Solution:
We know that if two matrices are equal then the elements of each matrices are also equal.
Given that two matrices are equal.
Therefore by equating them we get,
\[2a\text{ }+\text{ }b\text{ }=\text{ }4~\ldots \ldots ~\left( 1 \right)\]
And \[a\text{ }-\text{ }2b\text{ }=\text{ }\text{ }3~\ldots \ldots \text{ }\left( 2 \right)\]
And \[5c\text{ }-\text{ }d\text{ }=\text{ }11~\ldots \ldots ~\left( 3 \right)\]
\[4c\text{ }+\text{ }3d\text{ }=\text{ }24~\ldots \ldots ~\left( 4 \right)\]
Multiplying equation \[\left( 1 \right)\text{ }by\text{ }2\] and adding to equation (2)
\[4a\text{ }+\text{ }2b\text{ }+\text{ }a\text{ }-\text{ }2b\text{ }=\text{ }8\text{ }-\text{ }3\]
\[\Rightarrow ~5a\text{ }=\text{ }5\]
\[\Rightarrow ~a\text{ }=\text{ }1\]
Now, substituting the value of a in equation (1)
\[2\text{ }\times \text{ }1\text{ }+\text{ }b\text{ }=\text{ }4\]
\[\Rightarrow ~2\text{ }+\text{ }b\text{ }=\text{ }4\]
\[\Rightarrow ~b\text{ }=\text{ }4\text{ }-\text{ }2\]
\[\Rightarrow ~b\text{ }=\text{ }2\]
Multiplying equation \[\left( 3 \right)\text{ }by\text{ }3\] and adding to equation (4)
\[15c\text{ }-\text{ }3d\text{ }+\text{ }4c\text{ }+\text{ }3d\text{ }=\text{ }33\text{ }+\text{ }24\]
\[\Rightarrow ~19c\text{ }=\text{ }57\]
\[\Rightarrow ~c\text{ }=\text{ }3\]
Now, substituting the value of \[c\] in equation (4)
\[4\text{ }\times \text{ }3\text{ }+\text{ }3d\text{ }=\text{ }24\]
\[\Rightarrow ~12\text{ }+\text{ }3d\text{ }=\text{ }24\]
\[\Rightarrow ~3d\text{ }=\text{ }24\text{ }\text{ }12\]
\[\Rightarrow ~3d\text{ }=\text{ }12\]
\[\Rightarrow ~d\text{ }=\text{ }4\]
\[\therefore ~a\text{ }=\text{ }1,\text{ }b\text{ }=\text{ }2,\text{ }c\text{ }=\text{ }3\text{ }and\text{ }d\text{ }=\text{ }4\]