# Question 5 & 6, Exercise 2.1

Solutions of Question 5 & 6 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.

Matrix $A= \begin{bmatrix} 0 & 2b & -2 \\ 3 & 1 & 3 \\ 3a & 3 & -1 \end{bmatrix}$ is given to be symmetric. Find the value of $a$ and $b$.

Given: $A=\begin{bmatrix} 0 & 2b & -2 \\ 3 & 1 & 3 \\ 3a & 3 & -1 \end{bmatrix}$ Then $$A^t=\left[ \begin{matrix} 0 & 3 & 3a \\ 2b & 1 & 3 \\ -2 & 3 & -1 \\ \end{matrix} \right]$$ Since $A$ is given to be symmetirc, $A^t=A$, implies $$\left[ \begin{matrix} 0 & 3 & 3a \\ 2b & 1 & 3 \\ -2 & 3 & -1 \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & 2b & -2 \\ 3 & 1 & 3 \\ 3a & 3 & -1 \\ \end{matrix} \right]$$ This gives $$3a=-2 \text{ and } 2b=3,$$ $$\implies a=-\dfrac{2}{3} \text{ and } b=\dfrac{3}{2}.$$

Solve the matrix equations for $X.$ Find $X-3A=2B$, if $A=\begin{bmatrix} 1 & 0 & 3 \\-2 & 2 & 1 \end{bmatrix}$ and $B=\begin{bmatrix}2 & 1 & 1 \\ 3 & -1 & 4 \end{bmatrix}$.

Given $A=\left[ \begin{matrix} 1 & 0 & 3 \\ -2 & 2 & 1 \\ \end{matrix} \right]$ and $B=\left[ \begin{matrix} 2 & 1 & \quad 1 \\ 3 & -1 & 4 \\ \end{matrix} \right].$

As $$X-3A=2B$$ This gives $$X=3A+2B.$$ Now $$3A=\left[ \begin{matrix}3 & 0 & 9 \\-6 & 6 & 3 \\ \end{matrix} \right]$$ and $$2B=\left[ \begin{matrix}4 & 2 & \quad 2 \\6 & -2 & 8 \\ \end{matrix} \right].$$ Thus \begin{align}X&=\left[ \begin{matrix}3 & 0 & 9 \\6 & 6 & 3 \\ \end{matrix} \right]+\left[ \begin{matrix}4 & 2 & 2 \\ 6 & -2 & 8 \\ \end{matrix} \right] \\ &=\left[ \begin{matrix}3+4 & 0+2 & 9+2 \\-6+6 & 6-2 & 3+8 \\ \end{matrix} \right]\\ &=\left[ \begin{matrix}7 & 2 & 11 \\0 & 4 & 11 \\ \end{matrix} \right]\end{align}

Solve the matrix equations for $X.$ Find $2( X-A )=B$, if $A=\begin{bmatrix}1 & 2 & 2 \\ 3 & -1 & 2 \end{bmatrix}$ and $B=\begin{bmatrix} 4 & 6 & 2 \\ 0 & -4 & 2 \end{bmatrix}$.

Given $A=\left[ \begin{matrix}1 & 2 & 2 \\3 & -1 & 2 \\\end{matrix} \right]$ and $B=\left[ \begin{matrix}4 & 6 & \quad 2 \\0 & -4 & 2 \end{matrix} \right]$. As $$2( X-A )=B.$$ This gives $$X-A=\dfrac{B}{2}.$$ Now $$\dfrac{B}{2}=\left[ \begin{matrix}2 & 3 & \quad 1 \\0 & -2 & 1 \\ \end{matrix} \right]$$ So \begin{align}X&=\left[\begin{matrix}1 & 2 & 2 \\3 & -1 & 2 \\ \end{matrix} \right]+\left[ \begin{matrix}2 & 3 & \quad 1 \\0 & -2 & 1 \\\end{matrix} \right]\\ &=\left[ \begin{matrix}1+2 & 2+3 & 2+1 \\3+0 & -1-2 & 2+1 \\ \end{matrix} \right]\\ &=\left[ \begin{matrix}3 & \quad 5 & 3 \\3 & -3 & 3 \\ \end{matrix} \right].\end{align}