Question 11, Exercise 2.1

Solutions of Question 11 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 $A=\begin{bmatrix}0 & 1 & -2 \\-1 & 0 & 3 \\2 & -3 & 0 \end{bmatrix}$ and $B=\begin{bmatrix}0 & -6 & 11 \\6 & 0 & -7 \\-11 & 7 & 0 \end{bmatrix}$. Verify $A+B$ is skew symmetric.

$$A=\left[ \begin{matrix} 0 & 1 & -2 \\ -1 & 0 & 3 \\ 2 & -3 & 0 \\ \end{matrix} \right]$$ $$B=\left[ \begin{matrix} 0 & -6 & 11 \\ 6 & 0 & -7 \\ -11 & 7 & 0 \\ \end{matrix} \right]$$ For skew symmetric, we have $$( A+B )^t=-( A+B )$$ $$A+B=\left[ \begin{matrix} 0 & 1 & -2 \\ -1 & 0 & 3 \\ 2 & -3 & 0 \\ \end{matrix} \right]+\left[ \begin{matrix} 0 & -6 & 11 \\ 6 & 0 & -7 \\ -11 & 7 & 0 \\ \end{matrix} \right]$$ $$A+B=\left[ \begin{matrix} 0+0 & 1-6 & -2+11 \\ -1+6 & 0+0 & 3-7 \\ 2-11 & -3+7 & 0+0 \\ \end{matrix} \right]$$ $$A+B=\left[ \begin{matrix} 0 & -5 & 9 \\ 5 & 0 & -4 \\ -9 & 4 & 0 \\ \end{matrix} \right]$$ $$( A+B )^t=\left[ \begin{matrix} 0 & 5 & -9 \\ -5 & 0 & 4 \\ 9 & -4 & 0 \\ \end{matrix} \right]$$ $$( A+B )^t=-\left[ \begin{matrix} 0 & -5 & 9 \\ 5 & 0 & -4 \\ -9 & 4 & 0 \\ \end{matrix} \right]$$ $$( A+B )^t=-( A+B )$$