# Question 10, Exercise 2.1

Solutions of Question 10 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 $A=\begin{bmatrix}1 & -3 & 4 \\-3 & 2 & -5 \\4 & -5 & 0 \end{bmatrix}$ and $B=\begin{bmatrix}5 & 6 & 7 \\6 & -8 & 3 \\7 & 3 & 1 \end{bmatrix}$. Verify that $A$ and $B$ are symmetric. Also verify that $A+B$ is symmetric.

$$A=\left[ \begin{matrix} 1 & -3 & 4 \\ -3 & 2 & -5 \\ 4 & -5 & 0 \\ \end{matrix} \right]$$ $$B=\left[ \begin{matrix} 5 & 6 & 7 \\ 6 & -8 & 3 \\ 7 & 3 & 1 \\ \end{matrix} \right]$$ For symmetric, we have to find out, $$A=A^t$$ $$B=B^t$$ $$( A+B )^t=A^t+B^t$$ $$A^t=\left[ \begin{matrix} 1 & -3 & 4 \\ -3 & 2 & -5 \\ 4 & -5 & 0 \\ \end{matrix} \right]$$ $$B^t=\left[ \begin{matrix} 5 & 6 & 7 \\ 6 & -8 & 3 \\ 7 & 3 & 1 \\ \end{matrix} \right]$$ $$A=\left[ \begin{matrix} 1 & -3 & 4 \\ -3 & 2 & -5 \\ 4 & -5 & 0 \\ \end{matrix} \right]$$ $$A=A^t$$ $$B=\left[ \begin{matrix} 5 & 6 & 7 \\ 6 & -8 & 3 \\ 7 & 3 & 1 \\ \end{matrix} \right]$$ $$B=B^t$$ $$A+B=\left[ \begin{matrix} 1 & -3 & 4 \\ -3 & 2 & -5 \\ 4 & -5 & 0 \\ \end{matrix} \right]+\left[ \begin{matrix} 5 & 6 & 7 \\ 6 & -8 & 3 \\ 7 & 3 & 1 \\ \end{matrix} \right]$$ $$A+B=\left[ \begin{matrix} 1+5 & -3+6 & 4+7 \\ -3+6 & 2-8 & -5+3 \\ 4+7 & -5+3 & 0+1 \\ \end{matrix} \right]$$ $$A+B=\left[ \begin{matrix} 6 & 3 & 11 \\ 3 & -6 & -2 \\ 11 & -2 & 1 \\ \end{matrix} \right]$$ $$( A+B)^t=\left[ \begin{matrix} 6 & 3 & 11 \\ 3 & -6 & -2 \\ 11 & -2 & 1 \\ \end{matrix} \right]$$ $$A^t+B^t=\left[ \begin{matrix} 1 & -3 & 4 \\ -3 & 2 & -5 \\ 4 & -5 & 0 \\ \end{matrix} \right]+\left[ \begin{matrix} 5 & 6 & 7 \\ 6 & -8 & 3 \\ 7 & 3 & 1 \\ \end{matrix} \right]$$ $$A^t+B^t=\left[ \begin{matrix} 1+5 & -3+6 & 4+7 \\ -3+6 & 2-8 & -5+3 \\ 4+7 & -5+3 & 0+1 \\ \end{matrix} \right]$$ $$A^t+B^t=\left[ \begin{matrix} 6 & 3 & 11 \\ 3 & -6 & -2 \\ 11 & -2 & 1 \\ \end{matrix} \right]$$ $$( A+B )^t=A^t+B^t$$