Chapter 11.4, Problem 14AYU

In Problems 9-26, use the following matrices. Determine whether the given expression is defined. If it is defined, express the result as a single matrix; if it is not, say "not defined.�

$A=\left[\begin{array}{l}\begin{array}{ccc}0& 3& -5\end{array}\\ \begin{array}{ccc}1& 2& 6\end{array}\end{array}\right]$ $B=\left[\begin{array}{l}\begin{array}{ccc}4& 1& 0\end{array}\\ \begin{array}{ccc}-2& 3& -2\end{array}\end{array}\right]$ $C=\left[\begin{array}{c}4\\ 6\\ -2\end{array}\begin{array}{c}1\\ 2\\ 3\end{array}\right]$

$2A+4B$

To determine

The matrix is $2A+4B$ defined and if so, evaluate the matrix expression.

Answer to Problem 14AYU

Solution:

$2A+4B$ is defined and $2A+4B=\left[\begin{array}{ccc}16& 10& -10\\ -6& 16& 4\end{array}\right]$.

Explanation of Solution

Given:

$A=\left[\begin{array}{ccc}0& 3& -5\\ 1& 2& 6\end{array}\right]\&\text{B}=\left[\begin{array}{ccc}4& 1& 0\\ -2& 3& -2\end{array}\right]$

Formula used:

$mA+nB=\left[\begin{array}{ccc}m{a}_{11}+ n{b}_{11}& m{a}_{12}+n{b}_{12}& m{a}_{13}+n{b}_{13}\\ m{a}_{21}+n{b}_{21}& m{a}_{22}+n{b}_{22}& m{a}_{23}+n{b}_{23}\end{array}\right]$

Calculation:

If a scalar is multiplied with a matrix, the resultant matrix is defined. Hence $mA$ and $nB$ are defined of size $2\text{ x }3$. For $mA+nB$ to be defined, $mA$ and $mB$ should have same number of rows and columns. Since $mA$ is $2\text{ x }3$ and $nB$ is $2\text{ x }3$ matrix, $mA+nB$ is defined.

$A=\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\end{array}\right]=\left[\begin{array}{ccc}0& 3& -5\\ 1& 2& 6\end{array}\right]$

$B=\left[\begin{array}{ccc}{b}_{11}& b_{12}& b_{13}\\ b_{21}& b_{22}& b_{23}\end{array}\right]=\left[\begin{array}{ccc}4& 1& 0\\ -2& 3& -2\end{array}\right]$

$mA+nB=\left[\begin{array}{ccc}m{a}_{11}+ n{b}_{11}& m{a}_{12}+n{b}_{12}& m{a}_{13}+n{b}_{13}\\ m{a}_{21}+n{b}_{21}& m{a}_{22}+n{b}_{22}& m{a}_{23}+n{b}_{23}\end{array}\right]$

Where $m=2$ and $n=4$.

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

$=\left[\begin{array}{ccc}0+16& 6+4& -10+0\\ 2-8& 4+12& 12-8\end{array}\right]$

$=\left[\begin{array}{ccc}16& 10& -10\\ -6& 16& 4\end{array}\right]$