Chapter 7.6, Problem 57E

To determine

To solve: The given system of linear equations using inverse of the coefficient matrix.

Answer to Problem 57E

The person invests $\$4000$ in AAA bonds, $\$2000$ in A bonds and $\$4000$ in B bonds.

Explanation of Solution

Given information :

The system of linear equations:

  $\begin{array}{l}x+y+z=(\text{total investment)}\\ 0.045x+0.05y+0.07z=(\text{annual return)}\\ 2y-z=\text{ 0}\end{array}$

Total investment $=\$10,000$

Annual return $=\$560$

Formula used :

To solve a system of linear equations:

If $A$ is an invertible matrix, the system of linear equations represented by $AX=B$ has a unique solution given by $X=A^{-1}B$ .

Calculation :

As per the problem,

Put the given values of total investment and annual return in the given system of linear equations:

  $\begin{array}{l}x+y+z=10,000\\ 0.045x+0.05y+0.07z=560\\ 2y-z=\text{ 0}\end{array}$

Write the system in the matrix form $AX=B$

  $\left[\begin{array}{ccc}1& 1& 1\\ 0.045& 0.05& 0.07\\ 0& 2& -1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}10,000\\ 560\\ 0\end{array}\right]$

Where $A=\left[\begin{array}{ccc}1& 1& 1\\ 0.045& 0.05& 0.07\\ 0& 2& -1\end{array}\right]$ , $X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ and $B=\left[\begin{array}{c}10,000\\ 560\\ 0\end{array}\right]$

Then, use Gauss-Jordan elimination to find $A^{-1}$ .

  $A^{-1}=\left[\begin{array}{ccc}3.\overline{45}& -54.\overline{54}& -0.\overline{36}\\ -0.\overline{81}& 18.\overline{18}& 0.\overline{45}\\ -1.\overline{63}& 36.\overline{36}& -0.\overline{09}\end{array}\right]$

Finally, multiply $B$ by $A^{-1}$ on the left to obtain the solution:

  $\begin{array}{l}X=A^{-1}B\\ =\left[\begin{array}{ccc}3.\overline{45}& -54.\overline{54}& -0.\overline{36}\\ -0.\overline{81}& 18.\overline{18}& 0.\overline{45}\\ -1.\overline{63}& 36.\overline{36}& -0.\overline{09}\end{array}\right]\left[\begin{array}{c}10,000\\ 560\\ 0\end{array}\right]=\left[\begin{array}{c}4000\\ 2000\\ 4000\end{array}\right]\end{array}$

The solution of the system is $x=4000$ , $y=2000$ and $z=4000$ .So, the person invests $\$4000$ in AAA bonds, $\$2000$ in A bonds and $\$4000$ in B bonds.