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Question

Find the matrix $$X$$ so that $$\displaystyle X\left[ \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix} \right] =\left[ \begin{matrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{matrix} \right] $$


Solution

It is given that:
$$\displaystyle X\left[ \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix} \right] =\left[ \begin{matrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{matrix} \right] $$

The matrix given on the R.H.S. of the equation is a $$2 \times 3$$ matrix and the one given on the L.H.S. of the equation is a $$2 \times  3$$ matrix. 

Therefore, $$X$$  has to be a $$2 \times 2$$ matrix.
Now, let $$\displaystyle X=\begin{bmatrix} a & c \\ b & d \end{bmatrix}$$

Therefore, we have:
$$\displaystyle \begin{bmatrix} a & c \\ b & d \end{bmatrix}\left[ \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix} \right] =\left[ \begin{matrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{matrix} \right] $$
$$\displaystyle \Rightarrow \left[ \begin{matrix} a+4c & 2a+5c & 3a+6c \\ b+4d & 2b+5d & 3b+6d \end{matrix} \right] =\left[ \begin{matrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{matrix} \right] $$
Equating the corresponding elements of the two matrices, we have:
$$\displaystyle \begin{matrix} a+4c=-7 & 2a+5c=-8 & 3a+6c=-9 \\ b+4d=2 & 2b+5d=4 & 3b+6d=6 \end{matrix}$$
Now, $$\displaystyle a+4c=-7\Rightarrow a=-7-4c$$
$$\displaystyle \therefore 2a+5c=-8\Rightarrow -14-8c+5c=-8$$
$$\displaystyle \Rightarrow -3c=6 \Rightarrow c=-2$$
$$\displaystyle \therefore a=-7-4\left( -2 \right) =-7+8=1$$
Now, $$\displaystyle b+4d=2\Rightarrow b=2-4d$$
$$\displaystyle \therefore 2b+5d=4\Rightarrow 4-8d+5d=4$$
$$\displaystyle \Rightarrow -3d=0$$
$$\displaystyle \Rightarrow d=0$$
$$\displaystyle \therefore b=2-4\left( 0 \right) =2$$
Thus, $$a=1, b=2, c=-2, d=0$$
Hence, the required matrix $$X$$ is $$\displaystyle \begin{bmatrix} 1 & -2 \\ 2 & 0 \end{bmatrix}$$

Mathematics
NCERT
Standard XII

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