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Triangular Matrix

For the matri...

Question

For the matrices $A$ and $B$ , verify that $(A B)^{'} = B^{'} A^{'}$ where:
$(i)$ $A = ⎡ ⎢ ⎣ \begin{matrix} 1 - 4 3 \end{matrix} ⎤ ⎥ ⎦, B = [\begin{matrix} - 1 & 2 & 1 \end{matrix}]$
$(i i)$ $A = ⎡ ⎢ ⎣ \begin{matrix} 012 \end{matrix} ⎤ ⎥ ⎦, B = [\begin{matrix} 1 & 5 & 7 \end{matrix}]$

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Solution

$(i)$ Given: $A = ⎡ ⎢ ⎣ \begin{matrix} 1 - 4 3 \end{matrix} ⎤ ⎥ ⎦, B = [\begin{matrix} - 1 & 2 & 1 \end{matrix}]$
To verify: $(A B)^{'} = B^{'} A^{'}$
Solving $L . H . S .$
$(A B)^{'} = {⎛ ⎜ ⎝ {⎡ ⎢ ⎣ \begin{matrix} 1 - 4 3 \end{matrix} ⎤ ⎥ ⎦}_{3 \times 1} {[\begin{matrix} - 1 & 2 & 1 \end{matrix}]}_{1 \times 3} ⎞ ⎟ ⎠}_{3 \times 1}^{'}$
$\Rightarrow (A B)^{'} = {⎛ ⎜ ⎝ {⎡ ⎢ ⎣ \begin{matrix} 1 \times (- 1) & 1 \times 2 & 1 \times 1 - 4 \times (- 1) & - 4 \times 2 & - 4 \times 1 3 \times (- 1) & 3 \times 2 & 3 \times 1 \end{matrix} ⎤ ⎥ ⎦}_{3 \times 3} ⎞ ⎟ ⎠}_{3 \times 3}^{'}$
$\Rightarrow (A B)^{'} = {⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} - 1 & 2 & 1 4 & - 8 & - 4 - 3 & 6 & 3 \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠}^{'}$
$\Rightarrow (A B)^{'} = ⎡ ⎢ ⎣ \begin{matrix} - 1 & 4 & - 3 2 & - 8 & 6 1 & - 4 & 3 \end{matrix} ⎤ ⎥ ⎦$

Solving $R . H . S .$
$B^{'} A^{'} = {[\begin{matrix} - 1 & 2 & 1 \end{matrix}]}^{'} {⎡ ⎢ ⎣ \begin{matrix} 1 - 4 3 \end{matrix} ⎤ ⎥ ⎦}^{'}$
$\Rightarrow B^{'} A^{'} = ⎡ ⎢ ⎣ \begin{matrix} - 1 21 \end{matrix} ⎤ ⎥ ⎦ [\begin{matrix} 1 & - 4 & 3 \end{matrix}]$
$\Rightarrow B^{'} A^{'} = {⎡ ⎢ ⎣ \begin{matrix} - 1 \times 1 & - 1 \times (- 4) & - 1 \times 3 2 \times 1 & 2 \times (- 4) & 2 \times 3 1 \times 1 & 1 \times (- 4) & 1 \times 3 \end{matrix} ⎤ ⎥ ⎦}_{3 \times 3}$
$\Rightarrow B^{'} A^{'} = ⎡ ⎢ ⎣ \begin{matrix} - 1 & 4 & - 3 2 & - 8 & 6 1 & - 4 & 3 \end{matrix} ⎤ ⎥ ⎦$
Hence, $L . H . S . = R . H . S .$
Hence verified.
$(i i)$ Given :
$A = ⎡ ⎢ ⎣ \begin{matrix} 012 \end{matrix} ⎤ ⎥ ⎦, B = [\begin{matrix} 1 & 5 & 7 \end{matrix}]$
To verify: $(A B)^{'} = B^{'} A^{'}$
Solving $L . H . S .$
$(A B)^{'} = {⎛ ⎜ ⎝ {⎡ ⎢ ⎣ \begin{matrix} 012 \end{matrix} ⎤ ⎥ ⎦}_{3 \times 1} {[\begin{matrix} 1 & 5 & 7 \end{matrix}]}_{1 \times 3} ⎞ ⎟ ⎠}_{3 \times 1}^{'}$
$\Rightarrow (A B)^{'} = {⎛ ⎜ ⎝ {⎡ ⎢ ⎣ \begin{matrix} 0 \times 1 & 0 \times 5 & 0 \times 7 1 \times 1 & 1 \times 5 & 1 \times 7 2 \times 1 & 2 \times 5 & 2 \times 7 \end{matrix} ⎤ ⎥ ⎦}_{3 \times 3} ⎞ ⎟ ⎠}_{3 \times 3}^{'}$
$\Rightarrow (A B)^{'} = {⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 0 & 0 & 0 1 & 5 & 7 2 & 10 & 14 \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠}^{'}$
$\Rightarrow (A B)^{'} = ⎡ ⎢ ⎣ \begin{matrix} 0 & 1 & 2 0 & 5 & 10 0 & 7 & 14 \end{matrix} ⎤ ⎥ ⎦$

Solving $R . H . S .$
$B^{'} A^{'} = {[\begin{matrix} 1 & 5 & 7 \end{matrix}]}^{'} {⎡ ⎢ ⎣ \begin{matrix} 012 \end{matrix} ⎤ ⎥ ⎦}^{'}$
$\Rightarrow B^{'} A^{'} = {⎡ ⎢ ⎣ \begin{matrix} 157 \end{matrix} ⎤ ⎥ ⎦}_{3 \times 1} {[\begin{matrix} 0 & 1 & 2 \end{matrix}]}_{1 \times 3}$
$\Rightarrow B^{'} A^{'} = {⎡ ⎢ ⎣ \begin{matrix} 1 \times 0 & 1 \times 1 & 1 \times 2 5 \times 0 & 5 \times 1 & 5 \times 2 7 \times 0 & 7 \times 1 & 7 \times 2 \end{matrix} ⎤ ⎥ ⎦}_{3 \times 3}$
$\Rightarrow B^{'} A^{'} = ⎡ ⎢ ⎣ \begin{matrix} 0 & 1 & 2 0 & 5 & 10 0 & 7 & 14 \end{matrix} ⎤ ⎥ ⎦$
Hence, $L . H . S . = R . H . S .$
Hence verified.

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