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# Given $A=\left( \begin{matrix} 3 & 6 \\ -2 & -8 \\\end{matrix} \right)$ and $B=\left( \begin{matrix} 2 & 16 \\\end{matrix} \right)$ , find the matrix X such that $XA=B$ .

Last updated date: 13th Jul 2024
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Hint: According to the question we find the unknown matrix X such that it satisfies the given equation $XA=B$. Now, firstly we need to identify the order of the matrix X using the product rule property of the matrices and then continue to solve for X in order to get X.

Now, we know that order of matrix A is $2\times 2$ and order or matrix B is $1\times 2$ and B is equal to the product of XA. Therefore, for this product to occur we must have the number of columns of X equal to the number of rows of A and also since the number of rows of B is 1 therefore, the number of rows of matrix X is 1.
Therefore, we now know that the order of X is $1\times 2$ .
\begin{align} & XA=\left( \begin{matrix} x & y \\ \end{matrix} \right)\left( \begin{matrix} 3 & 6 \\ -2 & -8 \\ \end{matrix} \right) \\ & \Rightarrow \left( \begin{matrix} 3x-2y & 6x-8y \\ \end{matrix} \right) \\ \end{align}
And now this is equal to B which is $B=\left( \begin{matrix} 2 & 16 \\ \end{matrix} \right)$.
\begin{align} & 3x-2y=2 \\ & 6x-8y=16 \\ \end{align}
Now, dividing second equation by 2 we get $3x-4y=8$ and now subtracting first equation from this we get $y=-3$ and now putting this value in equation one we get $x=-\dfrac{4}{3}$ .
Therefore, the X matrix attained is $X=\left( \begin{matrix} -\dfrac{4}{3} & -3 \\ \end{matrix} \right)$ .