
Jason T. answered 04/21/20
MS in Chemical Engineering with 5 years of teaching experience
Good question!
So we will denote P for principal and use 9, 10, and 13 as subscripts on P denoting amounts borrowed at that interest rate.
So
P = $2,000,000 = P9 + P10 + P13
Given from problem statement: P9 = 2 * P13
Simple interest: I = Prt
Since we are just given an annual interest, t = 1 year
I9 = 0.09*P9
I10 = 0.1*P10
I13 = 0.13*P13
Given from problem statement: I9 + I10 + I13 = $204,500
I will solve here using substitution and will post matrices in a comment below if the formatting allows.
First, substitute interest formulas into sum of interests.
0.09*P9 + 0.1*P10 + 0.13*P13 = $204,500
Then substitute given, P9 = 2 * P13
0.09*2*P13 + 0.1*P10 + 0.13*P13 = $204,500
0.18*P13 + 0.1*P10 + 0.13*P13 = $204,500
0.1*P10 + 0.31*P13 = $204,500
From earlier, P = $2,000,000 = P9 + P10 + P13
Given from problem statement: P9 = 2 * P13
So, P10 + 3*P13 = $2,000,000
P10 = $2,000,000 - 3*P13
Substitute this into our last equation, 0.1*P10 + 0.31*P13 = $204,500
0.1*($2,000,000 - 3*P13) + 0.31*P13 = $204,500
$200,000 - 0.3*P13 + 0.31*P13 = $204,500
0.01*P13 = $4,500
P13 = $450,000
Now, substitute into earlier equations.
P10 = $2,000,000 - 3*P13 = $2,000,000 - 3*$450,000 = $2,000,000 - $1,350,000 = $650,000 = P10
P9 = 2 * P13 = 2*$450,000 = $900,000 = P9
Now if you add all the principals, it adds up to the full principle of $2 million. Yay, we solved it!
Again, I'll look into how easy it is to do matrix notation in this format in a comment below. But I believe the substitution will still be useful for most.
Update: I don't see any way here to show matrix notation, please follow an online guide or book a session with myself or someone else who is good with matrices.