SOLUTION: A security firm borrowed $900,000 for business expansion. Some of the money was borrowed at 5%, some at 6%,and some at 7% simple annual interest.The total annual interest is $52,00

Question 918070: A security firm borrowed $900,000 for business expansion. Some of the money was borrowed at 5%, some at 6%,and some at 7% simple annual interest.The total annual interest is $52,000 and the amount borrowed at 6% is one and a half of that borrowed at 7%.
a. Create a system of equations which captures the essence of the information given.
b. Write the system is matrix form?
c. Using the inverse method,find the amount borrowed at each rate.
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
x+y+z=900000,
0.06*x+0.05*y+0.07*z=52000,
x=1.5z
1,1,1,900000
.06,.05,.07,52000
1,0,-1.5,0

900000 52000 0
1 1 1
0.06 0.05 0.07
1 0 -1.5
1st=0.05 0.07 0 -1.5
2nd=0.06 0.07 1 -1.5
3rd=0.06 0.05 1 0
4th=1 1 0 -1.5
5th=1 1 1 -1.5
6th=1 1 1 0
7th=1 1 0.05 0.07
8th=1 1 0.06 0.07
9th=1 1 0.06 0.05
all minors
1=0.05 0.07 0 -1.5
2=0.06 0.07 1 -1.5
3=0.06 0.05 1 0
4=1 1 0 -1.5
5=1 1 1 -1.5
6=1 1 1 0
7=1 1 0.05 0.07
8=1 1 0.06 0.07
9=1 1 0.06 0.05
mult ad
ad-bc 1*4-2*3
1--
0.05 0.07 0 -1.5
(0.05*-1.5)-(0.07*0)
-0.075-(0.0) =-0.075
2--
0.06 0.07 1 -1.5
(0.06*-1.5)-(0.07*1)
-0.09-(0.07) =-0.16
3--
0.06 0.05 1 0
(0.06*0)-(0.05*1)
0.0-(0.05) =-0.05
4--
1 1 0 -1.5
(1*-1.5)-(1*0)
-1.5-(0) =-1.5
5--
1 1 1 -1.5
(1*-1.5)-(1*1)
-1.5-(1) =-2.5
6--
1 1 1 0
(1*0)-(1*1)
0-(1) =-1
7--
1 1 0.05 0.07
(1*0.07)-(1*0.05)
0.07-(0.05) =0.02
8--
1 1 0.06 0.07
(1*0.07)-(1*0.06)
0.07-(0.06) =0.01
9--
1 1 0.06 0.05
(1*0.05)-(1*0.06)
0.05-(0.06) =-0.01
matrix of minors
-0.075 -0.16 -0.05
-1.5 -2.5 -1
0.02 0.01 -0.01
original
1 1 1
0.06 0.05 0.07
1 0 -1.5
matrix of cofactors-switching signs
-0.075 same -0.075
-0.16 switch 0.16
-0.05 same -0.05
-1.5 switch 1.5
-2.5 same -2.5
-1 switch 1
0.02 same 0.02
0.01 switch -0.01
-0.01 same -0.01
matrix of cofactors
-0.075 0.16 -0.05
1.5 -2.5 1
0.02 -0.01 -0.01
adjoint transpose
-0.075 1.5 0.02
0.16 -2.5 -0.01
-0.05 1 -0.01
original by row
(row 1)1 1 1
(row 2)0.06 0.05 0.07
(row 3)1 0 -1.5
original by column
(column 1)1 0.06 1
(column 2)1 0.05 0
(column 3)1 0.07 -1.5
1) determinant using row three
1*0.02=0.02
0*-0.01=0
-1.5*-0.01=0.015
determinant=0.035
and finally the inverse
-0.075/0.035 1.5/0.035 0.02/0.035
0.16/0.035 -2.5/0.035 -0.01/0.035
-0.05/0.035 1/0.035 -0.01/0.035
2) same determinant using column 2
1*0.16=0.16
0.05*-2.5=-0.125
0*-0.01=0
determinant=0.035
the inverse in fractions
-0.075/0.035 1.5/0.035 0.02/0.035
0.16/0.035 -2.5/0.035 -0.01/0.035
-0.05/0.035 1/0.035 -0.01/0.035
the inverse in decimals
-2.14285714 42.8571429 0.57142857
4.57142857 -71.4285714 -0.28571429
-1.42857143 28.5714286 -0.28571429
900000 52000 0
column one
-2.14285714 42.8571429 0.57142857
-1928571.43 2228571.43 0.0
solutions
x=300000.0 y= 400000 z= 200000
x = $300000 at 6%
y = $400000 at 5%
z = $200000 at 7%
Here is the gauss jordan method
1,1,1,900000
.06,.05,.07,52000
1,0,-1.5,0

add down (-0.06) *row 1 to row 2
1,1,1,900000
0,-0.01/1,0.01/1,-2000
1,0,-1.5/1,0
add down (-1) *row 1 to row 3
1,1,1,900000
0,-0.01/1,0.01/1,-2000
0,-1,-2.5/1,-900000
divide row 2 by -0.01
1,1,1,900000
0,1,0.01/-0.01,-2000/-0.01
0,-1,-2.5/1,-900000
add down (1) *row 2 to row 3
1,1,1,900000
0,1,0.01/-0.01,2000/0.01
0,0,0.035/-0.01,-700000
divide row 3 by 0.035/-0.01
1,1,1,900000
0,1,0.01/-0.01,2000/0.01
0,0,1,7000.0/0.035
We now have the value for the last variable.
We will work our way up and get the other solutions.
add up (-0.01/-0.01) *row 3 to row 2
1,1,1,900000
0,1,0,-1.4/-0.0000035
0,0,1,7000.0/0.035
add up (-1) *row 3 to row 1
1,1,0,700000
0,1,0,1.4/0.0000035
0,0,1,7000.0/0.035
add up (-1) *row 2 to row 1
1,0,0,1.05/0.0000035
0,1,0,1.4/0.0000035
0,0,1,7000.0/0.035
final
1,0,0,1.05/0.0000035
0,1,0,1.4/0.0000035
0,0,1,7000.0/0.035
1,0,0,1.05/0.0000035 = 300000.0
0,1,0,1.4/0.0000035 = 400000.0
0,0,1,7000.0/0.035 = 200000.0
"1.05/0.0000035","1.4/0.0000035","7000.0/0.035"
(1.05/0.0000035,1.4/0.0000035,7000.0/0.035)
(300000.0,400000.0,200000.0)

RELATED QUESTIONS
a small corporation borrowed $500,000 to expand its product line. Some of the money was... (answered by josmiceli)

A natural history museum borrows $2,000,000 at simple annual interest to purchase new... (answered by CPhill)

A small corporation borrowed $900,000 to expand its line of toys. Some of the money was... (answered by stanbon)

A security firm borrowed $900,000 for business... (answered by richwmiller)

A city zoo borrowed $2,000,000 at simple annual interest to construct a breeding... (answered by addingup)

A corporation borrowed $760,000 to expand its line of clothing. Some of the money was... (answered by richwmiller)

A SMALL CORPORATION BORROWED $800,000 TO EXPAND ITS LINE OF TOYS. SOME OF THE MONEY WAS... (answered by Fombitz)

I have been "racking" my brain and wasting a lot of scratch paper over this one. I... (answered by stanbon)

Jim borrowed some money at 13% interest and some money at 16% interest. The amount at 16% (answered by MathTherapy)