Q17. - Algebra 2

Question

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

 

17. $-2\left[\begin{array}{ccc}2& -4& 1\\ -3& 5& 8\\ 7& 6& -2\end{array}\right]$

Step-by-Step Solution

Answer

The final matrix after performing the indicated matrix operation is

$-2\left[\begin{array}{ccc}2& -4& 1\\ -3& 5& 8\\ 7& 6& -2\end{array}\right]=\left[\begin{array}{ccc}-4& 8& -2\\ 6& -10& -16\\ -14& -12& 4\end{array}\right]$.

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 $-2\left[\begin{array}{ccc}2& -4& 1\\ -3& 5& 8\\ 7& 6& -2\end{array}\right]$, multiply each element of the matrix $\left[\begin{array}{ccc}2& -4& 1\\ -3& 5& 8\\ 7& 6& -2\end{array}\right]$ by -2.

$-2\left[\begin{array}{ccc}2& -4& 1\\ -3& 5& 8\\ 7& 6& -2\end{array}\right]=\left[\begin{array}{ccc}2\left(2\right)& 2\left(-4\right)& 2\left(1\right)\\ 2\left(-3\right)& 2\left(5\right)& 2\left(8\right)\\ 2\left(7\right)& 2\left(6\right)& 2\left(-2\right)\end{array}\right]$

Step 3 - Simplify the matrix

Further simplify the matrix.

$\begin{array}{c}-2\left[\begin{array}{ccc}2& -4& 1\\ -3& 5& 8\\ 7& 6& -2\end{array}\right]=\left[\begin{array}{ccc}2\left(2\right)& 2\left(-4\right)& 2\left(1\right)\\ 2\left(-3\right)& 2\left(5\right)& 2\left(8\right)\\ 2\left(7\right)& 2\left(6\right)& 2\left(-2\right)\end{array}\right]\\ =\left[\begin{array}{ccc}-4& 8& -2\\ 6& -10& -16\\ -14& -12& 4\end{array}\right]\end{array}$

Therefore, $-2\left[\begin{array}{ccc}2& -4& 1\\ -3& 5& 8\\ 7& 6& -2\end{array}\right]=\left[\begin{array}{ccc}-4& 8& -2\\ 6& -10& -16\\ -14& -12& 4\end{array}\right]$.