Chapter 8.3, Problem 58E

To determine

To solve the following system of linear equations using inverse matrix

  $\{\begin{array}{l}x+y+z=(\text{Total investment})\\ 0.045+0.05y+0.09z=(\text{Annual return})\\ 2y-z=0\end{array}$

Where, $\begin{array}{l}\text{Total investment}=\$12,000\\ \text{Annual return}=\$835\end{array}$

Answer to Problem 58E

  $x=\frac{51000}{19},y=\frac{59000}{19},z=\frac{118000}{19}$

Explanation of Solution

Given:

  $\{\begin{array}{l}x+y+z=(\text{Total investment})\\ 0.045+0.05y+0.09z=(\text{Annual return})\\ 2y-z=0\end{array}$

Also, $\begin{array}{l}\text{Total investment}=\$12,000\\ \text{Annual return}=\$835\end{array}$

Concept used:

To find inverse of a square matrix A, following steps should be followed.

1. Augment given matrix with the identity matrix
2. Perform row operations to make the identity matrix to the left and the right matrix becomes inverse matrix.

In this way, inverse matrix of a matrix can be obtained.

Formula used:

  $AX=B\iff X=A^{-1}B$

Calculation:

Consider given system of equations,

  $\{\begin{array}{l}x+y+z=(\text{Total investment})\\ 0.045+0.05y+0.09z=(\text{Annual return})\\ 2y-z=0\end{array}$

But,

  $\begin{array}{l}\text{Total investment}=\$12,000\\ \text{Annual return}=\$835\end{array}$

  $\Rightarrow \{\begin{array}{l}x+y+z=12000\\ 0.045+0.05y+0.09z=835\\ 2y-z=0\end{array}$

Writing above system of linear equations in matrix form $(AX=B)$ ,

  $\left[\begin{array}{ccc}1& 1& 1\\ 0.045& 0.05& 0.09\\ 0& 2& -1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}12000\\ 835\\ 0\end{array}\right]$

Here,

  $\begin{array}{l}A=\left[\begin{array}{ccc}1& 1& 1\\ 0.045& 0.05& 0.09\\ 0& 2& -1\end{array}\right]\\ X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\\ B=\left[\begin{array}{c}10000\\ 650\\ 0\end{array}\right]\end{array}$

Now, find the inverse matrix of matrix A.

Step 1: Augmenting matrix A with the identity matrix

  $[\begin{array}{ccc}1& 1& 1\\ 0.045& 0.05& 0.09\\ 0& 2& -1\end{array}\begin{array}{c}\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\end{array}]$

Step 2: Performing row operations to make the identity matrix to the left.

So, inverse matrix is,

  $A^{-1}=\left[\begin{array}{ccc}\frac{46}{19}& -\frac{600}{19}& -\frac{8}{19}\\ -\frac{9}{19}& \frac{200}{19}& \frac{9}{19}\\ -\frac{18}{19}& \frac{400}{19}& -\frac{1}{19}\end{array}\right]$

Now, applying the formula $AX=B\iff X=A^{-1}B$ ,$\left[\begin{array}{ccc}1& 1& 1\\ 0.045& 0.05& 0.09\\ 0& 2& -1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}12000\\ 835\\ 0\end{array}\right]\iff \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{ccc}\frac{46}{19}& -\frac{600}{19}& -\frac{8}{19}\\ -\frac{9}{19}& \frac{200}{19}& \frac{9}{19}\\ -\frac{18}{19}& \frac{400}{19}& -\frac{1}{19}\end{array}\right]\left[\begin{array}{c}12000\\ 835\\ 0\end{array}\right]$

Finding the solutions,

  $\begin{array}{l}\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{ccc}\frac{46}{19}& -\frac{600}{19}& -\frac{8}{19}\\ -\frac{9}{19}& \frac{200}{19}& \frac{9}{19}\\ -\frac{18}{19}& \frac{400}{19}& -\frac{1}{19}\end{array}\right]\left[\begin{array}{c}12000\\ 835\\ 0\end{array}\right]\\ \Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}\frac{46}{19}\times 12000-\frac{600}{19}\times 835-\frac{8}{19}\times 0\\ -\frac{9}{19}\times 12000+\frac{200}{19}\times 835+\frac{9}{19}\times 0\\ -\frac{18}{19}\times 12000+\frac{400}{19}\times 835-\frac{1}{19}\times 0\end{array}\right]\\ \Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}\frac{552000}{19}-\frac{501000}{19}+0\\ -\frac{108000}{19}+\frac{167000}{19}+0\\ -\frac{216000}{14}+\frac{334000}{19}-0\end{array}\right]\\ \Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}\frac{51000}{19}\\ \frac{59000}{19}\\ \frac{118000}{19}\end{array}\right]\end{array}$

Conclusion:

Therefore, the solution to given system of equations is $x=\frac{51000}{19},y=\frac{59000}{19},z=\frac{118000}{19}.$