(ii)

Solution:

From L.H.S, we have

\[3\left[ 4\text{ }x \right]\text{ }+\text{ }2\left[ y\text{ }-3 \right]\]

\[=\text{ }\left[ 12\text{ }3x \right]\text{ }+\text{ }\left[ 2y\text{ }-6 \right]~\]

\[=\text{ }\left[ \left( 12\text{ }+\text{ }2y \right)\text{ }\left( 3x\text{ }-\text{ }6 \right) \right]\]

On equating with R.H.S we get

\[\left[ \left( 12\text{ }+\text{ }2y \right)\text{ }\left( 3x\text{ }-\text{ }6 \right) \right]~\]

\[=\text{ }\left[ 10\text{ }0 \right]\]

\[12\text{ }+\text{ }2y\text{ }=\text{ }10~\]

And,

\[3x\text{ }-\text{ }6\text{ }=\text{ }0\]

\[2y\text{ }=\text{ }-2\text{ }and\text{ }3x\text{ }=\text{ }6\]

\[y\text{ }=\text{ }-1\text{ }and\text{ }x\text{ }=\text{ }2\]

(ii) We have

On equating the matrices we get

\[-x\text{ }+\text{ }8\text{ }=\text{ }7~\]

And,

\[~2x\text{ }-\text{ }4y\text{ }=\text{ }-8\]

\[x\text{ }=\text{ }1~\]

And,

\[2\left( 1 \right)-\text{ }\text{ }4y\text{ }=\text{ }-8\]

\[2\text{ }-\text{ }4y\text{ }=\text{ }-8\]

\[4y\text{ }=\text{ }10\]

\[y\text{ }=\text{ }5/2\]