Q50. - Algebra 2

Question

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

$4\left[\begin{array}{ccc}3& 4& -7\\ 6& -9& -2\\ -3& 1& 3\end{array}\right]-\left[\begin{array}{ccc}-8& 6& -4\\ -7& 10& 1\\ -2& 1& 5\end{array}\right]$

Step-by-Step Solution

Answer

After performing matrix operations, resulting matrix is $\left[\begin{array}{ccc}20& 10& -24\\ 31& -46& -9\\ -10& 3& 7\end{array}\right]$

Step 1 - Write operation with matrices

Let A and B be two $m\times n$ matrices, then following operations are defined:

• Matrix Addition- $A+B$ is an $m\times n$ matrix in which each element is the sum of the corresponding elements of A and B
•  Matrix Subtraction- $A-B$ is an $m\times n$ matrix in which each element is the difference of the corresponding elements of A and B
• Scalar Multiplication- The product of a scalar k with $m\times n$ matrix is an $m\times n$ matrix in which each element equals k times the corresponding elements of the original matrix.

Step 2 - Perform Scalar Multiplication

First matrix operation applied in given expression is scalar multiplication. Multiply given  matrices with scalars as shown below

 $4\left[\begin{array}{ccc}3& 4& -7\\ 6& -9& -2\\ 3& 1& 3\end{array}\right]-\left[\begin{array}{ccc}-8& 6& -4\\ -7& 10& 1\\ -2& 1& 5\end{array}\right]=\left[\begin{array}{ccc}12& 16& -28\\ 24& -36& -8\\ -12& 4& 12\end{array}\right]-\left[\begin{array}{ccc}-8& 6& -4\\ -7& 10& 1\\ -2& 1& 5\end{array}\right]$

Step 2 - Perform Matrix Subtraction

As both the matrices are of same dimension, so apply matrix subtraction to get,

$\begin{array}{c}\left[\begin{array}{ccc}12& 16& -28\\ 24& -36& -8\\ -12& 4& 12\end{array}\right]-\left[\begin{array}{ccc}-8& 6& -4\\ -7& 10& 1\\ -2& 1& 5\end{array}\right]=\left[\begin{array}{ccc}12+8& 16-6& -28+4\\ 24+7& -36-10& -8-1\\ -12+2& 4-1& 12-5\end{array}\right]\\ =\left[\begin{array}{ccc}20& 10& -24\\ 31& -46& -9\\ -10& 3& 7\end{array}\right]\end{array}$