Q29. - Algebra 2

Question

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

 

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

 

29. $C+\frac{1}{3}D$

Step-by-Step Solution

Answer

The final matrix after performing the indicated matrix operations is

$C+\frac{1}{3}D=\left[\begin{array}{cc}2& \frac{14}{3}\\ 1& 5\\ 6& -1\end{array}\right]$.

Step 1 - Substitute the matrix

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

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

Step 2 - Perform the scalar multiplication

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 second matrix $\left[\begin{array}{cc}6& 2\\ 9& 0\\ -3& 0\end{array}\right]$by $\frac{1}{3}$.

$\begin{array}{c}C+\frac{1}{3}D=\left[\begin{array}{cc}0& 4\\ -2& 5\\ 7& -1\end{array}\right]+\frac{1}{3}\left[\begin{array}{cc}6& 2\\ 9& 0\\ -3& 0\end{array}\right]\\ =\left[\begin{array}{cc}0& 4\\ -2& 5\\ 7& -1\end{array}\right]+\left[\begin{array}{cc}\frac{1}{3}\left(6\right)& \frac{1}{3}\left(2\right)\\ \frac{1}{3}\left(9\right)& \frac{1}{3}\left(0\right)\\ \frac{1}{3}\left(-3\right)& \frac{1}{3}\left(0\right)\end{array}\right]\\ =\left[\begin{array}{cc}0& 4\\ -2& 5\\ 7& -1\end{array}\right]+\left[\begin{array}{cc}2& \frac{2}{3}\\ 3& 0\\ -1& 0\end{array}\right]\end{array}$

Step 3 - Perform the matrix addition

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

$\begin{array}{c}C+\frac{1}{3}D=\left[\begin{array}{cc}0& 4\\ -2& 5\\ 7& -1\end{array}\right]+\left[\begin{array}{cc}2& \frac{2}{3}\\ 3& 0\\ -1& 0\end{array}\right]\\ =\left[\begin{array}{cc}0+2& 4+\frac{2}{3}\\ -2+3& 5+0\\ 7-1& -1+0\end{array}\right]\\ =\left[\begin{array}{cc}2& \frac{14}{3}\\ 1& 5\\ 6& -1\end{array}\right]\end{array}$