Chapter 3.5, Problem 20PPS

To determine

The solution of the given expression.

Answer to Problem 20PPS

  $\left[\begin{array}{cc}-22& -57\\ 62& 118\end{array}\right]$

Explanation of Solution

Given:

  $A=\left[\begin{array}{cc}0& -7\\ 8& 12\end{array}\right],\text{ and }B=\left[\begin{array}{cc}11& 4\\ -3& -17\end{array}\right]$

Concept Used:

• To multiply a scalar ‘ k’ to a matrix, multiply each element of matrix A by ‘ k’ :

  $k\left[\begin{array}{ccccc}{a}_{11}& a_{12}& .& .& a_{1n}\\ a_{21}& a_{22}& .& .& a_{2n}\\ .& .& .& .& .\\ .& .& .& .& .\\ a_{m1}& a_{m2}& .& .& a_{mn}\end{array}\right]=\left[\begin{array}{ccccc}k{a}_{11}& k{a}_{12}& .& .& k{a}_{1n}\\ k{a}_{21}& k{a}_{22}& .& .& k{a}_{2n}\\ .& .& .& .& .\\ .& .& .& .& .\\ k{a}_{m1}& k{a}_{m2}& .& .& k{a}_{mn}\end{array}\right]$

• Addition/subtraction of two matrices is possible only if their order is same, and it is performed as:

  $\left[\begin{array}{ccccc}{a}_{11}& a_{12}& .& .& a_{1n}\\ a_{21}& a_{22}& .& .& a_{2n}\\ .& .& .& .& .\\ .& .& .& .& .\\ a_{m1}& a_{m2}& .& .& a_{mn}\end{array}\right]\pm \left[\begin{array}{ccccc}{b}_{11}& b_{12}& .& .& b_{1n}\\ b_{21}& b_{22}& .& .& b_{2n}\\ .& .& .& .& .\\ .& .& .& .& .\\ b_{m1}& b_{m2}& .& .& b_{mn}\end{array}\right]=\left[\begin{array}{ccccc}{a}_{11}\pm b_{11}& a_{12}\pm b_{12}& .& .& a_{1n}\pm b_{1n}\\ a_{21}\pm b_{21}& a_{22}\pm b_{22}& .& .& a_{2n}\pm b_{2n}\\ .& .& .& .& .\\ .& .& .& .& .\\ a_{m1}\pm b_{m1}& a_{m2}\pm b_{m2}& .& .& a_{mn}\pm b_{mn}\end{array}\right]$

Calculation:

In order to find:

  $7A-2B$

Where, $A=\left[\begin{array}{cc}0& -7\\ 8& 12\end{array}\right],\text{ and }B=\left[\begin{array}{cc}11& 4\\ -3& -17\end{array}\right]$ .

First, rewrite it as:

  $7A-2B=7\left[\begin{array}{cc}0& -7\\ 8& 12\end{array}\right]-2\left[\begin{array}{cc}11& 4\\ -3& -17\end{array}\right]$

Now, to evaluate above first perform the multiplication of scalar with each matrices, as

  $\begin{array}{c}7\left[\begin{array}{cc}0& -7\\ 8& 12\end{array}\right]-2\left[\begin{array}{cc}11& 4\\ -3& -17\end{array}\right]=\left[\begin{array}{cc}7\left(0\right)& 7\left(-7\right)\\ 7\left(8\right)& 7\left(12\right)\end{array}\right]-\left[\begin{array}{cc}2\left(11\right)& 2\left(4\right)\\ 2\left(-3\right)& 2\left(-17\right)\end{array}\right]\\ =\left[\begin{array}{cc}0& -49\\ 56& 84\end{array}\right]-\left[\begin{array}{cc}22& 8\\ -6& -34\end{array}\right]\end{array}$

Now, to perform the above matrix subtraction note that order of both matrices is same, equals to $\left(2\times 2\right)$ , thus above matrix subtraction is possible, and is performed by subtracting corresponding elements of matrices, as:

  $\begin{array}{c}\left[\begin{array}{cc}0& -49\\ 56& 84\end{array}\right]-\left[\begin{array}{cc}22& 8\\ -6& -34\end{array}\right]=\left[\begin{array}{cc}0-22& -49-8\\ 56-\left(-6\right)& 84-\left(-34\right)\end{array}\right]\\ =\left[\begin{array}{cc}-22& -57\\ 56+6& 84+34\end{array}\right]\\ =\left[\begin{array}{cc}-22& -57\\ 62& 118\end{array}\right]\end{array}$

Thus, for the given matrices A and B ,

  $7A-2B=\left[\begin{array}{cc}-22& -57\\ 62& 118\end{array}\right]$