====== Question 8, Exercise 2.1 ======
Solutions of Question 8 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 8(i)=====
If $A=\begin{bmatrix}1 & 2 & 0  \\3 & -1 & 4 \end{bmatrix}$, show that $( A^t )^t=A$.
====Solution====
Given
$$A=\left[ \begin{matrix}
   1 & 2 & 0  \\
   3 & -1 & 4  \\
\end{matrix}  \right]$$
Then
$$A^t=\left[  \begin{matrix}
   1 & 3  \\
   2 & -1  \\
   0 & 4  \\
\end{matrix} \right]$$
This gives
\begin{align}( A^t )^t&=\left[ \begin{matrix}1 & 2 & 0  \\3 & -1 & 4  \\
\end{matrix}  \right]\\
\implies( A^t)^t&=A. \end{align}

=====Question 8(ii)=====
If $A=\begin{bmatrix}1 & 2 & 0\\3 & -1 & 4\end{bmatrix}$, show that $AA^t\ne A^tA$.
====Solution====
$$A=\left[ \begin{matrix}
   1 & 2 & 0  \\
   3 & -1 & 4  \\
\end{matrix}  \right]$$
$$A^t=\left[  \begin{matrix}
   1 & 3  \\
   2 & -1  \\
   0 & 4  \\
\end{matrix}  \right]$$
$$AA^t=\left[ \begin{matrix}
   1 & 2 & 0  \\
   3 & -1 & 4  \\
\end{matrix}  \right]\left[  \begin{matrix}
   1 & 3  \\
   2 & -1  \\
   0 & 4  \\
\end{matrix}  \right]$$
$$AA^t=\left[ \begin{matrix}
   1+4+0 & 3-2+0  \\
   3-2+0 & 9+1+16  \\
\end{matrix} \right]$$
$$AA^t=\left[ \begin{matrix}
   5 & 1  \\
   1 & 26  \\
\end{matrix} \right]$$
$$A^tA=\left[  \begin{matrix}
   1 & 3  \\
   2 & -1  \\
   0 & 4  \\
\end{matrix}  \right]\left[ \begin{matrix}
   1 & 2 & 0  \\
   3 & -1 & 4  \\
\end{matrix}  \right]$$
$$A^tA=\left[ \begin{matrix}
   1+9 & 2-3 & 0+12  \\
   2-3 & 4+1 & 0-8  \\
   0+12 & 0-4 & 0+16  \\
\end{matrix} \right]$$
$$A^tA=\left[ \begin{matrix}
   10 & -1 & 12  \\
   -1 & 5 & -8  \\
   12 & -4 & 16  \\
\end{matrix} \right]$$
$$AA^t\ne A^tA$$


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