Chapter 10.4, Problem 20E

To determine

To calculate:

The solution of given matrix equation:

  $2A=B-3X$

Answer to Problem 20E

There is matrix solution of $X$ is $\left[\begin{array}{l}\begin{array}{cc}-2& -\frac{7}{3}\end{array}\\ \begin{array}{cc}\frac{1}{3}& \frac{1}{3}\end{array}\end{array}\right]$.

Explanation of Solution

Given information:

  $\begin{array}{l}A=\left[\begin{array}{c}4\\ 1\end{array}\begin{array}{c}6\\ 3\end{array}\right]\\ B=\left[\begin{array}{c}2\\ 3\end{array}\begin{array}{c}5\\ 7\end{array}\right]\\ C=\left[\begin{array}{l}\begin{array}{cc}2& 3\end{array}\\ \begin{array}{cc}1& 0\end{array}\\ \begin{array}{cc}0& 2\end{array}\end{array}\right]\\ D=\left[\begin{array}{l}\begin{array}{cc}10& 20\end{array}\\ \begin{array}{cc}30& 20\end{array}\\ \begin{array}{cc}10& 0\end{array}\end{array}\right]\end{array}$

Calculation:

Know that:

Properties of Matrices:

  $A+B=0$

Consider the following matrices:

  $\begin{array}{l}A=\left[\begin{array}{c}4\\ 1\end{array}\begin{array}{c}6\\ 3\end{array}\right]\\ B=\left[\begin{array}{c}2\\ 3\end{array}\begin{array}{c}5\\ 7\end{array}\right]\\ C=\left[\begin{array}{l}\begin{array}{cc}2& 3\end{array}\\ \begin{array}{cc}1& 0\end{array}\\ \begin{array}{cc}0& 2\end{array}\end{array}\right]\\ D=\left[\begin{array}{l}\begin{array}{cc}10& 20\end{array}\\ \begin{array}{cc}30& 20\end{array}\\ \begin{array}{cc}10& 0\end{array}\end{array}\right]\end{array}$

Here determines the unknown matrix $X$ from the equation.

  $2A=B-3X$

Here the dimensions of the matrix $B$ is $\left(2\times 2\right)$ then the matrix $X$ will also be a $\left(2\times 2\right)$ . The reason is two matrices cansubtract only when they have the same dimensions.

Now use the properties of matrices to solve for $X$ ,

  $2A=B-3X$ .

Now subtract the matrix $B$ from both sides, $2A-B=-3X$

After multiply by the scalar $\frac{1}{3}$ on both sides, the equation is $-\frac{1}{3}\left(2A-B\right)=-\frac{1}{3}\left(-3X\right)$ .

So, the final equation is $X=-\frac{1}{3}\left(2A-B\right)$

Now Substitute the matrices $A$ and $B$ in $X=-\frac{1}{3}\left(2A-B\right)$ , then

  $\begin{array}{l}X=-\frac{1}{3}\left(2A-B\right)\\ X=-\frac{1}{3}\left(2\left[\begin{array}{l}\begin{array}{cc}4& 6\end{array}\\ \begin{array}{cc}1& 3\end{array}\end{array}\right]-\left[\begin{array}{l}\begin{array}{cc}2& 5\end{array}\\ \begin{array}{cc}3& 7\end{array}\end{array}\right]\right)\end{array}$

Now multiply by the scalar $2$ ,

  $\begin{array}{l}X=-\frac{1}{3}\left(\left[\begin{array}{l}\begin{array}{cc}2\times 4& 2\times 6\end{array}\\ \begin{array}{cc}2\times 1& 2\times 3\end{array}\end{array}\right]-\left[\begin{array}{l}\begin{array}{cc}2& 5\end{array}\\ \begin{array}{cc}3& 7\end{array}\end{array}\right]\right)\\ X=-\frac{1}{3}\left(\left[\begin{array}{l}\begin{array}{cc}8& 12\end{array}\\ \begin{array}{cc}2& 6\end{array}\end{array}\right]-\left[\begin{array}{l}\begin{array}{cc}2& 5\end{array}\\ \begin{array}{cc}3& 7\end{array}\end{array}\right]\right)\end{array}$

Now simplify matrices,

First subtract the matrices

  $X=-\frac{1}{3}\left(\left[\begin{array}{l}\begin{array}{cc}6& 7\end{array}\\ \begin{array}{cc}-1& -1\end{array}\end{array}\right]\right)$

Now multiply by the scalar $\frac{1}{3}$

  $\begin{array}{l}=\left[\begin{array}{l}\begin{array}{cc}-\frac{6}{3}& -\frac{7}{3}\end{array}\\ \begin{array}{cc}-\frac{\left(-1\right)}{3}& -\frac{\left(-1\right)}{3}\end{array}\end{array}\right]\\ =\left[\begin{array}{l}\begin{array}{cc}-2& -\frac{7}{3}\end{array}\\ \begin{array}{cc}\frac{1}{3}& \frac{1}{3}\end{array}\end{array}\right]\end{array}$