====== 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.

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


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