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December 6th, 2017, 11:27 AM  #1 
Newbie Joined: Nov 2017 From: MD Posts: 6 Thanks: 0  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 12:04 PM. 
December 6th, 2017, 12:30 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 21,026 Thanks: 2257 
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 $\displaystyle \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 7 & 100 \\ 22 & 300 \end{pmatrix}\begin{pmatrix} 50000 \\ 3620 \end{pmatrix} = \begin{pmatrix} 12000 \\ 14000 \end{pmatrix}$. 
December 8th, 2017, 05:29 AM  #3 
Newbie Joined: Nov 2017 From: MD Posts: 6 Thanks: 0 
I'm still a little confused; shouldn't there be an x, y and z answer?
Last edited by skipjack; December 8th, 2017 at 06:48 AM. 
December 8th, 2017, 06:46 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 21,026 Thanks: 2257 
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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