Q49. - Algebra 2

Question

Perform the indicated matrix operations. If the matrix does not exist, write impossible.

 $2\left[\begin{array}{cc}6& 3\\ -8& -2\end{array}\right]-4\left[\begin{array}{cc}8& 1\\ 3& -4\end{array}\right]$

Step-by-Step Solution

Answer

The answer is $\left[\begin{array}{cc}-20& 2\\ -28& 12\end{array}\right]$

Step 1 - State the definition of a matrix.

A rectangular array of numbers arranged in rows and columns is called a matrix.

Step 2 - Solve the matrix.

To evaluate this matrix first perform scalar multiplication and then subtraction.

 Scalar multiplication is the method in which matrix is multiplied by a constant called scalar.

 The matrix is:$2\left[\begin{array}{cc}6& 3\\ -8& -2\end{array}\right]-4\left[\begin{array}{cc}8& 1\\ 3& -4\end{array}\right]$

Multiply each element of the first matrix by 2 and each element of the second matrix by 4.

$2\left[\begin{array}{cc}6& 3\\ -8& -2\end{array}\right]-4\left[\begin{array}{cc}8& 1\\ 3& -4\end{array}\right]=\left[\begin{array}{cc}12& 6\\ -16& -4\end{array}\right]-\left[\begin{array}{cc}32& 4\\ 12& -16\end{array}\right]$

Matrices can be subtracted if they are of the same dimension.

 Here, both the matrices are of dimension  . Then subtraction can be performed.

 Subtract the corresponding elements

 $\begin{array}{c}\left[\begin{array}{cc}12& 6\\ -16& -4\end{array}\right]-\left[\begin{array}{cc}32& 4\\ 12& -16\end{array}\right]=\left[\begin{array}{cc}12-32& 6-4\\ -16-12& -4-(-16)\end{array}\right]\\ =\left[\begin{array}{cc}-20& 2\\ -28& -4+16\end{array}\right]\\ =\left[\begin{array}{cc}-20& 2\\ -28& 12\end{array}\right]\end{array}$

As the dimensions are not same, so subtraction of matrices cannot be possible.

Step 3 - State the conclusion.

Therefore, the solution matrix is $\left[\begin{array}{cc}-20& 2\\ -28& 12\end{array}\right]$