Q22. - Algebra 2

Question

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

 

22. $\frac{1}{2}\left[\begin{array}{cc}4& 6\\ 3& 0\end{array}\right]-\frac{2}{3}\left[\begin{array}{cc}9& 27\\ 0& 3\end{array}\right]$

Step-by-Step Solution

Answer

The final matrix after performing the indicated matrix operations is

$\frac{1}{2}\left[\begin{array}{cc}4& 6\\ 3& 0\end{array}\right]-\frac{2}{3}\left[\begin{array}{cc}9& 27\\ 0& 3\end{array}\right]=\left[\begin{array}{cc}-4& -15\\ \frac{3}{2}& -2\end{array}\right]$.

Step 1 - Define the concept used

Matrix subtraction and scalar multiplication:

If A and B are two matrices of order $m\times n$, then the subtraction of the two matrices will also yield a matrix of order $m\times n$ wherein each element is the difference 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.

Step 2 - Perform the scalar multiplication first

Multiply each element in the first matrix $\left[\begin{array}{cc}4& 6\\ 3& 0\end{array}\right]$ by  $\frac{1}{2}$ and multiply each element in the second matrix $\left[\begin{array}{cc}9& 27\\ 0& 3\end{array}\right]$ by $\frac{2}{3}$.

$\begin{array}{c}\frac{1}{2}\left[\begin{array}{cc}4& 6\\ 3& 0\end{array}\right]-\frac{2}{3}\left[\begin{array}{cc}9& 27\\ 0& 3\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}\left(4\right)& \frac{1}{2}\left(6\right)\\ \frac{1}{2}\left(3\right)& \frac{1}{2}\left(0\right)\end{array}\right]-\left[\begin{array}{cc}\frac{2}{3}\left(9\right)& \frac{2}{3}\left(27\right)\\ \frac{2}{3}\left(0\right)& \frac{2}{3}\left(3\right)\end{array}\right]\\ =\left[\begin{array}{cc}2& 3\\ \frac{3}{2}& 0\end{array}\right]-\left[\begin{array}{cc}6& 18\\ 0& 2\end{array}\right]\end{array}$

Step 3 - Subtract corresponding elements

Subtract corresponding elements of both the matrices $\left[\begin{array}{cc}2& 3\\ \frac{3}{2}& 0\end{array}\right]-\left[\begin{array}{cc}6& 18\\ 0& 2\end{array}\right]$and simplify.

$\begin{array}{c}\frac{1}{2}\left[\begin{array}{cc}4& 6\\ 3& 0\end{array}\right]-\frac{2}{3}\left[\begin{array}{cc}9& 27\\ 0& 3\end{array}\right]=\left[\begin{array}{cc}2& 3\\ \frac{3}{2}& 0\end{array}\right]-\left[\begin{array}{cc}6& 18\\ 0& 2\end{array}\right]\\ =\left[\begin{array}{cc}2-6& 3-18\\ \frac{3}{2}-0& 0-2\end{array}\right]\\ =\left[\begin{array}{cc}-4& -15\\ \frac{3}{2}& -2\end{array}\right]\end{array}$