Chapter 16, Problem 5CT

To determine

To find: The value of single matrix $A-2B.$

Answer to Problem 5CT

The value of single matrix $A-2B$ is $\left(\begin{array}{ccc}6& -1& -7\\ -10& 9& -1\\ 4& -7& 2\end{array}\right)\left(\begin{array}{ccc}-4& 0& 2\\ 5& -4& 0\\ -4& 4& -1\end{array}\right)$.

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:

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.

  $A-2B\left(\begin{array}{ccc}2& 1& 3\\ 0& -1& 1\\ 4& -1& 0\end{array}\right)-2\left(\begin{array}{ccc}-2& 1& 5\\ 5& -5& 1\\ 0& 3& -1\end{array}\right)$.

We know that the scalar product of a real number $r$ and a matrix $A$ is the matrix $rA$ and each element of $rA$ is $r$ times its corresponding element in $A$ .

Multiply each of the second matrix by $2$ .

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

Simplify.

  $\left(\begin{array}{ccc}2& 1& 3\\ 0& -1& 1\\ 4& -1& 0\end{array}\right)-2\left(\begin{array}{ccc}2(-2)& 2(1)& 2(5)\\ 2(5)& 2(-5)& 2(1)\\ 2(0)& 2(3)& 2(-1)\end{array}\right)=\left(\begin{array}{ccc}2& 1& 3\\ 0& -1& 1\\ 4& -1& 0\end{array}\right)-\left(\begin{array}{ccc}-4& 2& 10\\ 10& -10& 2\\ 0& 6& -2\end{array}\right)$ .

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

  $\left(\begin{array}{ccc}2& 1& 3\\ 0& -1& 1\\ 4& -1& 0\end{array}\right)-\left(\begin{array}{ccc}-4& 2& 10\\ 10& -10& 2\\ 0& 6& -2\end{array}\right)=\left(\begin{array}{ccc}2& 1& 3\\ 0& -1& 1\\ 4& -1& 0\end{array}\right)+\left(-\left(\begin{array}{ccc}-4& 2& 10\\ 10& -10& 2\\ 0& 6& -2\end{array}\right)\right)$

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

  $\left(\begin{array}{ccc}2& 1& 3\\ 0& -1& 1\\ 4& -1& 0\end{array}\right)+\left(-\left(\begin{array}{ccc}-4& 2& 10\\ 10& -10& 2\\ 0& 6& -2\end{array}\right)\right)=\left(\begin{array}{ccc}2& 1& 3\\ 0& -1& 1\\ 4& -1& 0\end{array}\right)+\left(\begin{array}{ccc}-4& 2& 10\\ 10& -10& 2\\ 0& 6& -2\end{array}\right)$

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

  $\left(\begin{array}{ccc}2& 1& 3\\ 0& -1& 1\\ 4& -1& 0\end{array}\right)+\left(-\begin{array}{ccc}4& -2& -10\\ 10& 10& -2\\ 0& -6& 2\end{array}\right)=\left(\begin{array}{ccc}2+4& 1+(-2)& 3+(-10)\\ 0+(-10)& -1+10& 1+(-2)\\ 4+0& -1+(-6)& 0+2\end{array}\right)$

Simplify.

  $\left(\begin{array}{ccc}2+4& 1+(-2)& 3+(-10)\\ 0+(-10)& -1+10& 1+(-2)\\ 4+0& -1+(-6)& 0+2\end{array}\right)=\left(\begin{array}{ccc}6& -1& -7\\ -10& 9& -1\\ 4& -7& 2\end{array}\right)$

Therefore, the expression of matrix $A-2B$ is $\left(\begin{array}{ccc}6& -1& -7\\ -10& 9& -1\\ 4& -7& 2\end{array}\right)\left(\begin{array}{ccc}-4& 0& 2\\ 5& -4& 0\\ -4& 4& -1\end{array}\right)$.