Use matrices$A$, $B$, $C$and $D$to find the following.

$A=\left[\begin{array}{cc}5& 7\\ -1& 6\\ 3& -9\end{array}\right],B=\left[\begin{array}{cc}8& 3\\ 5& 1\\ 4& 4\end{array}\right],C=\left[\begin{array}{cc}0& 4\\ -2& 5\\ 7& -1\end{array}\right],D=\left[\begin{array}{cc}6& 2\\ 9& 0\\ -3& 0\end{array}\right]$

28. $3C-4A+B$

Substitute$\left[\begin{array}{cc}5& 7\\ -1& 6\\ 3& -9\end{array}\right]$ for A, $\left[\begin{array}{cc}8& 3\\ 5& 1\\ 4& 4\end{array}\right]$for B and$\left[\begin{array}{cc}0& 4\\ -2& 5\\ 7& -1\end{array}\right]$ for C into $3C-4A+B$.

$3C-4A+B=3\left[\begin{array}{cc}0& 4\\ -2& 5\\ 7& -1\end{array}\right]-4\left[\begin{array}{cc}5& 7\\ -1& 6\\ 3& -9\end{array}\right]+\left[\begin{array}{cc}8& 3\\ 5& 1\\ 4& 4\end{array}\right]$

Scalar multiplication -The product of a scalar k and an $m\times n$ matrix is an $m\times n$ matrix in which each element equals k times the corresponding elements of the original matrix.

Multiply each element of the first matrix $\left[\begin{array}{cc}0& 4\\ -2& 5\\ 7& -1\end{array}\right]$by 3 and multiply each element of the second matrix $\left[\begin{array}{cc}5& 7\\ -1& 6\\ 3& -9\end{array}\right]$by 4.

$\begin{array}{c}3C-4A+B=3\left[\begin{array}{cc}0& 4\\ -2& 5\\ 7& -1\end{array}\right]-4\left[\begin{array}{cc}5& 7\\ -1& 6\\ 3& -9\end{array}\right]+\left[\begin{array}{cc}8& 3\\ 5& 1\\ 4& 4\end{array}\right]\\ =\left[\begin{array}{cc}3\left(0\right)& 3\left(4\right)\\ 3\left(-2\right)& 3\left(5\right)\\ 3\left(7\right)& 3\left(-1\right)\end{array}\right]-\left[\begin{array}{cc}4\left(5\right)& 4\left(7\right)\\ 4\left(-1\right)& 4\left(6\right)\\ 4\left(3\right)& 4\left(-9\right)\end{array}\right]+\left[\begin{array}{cc}8& 3\\ 5& 1\\ 4& 4\end{array}\right]\\ =\left[\begin{array}{cc}0& 12\\ -6& 15\\ 21& -3\end{array}\right]-\left[\begin{array}{cc}20& 28\\ -4& 24\\ 12& -36\end{array}\right]+\left[\begin{array}{cc}8& 3\\ 5& 1\\ 4& 4\end{array}\right]\end{array}$

Matrix addition/subtraction - If A and B are two matrices of order $m\times n$, then the addition/subtraction of the two matrices will also yield a matrix of order $m\times n$wherein each element is the sum/difference of the corresponding elements.

$\begin{array}{c}3C-4A+B=\left[\begin{array}{cc}0& 12\\ -6& 15\\ 21& -3\end{array}\right]-\left[\begin{array}{cc}20& 28\\ -4& 24\\ 12& -36\end{array}\right]+\left[\begin{array}{cc}8& 3\\ 5& 1\\ 4& 4\end{array}\right]\\ =\left[\begin{array}{cc}0-20+8& 12-28+3\\ -6-\left(-4\right)+5& 15-24+1\\ 21-12+4& -3-\left(-36\right)+4\end{array}\right]\\ =\left[\begin{array}{cc}-12& -13\\ 3& -8\\ 13& 37\end{array}\right]\end{array}$