Chapter 8, Problem 48RE

To determine

Find the difference of expression with scalar multiplication.

Answer to Problem 48RE

The differenceof theexpression is $\left[\begin{array}{ccc}2& 1& 12\\ -13& 1& -17\\ -30& -54& 40\end{array}\right]$ .

Explanation of Solution

Given information:

The given expression is

  $-\left[\begin{array}{ccc}8& -1& 8\\ -2& 4& 12\\ 0& -6& 0\end{array}\right]-5\left[\begin{array}{ccc}-2& 0& -4\\ 3& -1& 1\\ 6& 12& -8\end{array}\right]$

Calculation:

If $A=\left[a_{ij}\right]$ is an $m\times n$ matrix and $c$ is a scalar, then the scalar multiple of $A$ by $c$ is the $m\times n$ matrix given by

  $cA=\left[c{a}_{ij}\right]$

If $A=\left[a_{ij}\right]$ and $B=\left[b_{ij}\right]$ are matrices of order $m\times n$ , then their sum is the $m\times n$ matrix given by

  $A+B=\left[a_{ij}+b_{ij}\right]$

The sum of two matrices of different orders is undefined.

Now

  $-\left[\begin{array}{ccc}8& -1& 8\\ -2& 4& 12\\ 0& -6& 0\end{array}\right]-5\left[\begin{array}{ccc}-2& 0& -4\\ 3& -1& 1\\ 6& 12& -8\end{array}\right]$

  $=\left[\begin{array}{ccc}(-)\cdot 8& (-)\cdot -1& (-)\cdot 8\\ (-)\cdot -2& (-)\cdot 4& (-)\cdot 12\\ (-)\cdot 0& (-)\cdot -6& (-)\cdot 0\end{array}\right]-\left[\begin{array}{ccc}-2\cdot 5& 0\cdot 5& -4\cdot 5\\ 3\cdot 5& -1\cdot 5& 1\cdot 5\\ 6\cdot 5& 12\cdot 5& -8\cdot 5\end{array}\right]$

  $=\left[\begin{array}{ccc}-8& 1& -8\\ 2& -4& -12\\ 0& 6& 0\end{array}\right]-\left[\begin{array}{ccc}-10& 0& -20\\ 15& -5& 5\\ 30& 60& -40\end{array}\right]$

  $=\left[\begin{array}{ccc}-8-\left(-10\right)& 1-0& -8-\left(-20\right)\\ 2-15& -4-\left(-5\right)& -12-5\\ 0-30& 6-60& 0-\left(-40\right)\end{array}\right]$

  $=\left[\begin{array}{ccc}2& 1& 12\\ -13& 1& -17\\ -30& -54& 40\end{array}\right]$

Hence the difference of expressionis $\left[\begin{array}{ccc}2& 1& 12\\ -13& 1& -17\\ -30& -54& 40\end{array}\right]$ .