Chapter 8.3, Problem 62E

To determine

The amount invested in each type of bond by using the inverse of the coefficient matrix.

Answer to Problem 62E

  $x_1=4000,x_2=2000,$ and $x_3=4000$ .

Explanation of Solution

Given information:

The average yields are $6.5\%$ on AAA bonds, $7\%$ on A bonds, and $9\%$ on B bonds.

  $\{\begin{array}{l}x+y+z=(\text{total investment})\\ 0.065x+0.07y+0.09z=(\text{annual return})\\ 2y-z=0\end{array}$

Use the inverse of the coefficient matrix of this system to find the amount invested in each type of bond.

  $\begin{array}{l}\text{Total investment Annual Return}\\ \$10,000\text{ \$760}\end{array}$

Calculation:

The given system of equation is:

  $\{\begin{array}{l}x+y+z=(\text{total investment})\\ 0.065x+0.07y+0.09z=(\text{annual return})\\ 2y-z=0\end{array}$

Convert the augmented matrix into the row-echelon form by gauss elimination method:

Firstly find the pivot in the first column in the first row:

  $\left[\begin{array}{ccc}1& 1& 1\\ 0.065& 0.07& 0.09\\ 0& 2& -1\end{array}|\begin{array}{c}10,000\\ 760\\ 0\end{array}\right]$

Multiply the first row by $0.065$ and after that subtract the first row from the second row and restore it and make the pivot in the second column by dividing the second row by $0.005$ :

  $\left[\begin{array}{ccc}1& 1& 1\\ 0& 1& 5\\ 0& 2& -1\end{array}|\begin{array}{c}10,000\\ 22,000\\ 0\end{array}\right]$

Subtract the second row from the first row and multiply the second row by $2$ :

  $\left[\begin{array}{ccc}1& 0& -4\\ 0& 2& 10\\ 0& 2& -1\end{array}|\begin{array}{c}2\\ 44000\\ 0\end{array}\right]$

Subtract the second row from the third row and restore it and make the pivot in the third column by dividing the third row by $-11$ .

  $\left[\begin{array}{ccc}1& 0& -4\\ 0& 1& 5\\ 0& 0& 1\end{array}|\begin{array}{c}-12000\\ 22000\\ 4000\end{array}\right]$

Multiply the third row by $-4$ and after subtract the third row from the first row and restore it:

  $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 5\\ 0& 0& 1\end{array}|\begin{array}{c}4000\\ 22000\\ 4000\end{array}\right]$

Multiply the third row by $5$ and subtract the third row from the second row and restore it:

  $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}|\begin{array}{c}4000\\ 2000\\ 4000\end{array}\right]$

Therefore, the solutions are $x_1=4000,x_2=2000,$ and $x_3=4000$ .