====== 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.

=====Question 10=====
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.
====Solution====
$$A=\left[ \begin{matrix}
   1 & -3 & 4  \\
   -3 & 2 & -5  \\
   4 & -5 & 0  \\
\end{matrix} \right]$$
$$B=\left[ \begin{matrix}
   5 & 6 & 7  \\
   6 & -8 & 3  \\
   7 & 3 & 1  \\
\end{matrix} \right]$$
For symmetric, we have to find out, 
$$A=A^t$$ 
$$B=B^t$$ 
$$( A+B )^t=A^t+B^t$$ 
$$A^t=\left[ \begin{matrix}
   1 & -3 & 4  \\
   -3 & 2 & -5  \\
   4 & -5 & 0  \\
\end{matrix} \right]$$
$$B^t=\left[ \begin{matrix}
   5 & 6 & 7  \\
   6 & -8 & 3  \\
   7 & 3 & 1  \\
\end{matrix} \right]$$
$$A=\left[ \begin{matrix}
   1 & -3 & 4  \\
   -3 & 2 & -5  \\
   4 & -5 & 0  \\
\end{matrix} \right]$$
$$A=A^t$$
$$B=\left[ \begin{matrix}
   5 & 6 & 7  \\
   6 & -8 & 3  \\
   7 & 3 & 1  \\
\end{matrix} \right]$$
$$B=B^t$$ 
$$A+B=\left[ \begin{matrix}
   1 & -3 & 4  \\
   -3 & 2 & -5  \\
   4 & -5 & 0  \\
\end{matrix} \right]+\left[ \begin{matrix}
   5 & 6 & 7  \\
   6 & -8 & 3  \\
   7 & 3 & 1  \\
\end{matrix} \right]$$
$$A+B=\left[ \begin{matrix}
   1+5 & -3+6 & 4+7  \\
   -3+6 & 2-8 & -5+3  \\
   4+7 & -5+3 & 0+1  \\
\end{matrix} \right]$$
$$A+B=\left[ \begin{matrix}
   6 & 3 & 11  \\
   3 & -6 & -2  \\
   11 & -2 & 1  \\
\end{matrix} \right]$$
$$( A+B)^t=\left[ \begin{matrix}
   6 & 3 & 11  \\
   3 & -6 & -2  \\
   11 & -2 & 1  \\
\end{matrix} \right]$$
$$A^t+B^t=\left[ \begin{matrix}
   1 & -3 & 4  \\
   -3 & 2 & -5  \\
   4 & -5 & 0  \\
\end{matrix} \right]+\left[ \begin{matrix}
   5 & 6 & 7  \\
   6 & -8 & 3  \\
   7 & 3 & 1  \\
\end{matrix} \right]$$
$$A^t+B^t=\left[ \begin{matrix}
   1+5 & -3+6 & 4+7  \\
   -3+6 & 2-8 & -5+3  \\
   4+7 & -5+3 & 0+1  \\
\end{matrix} \right]$$
$$A^t+B^t=\left[ \begin{matrix}
   6 & 3 & 11  \\
   3 & -6 & -2  \\
   11 & -2 & 1  \\
\end{matrix} \right]$$
$$( A+B )^t=A^t+B^t$$


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