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.

If $A=\begin{bmatrix}1 & 2 & 0 \\3 & -1 & 4 \end{bmatrix}$, show that $( A^t )^t=A$.

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 $$\begin{align}( A^t )^t&=\left[ \begin{matrix}1 & 2 & 0 \\3 & -1 & 4 \\ \end{matrix} \right]\\ \implies( A^t)^t&=A. \end{align}$$

If $A=\begin{bmatrix}1 & 2 & 0\\3 & -1 & 4\end{bmatrix}$, show that $AA^t\ne A^tA$.

$$A=\left[ \begin{matrix} 1 & 2 & 0 \\ 3 & -1 & 4 \\ \end{matrix} \right]$$ $$A^t=\left[ \begin{matrix} 1 & 3 \\ 2 & -1 \\ 0 & 4 \\ \end{matrix} \right]$$ $$AA^t=\left[ \begin{matrix} 1 & 2 & 0 \\ 3 & -1 & 4 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 3 \\ 2 & -1 \\ 0 & 4 \\ \end{matrix} \right]$$ $$AA^t=\left[ \begin{matrix} 1+4+0 & 3-2+0 \\ 3-2+0 & 9+1+16 \\ \end{matrix} \right]$$ $$AA^t=\left[ \begin{matrix} 5 & 1 \\ 1 & 26 \\ \end{matrix} \right]$$ $$A^tA=\left[ \begin{matrix} 1 & 3 \\ 2 & -1 \\ 0 & 4 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 & 0 \\ 3 & -1 & 4 \\ \end{matrix} \right]$$ $$A^tA=\left[ \begin{matrix} 1+9 & 2-3 & 0+12 \\ 2-3 & 4+1 & 0-8 \\ 0+12 & 0-4 & 0+16 \\ \end{matrix} \right]$$ $$A^tA=\left[ \begin{matrix} 10 & -1 & 12 \\ -1 & 5 & -8 \\ 12 & -4 & 16 \\ \end{matrix} \right]$$ $$AA^t\ne A^tA$$

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