Chapter 2.3, Problem 8CFU

To determine

To find: value for the given case by using the matrices X,Y and Z. Write impossible if the matrix does not exist.

Answer to Problem 8CFU

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

Explanation of Solution

Given information:

The given matrices are as follows:

  $X=\left[\begin{array}{cc}4& 1\\ -2& 6\end{array}\right]Y=\left[\begin{array}{cc}0& -3\end{array}\right]\text{ Z}=\left[\begin{array}{cc}-1& 3\\ 0& -2\end{array}\right]$

The given case is $X+Z$ .

Calculation:

Solve $X+Z$ as follows:

  $\begin{array}{c}X+Z=\left[\begin{array}{cc}4& 1\\ -2& 6\end{array}\right]+\left[\begin{array}{cc}-1& 3\\ 0& -2\end{array}\right]\\ =\left[\begin{array}{cc}4& 4\\ -2& 4\end{array}\right]\end{array}$

Thus, the values of $X+Z$ is $\left[\begin{array}{cc}4& 4\\ -2& 4\end{array}\right]$ .