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

The matrix operation to be performed is addition.

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.

Here for the matrix $\frac{2}{3}\left[\begin{array}{cc}9& 0\\ 12& 15\end{array}\right]$, $\frac{2}{3}$ is a scalar quantity.

Multiply each element of the this matrix by $\frac{2}{3}$.

That is equal to,

$\left[\begin{array}{cc}9×\frac{2}{3}& 0×\frac{2}{3}\\ 12×\frac{2}{3}& 15×\frac{2}{3}\end{array}\right]=\left[\begin{array}{cc}6& 0\\ 8& 10\end{array}\right]$

Now, Matrices can be added if they are of the same dimension.

Here, both the matrices are of dimension $(2×2)$ that is both are of same dimension. Then addition can be performed.

Add the corresponding elements

$\begin{array}{c}\frac{2}{3}\left[\begin{array}{cc}9& 0\\ 12& 15\end{array}\right]+\left[\begin{array}{cc}−2& 3\\ −7& −7\end{array}\right]=\left[\begin{array}{cc}6& 0\\ 8& 10\end{array}\right]+\left[\begin{array}{cc}−2& 3\\ −7& −7\end{array}\right]\\ =\left[\begin{array}{cc}6+(−2)& 0+3\\ 8+(−7)& 10+(−7)\end{array}\right]\\ =\left[\begin{array}{cc}6−2& 3\\ 8−7& 10−7\end{array}\right]\\ =\left[\begin{array}{cc}4& 3\\ 1& 3\end{array}\right]\end{array}$

Therefore, the resultant matrix is$\left[\begin{array}{cc}4& 3\\ 1& 3\end{array}\right]$