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Matrices — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSMatrices2 Marks
Question
For the matrices $, A$ and $B,$ verify that $(AB)\ ' = B\ 'A\ ',$ where $A=\left[\begin{array}{l} {0} \\ {1} \\ {2} \end{array}\right], B=\left[\begin{array}{lll} {1} & {5} & {7} \end{array}\right]$
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Answer

Given that, $A=\left[\begin{array}{l} {0} \\ {1} \\ {2} \end{array}\right]$ and  $B=\left[\begin{array}{lll} {1} & {5} & {7} \end{array}\right]$
Explanation: The product of two matrices is defined or possible only if the number of columns of the former matrix is equal to number of rows of the latter matrix.
$A B=\left[\begin{array}{l} {0} \\ {1} \\ {2} \end{array}\right] \left[\begin{array}{lll} {1} & {5} & {7} \end{array}\right]$
$\Rightarrow \mathrm{AB}=\left[\begin{array}{ccc} {0 \times 1} & {0 \times 5} & {0 \times 7} \\ {1 \times 1} & {1 \times 5} & {1 \times 7} \\ {2 \times 1} & {2 \times 5} & {2 \times 7} \end{array}\right]$
$\Rightarrow \mathrm{AB}=\left[\begin{array}{ccc} {0} & {0} & {0} \\ {1} & {5} & {7} \\ {2} & {10} & {14} \end{array}\right]$
Therefore, $(\mathrm{AB})^\prime=\left[\begin{array}{ccc} {0} & {1} & {2} \\ {0} & {5} & {10} \\ {0} & {7} & {14} \end{array}\right] ...(1)$
Now, $A^{\prime}=\left[\begin{array}{lll} {0} & {1} & {2} \end{array}\right] $ and $ B^{\prime}=\left[\begin{array}{l} {1} \\ {5} \\ {7} \end{array}\right]$
Therefore, $\mathrm{B}^{\prime} \mathrm{A}^{\prime}=\left[\begin{array}{l} {1} \\ {5} \\ {7} \end{array}\right] \left[\begin{array}{lll} {0} & {1} & {2} \end{array}\right]$
$\Rightarrow \mathrm{B}^{\prime} \mathrm{A}^{\prime}=\left[\begin{array}{ccc} {1 \times 0} & {1 \times 1} & {1 \times 2} \\ {5 \times 0} & {5 \times 1} & {5 \times 2} \\ {7 \times 0} & {7 \times 1} & {7 \times 2} \end{array}\right]$
$\Rightarrow \mathrm{B}^{\prime} \mathrm{A}^{\prime}=\left[\begin{array}{ccc} {0} & {1} & {2} \\ {0} & {5} & {10} \\ {0} & {7} & {14} \end{array}\right] ...(2)$
From equation $(1) \ (2)$ we see that
$(AB)\ ' = B\ 'A\ '.$ Hence verified.

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