Question

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

7.

$4\left[\begin{array}{cc}2& 7\\ -3& 6\end{array}\right]+5\left[\begin{array}{cc}-6& -4\\ 3& 0\end{array}\right]$

Short answer

The final matrix after performing the indicated matrix operations is

$4\left[\begin{array}{cc}2& 7\\ -3& 6\end{array}\right]+5\left[\begin{array}{cc}-6& -4\\ 3& 0\end{array}\right]=\left[\begin{array}{cc}-22& 8\\ 3& 24\end{array}\right]$

Step-by-step solution

- Define the concept used

Matrix subtraction and scalar multiplication:

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.

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.

- Perform the scalar multiplication first

Multiply each element in the first matrix $\left[\begin{array}{cc}2& 7\\ -3& 6\end{array}\right]$

by 4 and multiply each element in the second matrix$\left[\begin{array}{cc}-6& -4\\ 3& 0\end{array}\right]$

by 5.

$\begin{array}{c}4\left[\begin{array}{cc}2& 7\\ -3& 6\end{array}\right]+5\left[\begin{array}{cc}-6& -4\\ 3& 0\end{array}\right]=\left[\begin{array}{cc}4\left(2\right)& 4\left(7\right)\\ 4\left(-3\right)& 4\left(6\right)\end{array}\right]+\left[\begin{array}{cc}5\left(-6\right)& 5\left(-4\right)\\ 5\left(3\right)& 5\left(0\right)\end{array}\right]\\ =\left[\begin{array}{cc}8& 28\\ -12& 24\end{array}\right]+\left[\begin{array}{cc}-30& -20\\ 15& 0\end{array}\right]\end{array}$

- Add corresponding elements

Add corresponding elements of both the matrices $\left[\begin{array}{cc}8& 28\\ -12& 24\end{array}\right]+\left[\begin{array}{cc}-30& -20\\ 15& 0\end{array}\right]$and simplify.

$\begin{array}{c}4\left[\begin{array}{cc}2& 7\\ -3& 6\end{array}\right]+5\left[\begin{array}{cc}-6& -4\\ 3& 0\end{array}\right]=\left[\begin{array}{cc}8& 28\\ -12& 24\end{array}\right]+\left[\begin{array}{cc}-30& -20\\ 15& 0\end{array}\right]\\ =\left[\begin{array}{cc}8-30& 28-20\\ -12+15& 24+0\end{array}\right]\\ =\left[\begin{array}{cc}-22& 8\\ 3& 24\end{array}\right]\end{array}$