91Ó°ÊÓ

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 $m×n$ matrix is an $m×n$ 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 $m×n$, then the addition/subtraction of the two matrices will also yield a matrix of order $m×n$wherein each element is the sum/difference of the corresponding elements.

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