Chapter 16, Problem 3CT

To determine

To find: The value of matrix $B-A$ .

Answer to Problem 3CT

The value of matrix $B-A=\left[\begin{array}{ccc}-4& 0& 2\\ 5& -4& 0\\ -4& 4& -1\end{array}\right]$ .

Explanation of Solution

Given information:

The given expression is $A=\left[\begin{array}{ccc}2& 1& 3\\ 0& -1& 1\\ 4& -1& 0\end{array}\right]$ and $B=\left[\begin{array}{ccc}-2& 1& 5\\ 5& -5& 1\\ 0& 3& -1\end{array}\right]$

Calculation:

Consider the expression.

Substitute $\left[\begin{array}{ccc}2& 1& 3\\ 0& -1& 1\\ 4& -1& 0\end{array}\right]$ for $A$ , and $\left[\begin{array}{ccc}-2& 1& 5\\ 5& -5& 1\\ 0& 3& -1\end{array}\right]$ for $B$ .

  $B-A=\left[\begin{array}{ccc}-2& 1& 5\\ 5& -5& 1\\ 0& 3& -1\end{array}\right]-\left[\begin{array}{ccc}2& 1& 3\\ 0& -1& 1\\ 4& -1& 0\end{array}\right]$

We know that subtract is defined in terms of addition. Subtraction is defined for each set of $m\times n$ matrices as $A_{m\times n}-B_{m\times n}=A_{m\times n}+\left(-B_{m\times n}\right).$

  $\left[\begin{array}{ccc}-2& 1& 5\\ 5& -5& 1\\ 0& 3& -1\end{array}\right]-\left[\begin{array}{ccc}2& 1& 3\\ 0& -1& 1\\ 4& -1& 0\end{array}\right]=\left[\begin{array}{ccc}-2& 1& 5\\ 5& -5& 1\\ 0& 3& -1\end{array}\right]+\left(-\left[\begin{array}{ccc}2& 1& 3\\ 0& -1& 1\\ 4& -1& 0\end{array}\right]\right)$

Each element of $-A$ is the opposite of its corresponding element in A.

  $\left[\begin{array}{ccc}-2& 1& 5\\ 5& -5& 1\\ 0& 3& -1\end{array}\right]+\left(-\left[\begin{array}{ccc}2& 1& 3\\ 0& -1& 1\\ 4& -1& 0\end{array}\right]\right)=\left[\begin{array}{ccc}-2& 1& 5\\ 5& -5& 1\\ 0& 3& -1\end{array}\right]+\left[\begin{array}{ccc}-2& -1& -3\\ 0& 1& -1\\ -4& 1& 0\end{array}\right]$

The dimensions of the given matrices are the same. Add the corresponding elements.

  $\begin{array}{c}\left[\begin{array}{ccc}-2& 1& 5\\ 5& -5& 1\\ 0& 3& -1\end{array}\right]+\left[\begin{array}{ccc}-2& -1& -3\\ 0& 1& -1\\ -4& 1& 0\end{array}\right]=\left[\begin{array}{ccc}-2+\left(-2\right)& 1+\left(-1\right)& 5+\left(-3\right)\\ 5+0& -5+1& 1+\left(-1\right)\\ 0+\left(-4\right)& 3+1& -1+0\end{array}\right]\\ =\left[\begin{array}{ccc}-4& 0& 2\\ 5& -4& 0\\ -4& 4& -1\end{array}\right]\end{array}$

Therefore, the value of matrix $B-A$ is $\left[\begin{array}{ccc}-4& 0& 2\\ 5& -4& 0\\ -4& 4& -1\end{array}\right]$ .