Chapter 10.4, Problem 12E

To determine

To find: Perform the matrix operation and in case it is impossible then state reason.

Answer to Problem 12E

Solution for given matrix is $\left[\begin{array}{cc}1& 2\\ 7& 6\end{array}\right]$ .

Explanation of Solution

Given information:

  $\left[\begin{array}{ccc}2& 1& 2\\ 6& 3& 4\end{array}\right]\left[\begin{array}{cc}1& -2\\ 3& 6\\ -2& 0\end{array}\right]$

Calculation:

Given matrix is :

  $\left[\begin{array}{ccc}2& 1& 2\\ 6& 3& 4\end{array}\right]\left[\begin{array}{cc}1& -2\\ 3& 6\\ -2& 0\end{array}\right]$

Multiplication of the matrix is possible only if number of columns in first matrix is equal to the number of rows in the second matrix.

In the above case, number of columns in the first matrix is equal to $3$ while number of rows in the second matrix is equal to $2$ .

Because of this fact, conclusion is given operation on matrix is possible.

Dimension of first matrix is $2\times 3$ and second matrix is $3\times 2$

Product dimension will be $2\times 2$ .

Consider $C=AB$ , then entry $C_{ij}$ is inner product of the $i^{th}$ row of matrix $A$ and $j^{th}$ column of the matrix $B$ .

Matrix $C$ will be,

  $\left[\begin{array}{cc}{C}_{11}& C_{12}\\ C_{21}& C_{22}\end{array}\right]$

For calculating entries of matrix $C$ is as below,

  $\begin{array}{l}\left[\begin{array}{ccc}2& 1& 2\\ 6& 3& 4\end{array}\right]\left[\begin{array}{cc}1& -2\\ 3& 6\\ -2& 0\end{array}\right]\\ =\left[\begin{array}{cc}2.1+1.3+2.\left(-2\right)& 2.\left(-2\right)+1.6+2.0\\ 6.1+3.3+4.\left(-2\right)& 6.\left(-2\right)+3.6+4.0\end{array}\right]\\ =\left[\begin{array}{cc}2+3-4& -4+6+0\\ 6+9-8& -12+18+0\end{array}\right]\\ =\left[\begin{array}{cc}1& 2\\ 7& 6\end{array}\right]\end{array}$