For the matrices, A and B, verify that (AB)′ = B′A′, where A= [ l 0 1 2 ], B= [ lll 1 & 5 & 7 ]

Given that, A= [ l 0 1 2 ] and B= [ lll 1 & 5 & 7 ] 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= [ l 0 1 2 ] [ lll 1 & 5 & 7 ] AB= [ ccc 0 1 & 0 5 & 0 7 1 1 & 1 5 & 1 7 2 1 & 2 5 & 2 7 ] AB= [ ccc 0 & 0 & 0 1 & 5 & 7 2 & 10 & 14 ] Therefore, (AB)^ = [ ccc 0 & 1 & 2 0 & 5 & 10 0 & 7 & 14 ] ~~~~~~...(1) Now, A^ = [ lll 0 & 1 & 2 ] and B^ = [ l 1 5 7 ] Therefore, B^ A^ = [ l 1 5 7 ] [ lll 0 & 1 & 2 ] B^ A^ = [ ccc 1 0 & 1 1 & 1 2 5 0 & 5 1 & 5 2 7 0 & 7 1 & 7 2 ] B^ A^ = [ ccc 0 & 1 & 2 0 & 5 & 10 0 & 7 & 14 ] ~~~~~~~...(2) From equation (1) & (2) we see that (AB)' = B'A'. Hence verified.

For the matrices, A and B, verify that (AB)′ = B′A′, where A= [ l…

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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, $A=\left[\begin{array}{l} {0} \\ {1} \\ {2} \end{array}\right] \text { and } B=\left[\begin{array}{lll} {1} & {5} & {7} \end{array}\right]$
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, $A^{\prime}=\left[\begin{array}{lll} {0} & {1} & {2} \end{array}\right] \text { and } B^{\prime}=\left[\begin{array}{l} {1} \\ {5} \\ {7} \end{array}\right]$
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.