Q26. - Algebra 2

Question

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

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

 

$4C$

Step-by-Step Solution

Answer

The final matrix after performing the indicated matrix operations is $4C=\left[\begin{array}{cc}0& 16\\ -8& 20\\ 28& -4\end{array}\right]$.

Step 1 - Substitute the matrix

Substitute $\left[\begin{array}{cc}0& 4\\ -2& 5\\ 7& -1\end{array}\right]$ for C into $4C$.

 

$4C=4\left[\begin{array}{cc}0& 4\\ -2& 5\\ 7& -1\end{array}\right]$

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 matrix $\left[\begin{array}{cc}0& 4\\ -2& 5\\ 7& -1\end{array}\right]$ by 4.

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

Step 3 - Simplify

Further simplify the matrix.

 

 $\begin{array}{c}4C=\left[\begin{array}{cc}4\left(0\right)& 4\left(4\right)\\ 4\left(-2\right)& 4\left(5\right)\\ 4\left(7\right)& 4\left(-1\right)\end{array}\right]\\ =\left[\begin{array}{cc}0& 16\\ -8& 20\\ 28& -4\end{array}\right]\end{array}$