Q21. - Algebra 2

Question

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

 

21. $8\left[\begin{array}{cc}0.25& 0.5\\ 0.75& 1.5\end{array}\right]-2\left[\begin{array}{cc}0.25& 0.5\\ 0.75& 1.5\end{array}\right]$

Step-by-Step Solution

Answer

The final matrix after performing the indicated matrix operations is

$8\left[\begin{array}{cc}0.25& 0.5\\ 0.75& 1.5\end{array}\right]-2\left[\begin{array}{cc}0.25& 0.5\\ 0.75& 1.5\end{array}\right]=\left[\begin{array}{cc}1.5& 3\\ 4.5& 9\end{array}\right]$.

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 $\left[\begin{array}{cc}0.25& 0.5\\ 0.75& 1.5\end{array}\right]$ by 8 and multiply each element in the second matrix $\left[\begin{array}{cc}0.25& 0.5\\ 0.75& 1.5\end{array}\right]$ by 2.

$\begin{array}{c}8\left[\begin{array}{cc}0.25& 0.5\\ 0.75& 1.5\end{array}\right]-2\left[\begin{array}{cc}0.25& 0.5\\ 0.75& 1.5\end{array}\right]=\left[\begin{array}{cc}8\left(0.25\right)& 8\left(0.5\right)\\ 8\left(0.75\right)& 8\left(1.5\right)\end{array}\right]-\left[\begin{array}{cc}2\left(0.25\right)& 2\left(0.5\right)\\ 2\left(0.75\right)& 2\left(1.5\right)\end{array}\right]\\ =\left[\begin{array}{cc}2& 4\\ 6& 12\end{array}\right]-\left[\begin{array}{cc}0.5& 1\\ 1.5& 3\end{array}\right]\end{array}$

Step 3 - Subtract corresponding elements

Subtract corresponding elements of both the matrices $\left[\begin{array}{cc}2& 4\\ 6& 12\end{array}\right]-\left[\begin{array}{cc}0.5& 1\\ 1.5& 3\end{array}\right]$ and simplify.

$\begin{array}{c}8\left[\begin{array}{cc}0.25& 0.5\\ 0.75& 1.5\end{array}\right]-2\left[\begin{array}{cc}0.25& 0.5\\ 0.75& 1.5\end{array}\right]=\left[\begin{array}{cc}2& 4\\ 6& 12\end{array}\right]-\left[\begin{array}{cc}0.5& 1\\ 1.5& 3\end{array}\right]\\ =\left[\begin{array}{cc}2-0.5& 4-1\\ 6-1.5& 12-3\end{array}\right]\\ =\left[\begin{array}{cc}1.5& 3\\ 4.5& 9\end{array}\right]\end{array}$