Linear Algebra Problem Solving

Question

The owner of J3K invested an amount of P30 000.00 separately. The first at 6%, the second at 7% and the third at 9%. In one year, the profit was P2210.00. However, if the first amount was invested at 9%, the second at 7% and the third at 6%, the profit would only be P2180. Find the three amounts invested.

Answer 1

Let x, y and z be the three amounts invested. We have 3 unknowns, so we need 3 equations.

x + y + z = 30000 Eq1

.06x + .07y + .09z = 2210 Eq2

.09x + .07y + .06z = 2180 Eq3

Multiply the last two equations by 100 and put into a matrix

[1 1 1 30000]

[6 7 9 221000]

[9 7 6 218000]

Row2 = Row2 - 6Row1

Row3 = Row3 - 9Row1

[1 1 1 30000]

[0 1 3 41000]

[0 -2 -3 -52000]

Row3 = Row3+2Row2

[1 1 1 30000]

[0 1 3 41000]

[0 0 3 30000]

Divide Row 3 by 3

[1 1 1 30000]

[0 1 3 41000]

[0 0 1 10000]

Back solve

z = 10000

y + 30000 = 41000 --> y = 11000

x + 11000 + 10000 = 30000 --> x = 30000-21000 = 9000

The three amounts are P9000.00, P11000.00 and P10000.00