Chapter 8.1, Problem 94E

To determine

To find:solution to the system of linear equation using matrices

Answer to Problem 94E

Amount borrowed at 7% is $\boxed{\$53125}$

Amount borrowed by 8.5% is $\boxed{\$1557500}$

Amount borrowed by 9.5% is $\boxed{\$389375}$

Explanation of Solution

Given information:(i) A natural history museum borrows $2,000,000 at simple annual interest

(ii) Some of the money is borrowed at 7%, some at 8.5% and some at 9.5%

(iii) Total annual interest is $169,750

(iii) The amount borrowed at 8.5% is four times the amount borrowed at 9.5%.

Concept Involved:

Assign variable for the unknown that we need to find, then based on the given information, we are expected to write three equations.

A matrix derived from a system of linear equations (each written in standard form with the constant term on the right) is the augmented matrix of the system.

Elementary Row Operation:

The three operations that can be used on a system of linear equations to produce an equivalent system.

Operation	Notation
1.Interchange two equations	$R_a\leftrightarrow R_b$
2. Multiply an equation by a nonzero constant	$c{R}_a\left(c\ne 0\right)$
3. Add a multiple of an equation to another equation.	$c{R}_a+R_b$

In matrix terminology, these three operations correspond to elementary row operations.

An elementary row operation on an augmented matrix of a given system of linear equations produces a new augmented matrix corresponding to a new (but equivalent) system of linear equations.

Two matrices are row-equivalent when one can be obtained from the other by a sequence of elementary row operations.

Row-Echelon Form and Reduced Row-Echelon Form:

A matrix in row-echelon form has the following properties.

1. Any rows consisting entirely of zeros occur at the bottom of the matrix.

2. For each row that does not consist entirely of zeros, the first nonzero entryis 1 (called a leading 1).

3. For two successive (nonzero) rows, the leading 1 in the higher row is fartherto the left than the leading 1 in the lower row.

A matrix in row-echelon form is in reduced row-echelon form when every columnthat has a leading 1 has zeros in every position above and below its leading 1.

Calculation:

Assign variable for the unknown that we need to find

Let x be amount borrowed at 7%, let y be amount borrowed by 8.5%, and let z be amount borrowed by 9.5%.

Write first equation based on total amount borrowed

(i) A natural history museum borrows $2,000,000 at simple annual interest

  $x+y+z=2000000$(1^st equation

Write second equation based on total interest amount accrued

(ii) Some of the money is borrowed at 7%, some at 8.5% and some at 9.5%

(iii) Total annual interest is $169,750

  $0.07x+0.085y+0.095z=169750$ (2^nd equation

Write third equation based on relationship between amount borrowed at different rates

(iii) The amount borrowed at 8.5% is four times the amount borrowed at 9.5%.

  $y=4z$(3^rd equation

Multiply 1000 to the 2^nd equation to get rid of decimal

  $7x+85y+95z=169750000$

Get the variables in the 3^rd equation in one side of the equation by subtracting 4z on both sides

  $y-4z=0$

Write the system of equation $\{\begin{array}{l}x+y+z=2000000\\ 7x+85y+95z=169750000\\ y-4z=0\end{array}$ in augmented form

  $\left[\begin{array}{ccccc}1& 1& 1& ⋮& 2000000\\ 7& 85& 95& ⋮& 169750000\\ 0& 1& -4& ⋮& 0\end{array}\right]$

Add -7 times the 1st row to the 2nd row

  $\begin{array}{c}\\ -7R_1+R_2\to \\ \end{array}\left[\begin{array}{ccccc}1& 1& 1& ⋮& 2000000\\ 0& 78& 88& ⋮& 155750000\\ 0& 1& -4& ⋮& 0\end{array}\right]$

Multiply the 2nd row by 1/78

  $\begin{array}{c}\\ 1/78R_2\to \\ \end{array}\left[\begin{array}{ccccc}1& 1& 1& ⋮& 2000000\\ 0& 1& 44/39& ⋮& 77875000/39\\ 0& 1& -4& ⋮& 0\end{array}\right]$

Add -1 times the 2nd row to the 3rd row

  $\begin{array}{c}\\ \\ -1R_2+R_3\to \end{array}\left[\begin{array}{ccccc}1& 1& 1& ⋮& 2000000\\ 0& 1& 44/39& ⋮& 77875000/39\\ 0& 0& -200/39& ⋮& -77875000/39\end{array}\right]$

Multiply the 3rd row by -39/200

  $\begin{array}{c}\\ \\ -39/200R_3\to \end{array}\left[\begin{array}{ccccc}1& 1& 1& ⋮& 2000000\\ 0& 1& 44/39& ⋮& 77875000/39\\ 0& 0& 1& ⋮& 389375\end{array}\right]$

Add -44/39 times the 3rd row to the 2nd row

  $\begin{array}{c}\\ -44/39R_3+R_2\to \\ \end{array}\left[\begin{array}{ccccc}1& 1& 1& ⋮& 2000000\\ 0& 1& 0& ⋮& 1557500\\ 0& 0& 1& ⋮& 389375\end{array}\right]$

Add -1 times the 3rd row to the 1st row

  $\begin{array}{c}-R_3+R_1\to \\ \\ \end{array}\left[\begin{array}{ccccc}1& 1& 0& ⋮& 1610625\\ 0& 1& 0& ⋮& 1557500\\ 0& 0& 1& ⋮& 389375\end{array}\right]$

Add -1 times the 2nd row to the 1st row

  $\begin{array}{c}-R_2+R_1\to \\ \\ \end{array}\left[\begin{array}{ccccc}1& 0& 0& ⋮& 53125\\ 0& 1& 0& ⋮& 1557500\\ 0& 0& 1& ⋮& 389375\end{array}\right]$

The matrix is now in reduced row-echelon form. Converting back to a system of linear

equations, you have

  $\{\begin{array}{l}x=53125\\ y=1557500\\ z=389375\end{array}$

Conclusion:

Amount borrowed at 7% is $\$53125$

Amount borrowed by 8.5% is $\$1557500$

Amount borrowed by 9.5% is $\$389375$