Question

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Answer 1

$\large\underline{\sf{Given \:Question - }}$

$\sf \: A =\bigg[ \begin{matrix}3&1 \\ 1&2 \end{matrix} \bigg], \: B = \bigg[ \begin{matrix} - 1&2 \\ 1&4 \end{matrix} \bigg], \: C = \bigg[ \begin{matrix} 2&3 \\ 1&1 \end{matrix} \bigg], \: then \:$

find matrix D such that AB = CD

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:A = \bigg[ \begin{matrix}3&1 \\ 1&2 \end{matrix} \bigg]$

$\rm :\longmapsto\:B = \bigg[ \begin{matrix} - 1&2 \\ 1&4 \end{matrix} \bigg]$

$\rm :\longmapsto\:C = \bigg[ \begin{matrix} 2&3 \\ 1&1 \end{matrix} \bigg]$

Let assume that,

$\rm :\longmapsto\:D = \bigg[ \begin{matrix} a&b \\ c&d \end{matrix} \bigg]$

Now, given that,

$\rm :\longmapsto\:AB = CD$

So, on substituting the values of matrices, we get

$\rm :\longmapsto\:\bigg[ \begin{matrix}3&1 \\ 1&2 \end{matrix} \bigg]\bigg[ \begin{matrix} - 1&2 \\ 1&4 \end{matrix} \bigg] = \bigg[ \begin{matrix} 2&3 \\ 1&1 \end{matrix} \bigg]\bigg[ \begin{matrix} a&b \\ c&d \end{matrix} \bigg]$

$\rm :\longmapsto\:\bigg[ \begin{matrix} - 3 + 1&6 + 4 \\ - 1 + 2&2 + 8 \end{matrix} \bigg] = \bigg[ \begin{matrix} 2a + 3c&2b + 3d \\ a + c&b + d \end{matrix} \bigg]$

$\rm :\longmapsto\:\bigg[ \begin{matrix} - 2&10\\ 1&10 \end{matrix} \bigg] = \bigg[ \begin{matrix} 2a + 3c&2b + 3d \\ a + c&b + d \end{matrix} \bigg]$

So, on comparing, we get

$\rm :\longmapsto\:2a + 3c = - 2$

and

$\rm :\longmapsto\:a + c = 1\bf\implies \:2a + 2c = 2$

On Subtracting these equations, we get

$\bf :\longmapsto\:c = - \: 4$

$\bf\implies \:a = 5$

Also,

$\rm :\longmapsto\:2b + 3d = 10$

and

$\rm :\longmapsto\:b + d = 10\bf\implies \:2b + 2d = 20$

On Subtracting these 2 equations, we get

$\bf :\longmapsto\:d = - \: 10$

$\bf :\longmapsto\:b = \: 20$

Hence,

$\bf\implies \:D = \bigg[ \begin{matrix} a&b \\ c&d \end{matrix} \bigg] = \bigg[ \begin{matrix} 5&20 \\ - 4& - 10 \end{matrix} \bigg]$