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A rectangular array of numbers arranged in rows and columns is called a matrix.

To evaluate this matrix first perform scalar multiplication and then subtraction.

Scalar multiplication is the method in which matrix is multiplied by a constant called scalar.

The matrix is:$2\left[\begin{array}{cc}6& 3\\ −8& −2\end{array}\right]−4\left[\begin{array}{cc}8& 1\\ 3& −4\end{array}\right]$

Multiply each element of the first matrix by 2 and each element of the second matrix by 4.

$2\left[\begin{array}{cc}6& 3\\ −8& −2\end{array}\right]−4\left[\begin{array}{cc}8& 1\\ 3& −4\end{array}\right]=\left[\begin{array}{cc}12& 6\\ −16& −4\end{array}\right]−\left[\begin{array}{cc}32& 4\\ 12& −16\end{array}\right]$

Matrices can be subtracted if they are of the same dimension.

Here, both the matrices are of dimension . Then subtraction can be performed.

Subtract the corresponding elements

$\begin{array}{c}\left[\begin{array}{cc}12& 6\\ −16& −4\end{array}\right]−\left[\begin{array}{cc}32& 4\\ 12& −16\end{array}\right]=\left[\begin{array}{cc}12−32& 6−4\\ −16−12& −4−(−16)\end{array}\right]\\ =\left[\begin{array}{cc}−20& 2\\ −28& −4+16\end{array}\right]\\ =\left[\begin{array}{cc}−20& 2\\ −28& 12\end{array}\right]\end{array}$

As the dimensions are not same, so subtraction of matrices cannot be possible.

Therefore, the solution matrix is$\left[\begin{array}{cc}−20& 2\\ −28& 12\end{array}\right]$