# Solving equations - Matrix Help!

• Jan 10th 2010, 10:04 AM
Jiyongie
Solving equations - Matrix Help!
I just started learning about matrices in class, so this question is supposed to be fairly simple.. but I've never been any good at math, so I'm having a bit of trouble with it.

Mark had \$24,500 to invest. He divided the money into three different accounts. At the end of the year, he had made \$1,300 in interest. The annual yield on each of the three accounts was 4%, 5.5%, and 6%. If the amount of money in the 4% account was four times the amount of money in the 5.5% account, how much had he placed in each account?

I started with -

0.06 0.055 0.04 1300
1 1 1 24500
0 422 -104 0

then I stopped at
0 1 4 34000
1 0 -4 -9500
0 0 792 7174000

I'm stuck!!
• Jan 10th 2010, 10:32 AM
HallsofIvy
Quote:

Originally Posted by Jiyongie
I just started learning about matrices in class, so this question is supposed to be fairly simple.. but I've never been any good at math, so I'm having a bit of trouble with it.

Mark had \$24,500 to invest. He divided the money into three different accounts. At the end of the year, he had made \$1,300 in interest. The annual yield on each of the three accounts was 4%, 5.5%, and 6%. If the amount of money in the 4% account was four times the amount of money in the 5.5% account, how much had he placed in each account?

I started with -

0.06 0.055 0.04 1300
1 1 1 24500
0 422 -104 0

then I stopped at
0 1 4 34000
1 0 -4 -9500
0 0 792 7174000

I'm stuck!!

It would make more sense if you would explain what you are doing and what equations/matrices you are using rather than just giving a list of numbers!

Since he had three accounts, let the amount of money he invested in each account be A, B, and C. He had \$24,500 so A+ B+ C= 24500. He made 4%, 5.5%, and 6%, receiving \$1300 in interest so .04A+ .055B+ .06C= 1300. Finally, " If the amount of money in the 4% account was four times the amount of money in the 5.5% account" so A= 4B. Your three equations for A, B, and C are A+ B+ C= 24500, .04A+ .055B+ .06C= 1300 and A- 4B= 0.

But what you have done is find and you are almost finished.
You can simplify by dividing that third row by 792 to get
0 0 1 9058.08

Add four times that new third row to the second row to get rid of the "-4":
1 0 (-4)+ 4(1) -9500+ 4(9058.08)
1 0 0 26732.32

And subtract four times that new third row to the first row to get rid of the "4":
0 1 4- 4(1) 34000- 4(9058.08)
0 1 0 -5658.08
That comes up negative but that is what I get too. It might not be possible to meet the conditions of the problem.