Q6. - Algebra 2

Question

Perform the indicated matrix operation. If the matrix doesn’t exist, write impossible.

 

6. $3\left[\begin{array}{cccc}6& -1& 5& 2\\ 7& 3& -2& 8\end{array}\right]$

Step-by-Step Solution

Answer

The final matrix after performing the indicated matrix operation is

$3\left[\begin{array}{cccc}6& -1& 5& 2\\ 7& 3& -2& 8\end{array}\right]=\left[\begin{array}{cccc}18& -3& 15& 6\\ 21& 9& -6& 24\end{array}\right]$.

Step 1 - Define concept of 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.

Step 2 - Multiply each element by

In order to find $3\left[\begin{array}{cccc}6& -1& 5& 2\\ 7& 3& -2& 8\end{array}\right]$, multiply each element of the matrix $\left[\begin{array}{cccc}6& -1& 5& 2\\ 7& 3& -2& 8\end{array}\right]$ by 3.

$3\left[\begin{array}{cccc}6& -1& 5& 2\\ 7& 3& -2& 8\end{array}\right]=\left[\begin{array}{cccc}3\left(6\right)& 3\left(-1\right)& 3\left(5\right)& 3\left(2\right)\\ 3\left(7\right)& 3\left(3\right)& 3\left(-2\right)& 3\left(8\right)\end{array}\right]$

Step 3 - Simplify the matrix

Further simplify the matrix.

$\begin{array}{c}3\left[\begin{array}{cccc}6& -1& 5& 2\\ 7& 3& -2& 8\end{array}\right]=\left[\begin{array}{cccc}3\left(6\right)& 3\left(-1\right)& 3\left(5\right)& 3\left(2\right)\\ 3\left(7\right)& 3\left(3\right)& 3\left(-2\right)& 3\left(8\right)\end{array}\right]\\ =\left[\begin{array}{cccc}18& -3& 15& 6\\ 21& 9& -6& 24\end{array}\right]\end{array}$