1}{3}\left[ \begin{matrix}
1+16+16 & 4+4+16 & 4+16+4 \\
4+4+16 & 16+1+16 & 16+4+4 \\
4+16+4 & 16+4+4 & 16+16+1 \\
\end{matrix} \right] \\
&=\frac{1}{3}\left[ \begin{matrix}
33 & 24 & 24 \\
24 & 33 & 24 \\
24 & 24 & 33 \\
\end{matrix} \right]\\
&=\left[ \begin{matrix}
11 & 8 & 8 \\
8 & 11 & 8 \\
8 & 8 & 11 \\
\end{matrix} \right]\end{align}
Now we take
$$2A=\left[ \begin{matrix}
2 & 8 & 8 \\
8 & 2 & 8 \\
8 & 8 & 2 \\
\end{matrix} \right]$$
and
\begin{align}9I&=9\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\ \end{matrix} \right] \\
&=\left[ \begin{matrix}
9 & 0 & 0 \\
0 & 9 & 0 \\
0 & 0 & 9 \\
\end{matrix} \right]\end{align}
Now, we have
\begin{align}&\dfrac{1}{3}A^2-2A-9I \\
=&\left[ \begin{matrix}
11 & 8 & 8 \\
8 & 11 & 8 \\
8 & 8 & 11 \\
\end{matrix} \right]-\left[ \begin{matrix}
2 & 8 & 8 \\
8 & 2 & 8 \\
8 & 8 & 2 \\
\end{matrix} \right]-\left[ \begin{matrix}
9 & 0 & 0 \\
0 & 9 & 0 \\
0 & 0 & 9 \\
\end{matrix} \right] \\
=&\left[ \begin{matrix}
11-2-9 & 8-8-0 & 8-8-0 \\
8-8-0 & 11-2-9 & 8-8-0 \\
8-8-0 & 8-8-0 & 11-2-9 \\
\end{matrix} \right] \\
=&\left[ \begin{matrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix} \right] \\
\implies & \dfrac{1}{3}A^2-2A-9I=0 \end{align}
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