Chapter 7.4, Problem 95E

To determine

To solve: The system of equations using matrices.

Answer to Problem 95E

The solution is:

  $\begin{array}{l}{A}_1=200,000\\ A_2=500,000\\ A_3=800,000\end{array}$ .

Explanation of Solution

Given information: A system of equations, $\begin{array}{l}{A}_1+A_2+A_3=1,500,000\\ 0.03A_1+0.04A_2+0.06A_3=74,000\\ 4I_1-A_3=0\end{array}$

Calculation:

Rewriting the system of equations,

  $\begin{array}{l}{A}_1+A_2+A_3=1,500,000\\ 0.03A_1+0.04A_2+0.06A_3=74,000\\ 4I_1-A_3=0\end{array}$

Now the augmented matrix can be written as,

  $\left(\begin{array}{cccc}1& 1& 1& 1,500,000\\ 0.03& 0.04& 0.06& 74,000\\ 4& 0& -1& 0\end{array}\right)$

Using row transformations,

  $\begin{array}{l}{R}_2\to 100R_2\\ \left(\begin{array}{cccc}1& 1& 1& 1,500,000\\ 3& 4& 6& 7,400,000\\ 4& 0& -1& 0\end{array}\right)\end{array}$

  $\begin{array}{l}{R}_2\to R_2-3R_1\\ R_3\to R_3-4R_1\\ \left(\begin{array}{cccc}1& 1& 1& 1,500,000\\ 0& 1& 3& 2,900,000\\ 0& -4& -5& -6,000,000\end{array}\right)\end{array}$

  $\begin{array}{l}{R}_1\to R_1-R_2\\ R_3\to R_3+4R_2\\ \left(\begin{array}{cccc}1& 0& -2& -1,400,000\\ 0& 1& 3& 2,900,000\\ 0& 0& 7& 5,600,000\end{array}\right)\end{array}$

  $\begin{array}{l}{R}_3\to \frac{1}{7}{R}_3\\ \left(\begin{array}{cccc}1& 0& -2& -1,400,000\\ 0& 1& 3& 2,900,000\\ 0& 0& 1& 800,000\end{array}\right)\end{array}$

  $\begin{array}{l}{R}_2\to R_2-3R_3\\ R_1\to R_1+2R_3\\ \left(\begin{array}{cccc}1& 0& 0& 200,000\\ 0& 1& 0& 500,000\\ 0& 0& 1& 800,000\end{array}\right)\end{array}$

The matrix is now in reduced row echelon form.

Converting back to the system of linear equations,

  $\begin{array}{l}{A}_1=200,000\\ A_2=500,000\\ A_3=800,000\end{array}$ .