====== 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.

=====Question 5=====
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 $b$.
====Solution====
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
$$ 3a=-2 \text{ and } 2b=3,$$
$$\implies a=-\dfrac{2}{3} \text{ and }  b=\dfrac{3}{2}.$$

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


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


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