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December 6th, 2017, 12:27 PM   #1
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Inverse Matrix Problem Help

Bond Investment:
An investment company recommends that a client invest in AAA, AA, and A rated bonds. The average annual yield on AAA bonds is 6%, on AA bonds 7%, and on A bonds 10%. The client tells the company she wants to invest twice as much in AAA bonds as in A bonds. How much should be invested in each if the total investment is \$50,000 and the investor wants an annual income (that is, earned interest) of \$3,620? You have to use the inverse matrix to find your solution.

So this is what I have so far and I just wanted verification that my inverse matrix is correct
My equations are 3x+y=50,000 and 2x(0.6)+y(0.7)+x(0.10)=3620

My inverse matrix is {3 1 1 0 }
{.22 .07 0 1} Is this correct?

Last edited by skipjack; December 6th, 2017 at 01:04 PM.
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December 6th, 2017, 01:30 PM   #2
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Your inverse matrix is incorrect. Also, your equations are incorrect.

You should have used 3x + y = 50,000 and 2x(0.06) + y(0.07) + x(0.10) = 3620.

In matrix form, $\displaystyle \begin{pmatrix}
3 & 1 \\
0.22 & 0.07
\end{pmatrix}\begin{pmatrix}
x \\
y
\end{pmatrix} = \begin{pmatrix}
50000 \\
3620
\end{pmatrix}$.

The relevant inverse matrix is $\displaystyle \begin{pmatrix}
-7 & 100 \\
22 & -300
\end{pmatrix}$,

so $\displaystyle \begin{pmatrix}
x \\
y
\end{pmatrix} = \begin{pmatrix}
-7 & 100 \\
22 & -300
\end{pmatrix}\begin{pmatrix}
50000 \\
3620
\end{pmatrix} = \begin{pmatrix}
12000 \\
14000
\end{pmatrix}$.
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December 8th, 2017, 06:29 AM   #3
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I'm still a little confused; shouldn't there be an x, y and z answer?

Last edited by skipjack; December 8th, 2017 at 07:48 AM.
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December 8th, 2017, 07:46 AM   #4
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Yes, in which case your z is my x, and your x is 2z (i.e. twice my x).

Using a 3 × 3 matrix, $\displaystyle \begin{pmatrix}
1 & 1 & 1 \\
0.06 & 0.07 & 0.1 \\
1 & 0 & -2
\end{pmatrix}\begin{pmatrix}
x \\
y \\
z
\end{pmatrix} = \begin{pmatrix}
50000 \\
3620 \\
0
\end{pmatrix}$.

The relevant inverse matrix is then $\displaystyle \begin{pmatrix}
-14 & 200 & 3 \\
22 & -300 & -4 \\
-7 & 100 & 1
\end{pmatrix}$,

so $\displaystyle \begin{pmatrix}
x \\
y \\
z
\end{pmatrix} = \begin{pmatrix}
-14 & 200 & 3 \\
22 & -300 & -4 \\
-7 & 100 & 1
\end{pmatrix}\begin{pmatrix}
50000 \\
3620 \\
0
\end{pmatrix} = \begin{pmatrix}
24000 \\
14000 \\
12000
\end{pmatrix}$.

The methods are very similar, but the second would take a bit longer to do.
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