# Question 12, Exercise 2.1

Solutions of Question 12 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 $A=\begin{bmatrix}3 & 2 & 1 \\4 & 5 & 6 \\-2 & 3 & 4\end{bmatrix}$. Verify that$A+A^t$ is symmetric.

$$A=\left[ \begin{matrix} 3 & 2 & 1 \\ 4 & 5 & 6 \\ -2 & 3 & 4 \\ \end{matrix} \right]$$ $$A^t=\left[ \begin{matrix} 3 & 4 & -2 \\ 2 & 5 & 3 \\ 1 & 6 & 4 \\ \end{matrix} \right]$$ For symmetric, we have, $$( A+A^t )^t=A+A^t$$ $$A+A^t=\left[ \begin{matrix} 3 & 2 & 1 \\ 4 & 5 & 6 \\ -2 & 3 & 4 \\ \end{matrix} \right]+\left[ \begin{matrix} 3 & 4 & -2 \\ 2 & 5 & 3 \\ 1 & 6 & 4 \\ \end{matrix} \right]$$ $$=\left[ \begin{matrix} 3+3 & 2+4 & 1-2 \\ 4+2 & 5+5 & 6+3 \\ -2+1 & 3+6 & 4+4 \\ \end{matrix} \right]$$ $$A+A^t=\left[ \begin{matrix} 6 & 6 & -1 \\ 6 & 10 & 9 \\ -1 & 9 & 8 \\ \end{matrix} \right]$$ $$( A+A^t )^t=\left[ \begin{matrix} 6 & 6 & -1 \\ 6 & 10 & 9 \\ -1 & 9 & 8 \\ \end{matrix} \right]$$ $$( A+A^t )^t=( A+A^t )$$

Let $A=\begin{bmatrix}3 & 2 & 1 \\ 4 & 5 & 6 \\-2 & 3 & 4\end{bmatrix}$. Verify that$A-A^t$ is skew symmetric.

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