# Math Help - Another matrix problem

1. ## Another matrix problem

Hi I'm stuck in this problem again.

Using Gaussian elimination with back-substition or Gauss-Jordan Elimination solve the following problem.

A small corporation borrowed $500,000 to expand its product line. Some of the money was borrowed at 9%, some at 10%, and some at 12%. How much was borrowed at each rate if the annual interest was$52,000 and the amount borrowed at 10% was 2.5 times the amount borrowed at 9%?

2. Hello, cuteisa89!

I'll set up for you . . .

Using Gaussian elimination, solve the following problem.

A small corporation borrowed $500,000 to expand its product line. Some of the money was borrowed at 9%, some at 10%, and some at 12%. How much was borrowed at each rate if the annual interest was$52,000
and the amount borrowed at 10% was 2.5 times the amount borrowed at 9%?

Let: . $\begin{array}{c}x = \text{amount borrowed at 9\%} \\ y = \text{amount borrowed at 10\%} \\ z = \text{amount borrowed at 12\%}\end{array}$

The total borrowed is $500,000: . $x + y + z \:=\:500,000$ .(a) The interest on $x$ dollars at 9% is: $0.09x$ dollars. The interest on $y$ dollars at 10% is: $0.10y$ dollars. The interest on $z$ dollars at 12% is: $0.12z$ dollars. . . The total interest was$52,000: . $0.09x + 0.10y + 0.12z \:=\:52,000$
Multiply by 100: . $9x + 10y + 12z \:=\:5,200,000$ .(b)

The amount borrowed at 10% was 2.5 times the amount borrowed at 9%:
. . $y \:=\:2.5x\quad\Rightarrow\quad 2.5x - y \:=\:0$
Multiply by 2: . $5x - 2y \:=\:0$ .(c)

And we have the sytem: . $\begin{vmatrix}1 & 1 & 1 & | & 50,000 \\ 9 & 10 & 12 & | & 5,200,000 \\ 5 & \text{-}2 & 0 & | & 0\end{vmatrix}$

Now solve it . . .

3. So would it be:
x= 100,000
y = 250,000
z = 150,000

Is this correct?
By the way, thank you so much for helping me! Really appreciated it!