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.

Let $A= \begin{bmatrix}1 & 4 & 4 \\ 4 & 1 & 4 \\ 4 & 4 & 1 \end{bmatrix}$. Show that $\dfrac{1}{3}A^2-2A-9I=0$.

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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