Chapter 11.4, Problem 22AYU

To determine

To calculate: The value of the expression $CA+5I_3$ .

Answer to Problem 22AYU

The value of the expression $CA+5I_3$ is $\left[\begin{array}{ccc}6& 14& -14\\ 2& 27& -18\\ 3& 0& 33\end{array}\right]$ .

Explanation of Solution

Given information:

The matrices are $A=\left[\begin{array}{ccc}0& 3& -5\\ 1& 2& 6\end{array}\right],B=\left[\begin{array}{ccc}4& 1& 0\\ -2& 3& -2\end{array}\right],C=\left[\begin{array}{cc}4& 1\\ 6& 2\\ -2& 3\end{array}\right]$

Formula used:

To find the product AB of two matrices A and B, the number of column in matrix A must be equal to number of rows in matrix B.

Calculation:

Consider the matrices are $A=\left[\begin{array}{ccc}0& 3& -5\\ 1& 2& 6\end{array}\right],B=\left[\begin{array}{ccc}4& 1& 0\\ -2& 3& -2\end{array}\right],C=\left[\begin{array}{cc}4& 1\\ 6& 2\\ -2& 3\end{array}\right]$

First compute, $CA$ . Recall that to find the product AB of two matrices A and B, the number of column in matrix A must be equal to number of rows in matrix B.

  $\begin{array}{c}CA=\left[\begin{array}{cc}4& 1\\ 6& 2\\ -2& 3\end{array}\right]\left[\begin{array}{ccc}0& 3& -5\\ 1& 2& 6\end{array}\right]\\ =\left[\begin{array}{ccc}0+1& 12+2& -20+6\\ 0+2& 18+4& -30+12\\ 3& -6+6& 10+18\end{array}\right]\\ =\left[\begin{array}{ccc}1& 14& -14\\ 2& 22& -18\\ 3& 0& 28\end{array}\right]\end{array}$

Now, compute $CA+5I_3$ .

  $\begin{array}{c}CA+5I_3=\left[\begin{array}{ccc}1& 14& -14\\ 2& 22& -18\\ 3& 0& 28\end{array}\right]+5\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\\ =\left[\begin{array}{ccc}1& 14& -14\\ 2& 22& -18\\ 3& 0& 28\end{array}\right]+\left[\begin{array}{ccc}5& 0& 0\\ 0& 5& 0\\ 0& 0& 5\end{array}\right]\\ =\left[\begin{array}{ccc}6& 14& -14\\ 2& 27& -18\\ 3& 0& 33\end{array}\right]\end{array}$

Thus, the value of the expression $CA+5I_3$ is $\left[\begin{array}{ccc}6& 14& -14\\ 2& 27& -18\\ 3& 0& 33\end{array}\right]$ .