For the matrices, A and B, verify that (AB)′ = B′A′, where A= [ c 1 -4 3 ], B= [ ccc -1 & 2 & 1 ]

A= [ c 1 -4 3 ] and B= [ ccc -1 & 2 & 1 ] Explanation: The product of two matrices is defined or possible only if the number of columns of the former matrix is equal to a number of rows of the latter matrix. A B= [ c 1 -4 3 ] [ ccc -1 & 2 & 1 ] AB= [ ccc 1 (-1) & 1 2 & 1 1 -4 (-1) & -4 2 & -4 1 3 (-1) & 3 2 & 3 1 ] A B= [ ccc -1 & 2 & 1 4 & -8 & -4 -3 & 6 & 3 ] So, (AB)^ = [ ccc -1 & 4 & -3 2 & -8 & 6 1 & -4 & 3 ] ~~~~~~...(i) A^ = [ lll 1 & -4 & 3 ] and B^ = [ c -1 2 1 ] Therefore, B^ A^ = [ c -1 2 1 ] [ ccc 1 & -4 & 3 ] B^ A^ = [ ccc -1 1 & -1 (-4) & -1 3 2 1 & 2 (-4) & 2 3 1 1 & 1 (-4) & 1 3 ] B^ A^ = [ ccc -1 & 4 & -3 2 & -8 & 6 1 & -4 & 3 ] ....(ii) From equation (i) & (ii) we see that (AB)’ = B’A’. Hence verified.

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

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Question

For the matrices, A and B, verify that (AB)′ = B′A′, where $A=\left[\begin{array}{c} {1} \\ {-4} \\ {3} \end{array}\right], B=\left[\begin{array}{ccc} {-1} & {2} & {1} \end{array}\right]$

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Answer

$A=\left[\begin{array}{c} {1} \\ {-4} \\ {3} \end{array}\right] \text { and } B=\left[\begin{array}{ccc} {-1} & {2} & {1} \end{array}\right]$
Explanation: The product of two matrices is defined or possible only if the number of columns of the former matrix is equal to a number of rows of the latter matrix.
$A B=\left[\begin{array}{c} {1} \\ {-4} \\ {3} \end{array}\right] \left[\begin{array}{ccc} {-1} & {2} & {1} \end{array}\right]$
$\Rightarrow \mathrm{AB}=\left[\begin{array}{ccc} {1 \times(-1)} & {1 \times 2} & {1 \times 1} \\ {-4 \times(-1)} & {-4 \times 2} & {-4 \times 1} \\ {3 \times(-1)} & {3 \times 2} & {3 \times 1} \end{array}\right]$
$\Rightarrow A B=\left[\begin{array}{ccc} {-1} & {2} & {1} \\ {4} & {-8} & {-4} \\ {-3} & {6} & {3} \end{array}\right]$
So, $(\mathrm{AB})^\prime=\left[\begin{array}{ccc} {-1} & {4} & {-3} \\ {2} & {-8} & {6} \\ {1} & {-4} & {3} \end{array}\right] ~~~~~~...(i)$
$\mathrm{A}^{\prime}=\left[\begin{array}{lll} {1} & {-4} & {3} \end{array}\right] \text { and } \mathrm{B}^{\prime}=\left[\begin{array}{c} {-1} \\ {2} \\ {1} \end{array}\right]$
Therefore, $\mathrm{B}^{\prime} \mathrm{A}^{\prime}=\left[\begin{array}{c} {-1} \\ {2} \\ {1} \end{array}\right] \times\left[\begin{array}{ccc} {1} & {-4} & {3} \end{array}\right]$
$\Rightarrow \mathrm{B}^{\prime} \mathrm{A}^{\prime}=\left[\begin{array}{ccc} {-1 \times 1} & {-1 \times(-4)} & {-1 \times 3} \\ {2 \times 1} & {2 \times(-4)} & {2 \times 3} \\ {1 \times 1} & {1 \times(-4)} & {1 \times 3} \end{array}\right]$
$\Rightarrow \mathrm{B}^{\prime} \mathrm{A}^{\prime}=\left[\begin{array}{ccc} {-1} & {4} & {-3} \\ {2} & {-8} & {6} \\ {1} & {-4} & {3} \end{array}\right] ....(ii)$
From equation (i) & (ii) we see that
(AB)’ = B’A’. Hence verified.