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Multiplication of Matrices

If A = [ 1 1 ...

Question

If $A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & - 1 2 & 0 & 3 3 & - 1 & 2 \end{matrix} ⎤ ⎥ ⎦, B = ⎡ ⎢ ⎣ \begin{matrix} 1 & 3 0 & 2 - 1 & 4 \end{matrix} ⎤ ⎥ ⎦$ and $C = [\begin{matrix} 1 & 2 & 3 & - 4 2 & 0 & - 2 & 1 \end{matrix}]$ , find $A (B C), (A B) C$ and show that $(A B) C = A (B C) .$

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Solution

Calculating: $A (B C)$
Given: $A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & - 1 2 & 0 & 3 3 & - 1 & 2 \end{matrix} ⎤ ⎥ ⎦, B = ⎡ ⎢ ⎣ \begin{matrix} 1 & 3 0 & 2 - 1 & 4 \end{matrix} ⎤ ⎥ ⎦$ and $C = [\begin{matrix} 1 & 2 & 3 & - 4 2 & 0 & - 2 & 1 \end{matrix}]$
$B C = {⎡ ⎢ ⎣ \begin{matrix} 1 & 3 0 & 2 - 1 & 4 \end{matrix} ⎤ ⎥ ⎦}_{3 \times 2} {[\begin{matrix} 1 & 2 & 3 & - 4 2 & 0 & - 2 & 1 \end{matrix}]}_{2 \times 4}$
$\Rightarrow B C = ⎡ ⎢ ⎣ \begin{matrix} 1 + 6 & 2 + 0 & 3 - 6 & - 4 + 3 0 + 4 & 0 + 0 & 0 - 4 & 0 + 2 - 1 + 8 & - 2 + 0 & - 3 - 8 & 4 + 4 \end{matrix} ⎤ ⎥ ⎦$
$\Rightarrow B C = ⎡ ⎢ ⎣ \begin{matrix} 7 & 2 & - 3 & - 1 4 & 0 & - 4 & 2 7 & - 2 & - 11 & 8 \end{matrix} ⎤ ⎥ ⎦$
Now,
$A (B C) = {⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & - 1 2 & 0 & 3 3 & - 1 & 2 \end{matrix} ⎤ ⎥ ⎦}_{3 \times 3} {⎡ ⎢ ⎣ \begin{matrix} 7 & 2 & - 3 & - 1 4 & 0 & - 4 & 2 7 & - 2 & - 11 & 2 \end{matrix} ⎤ ⎥ ⎦}_{3 \times 4}$
$\Rightarrow A (B C) = ⎡ ⎢ ⎣ \begin{matrix} 4 & 4 & 4 & - 7 35 & - 2 & - 39 & 22 31 & 2 & - 27 & 11 \end{matrix} ⎤ ⎥ ⎦ \dots (1)$

Calculating $(A B) C$
$A B = {⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & - 1 2 & 0 & 3 3 & - 1 & 2 \end{matrix} ⎤ ⎥ ⎦}_{3 \times 3} {⎡ ⎢ ⎣ \begin{matrix} 1 & 3 0 & 2 - 1 & 4 \end{matrix} ⎤ ⎥ ⎦}_{3 t i m e s 2}$
$\Rightarrow A B = ⎡ ⎢ ⎣ \begin{matrix} 1 + 0 + 1 & 3 + 2 - 4 2 + 0 - 3 & 6 + 0 + 12 3 + 0 - 2 & 9 - 2 + 8 \end{matrix} ⎤ ⎥ ⎦$
$\Rightarrow A B = ⎡ ⎢ ⎣ \begin{matrix} 2 & 1 - 1 & 18 1 & 15 \end{matrix} ⎤ ⎥ ⎦$
$∴ (A B) C = {⎡ ⎢ ⎣ \begin{matrix} 2 & 1 - 1 & 18 1 & 15 \end{matrix} ⎤ ⎥ ⎦}_{3 \times 2} {[\begin{matrix} 1 & 2 & 3 & - 4 2 & 0 & - 2 & 1 \end{matrix}]}_{2 \times 4}$
$= ⎡ ⎢ ⎣ \begin{matrix} 2 + 2 & 4 + 0 & 6 - 2 & - 8 + 1 - 1 + 36 & - 2 + 0 & - 3 - 36 & 4 + 18 1 + 30 & 2 + 0 & 3 - 30 & - 4 + 15 \end{matrix} ⎤ ⎥ ⎦$
$\Rightarrow (A B) C = ⎡ ⎢ ⎣ \begin{matrix} 4 & 4 & 4 & - 7 35 & - 2 & - 39 & 22 31 & 2 & - 27 & 11 \end{matrix} ⎤ ⎥ ⎦ \dots (2)$
From equation $(1)$ and $(2)$
$∴ (A B) C = A (B C)$
Hence proved.

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