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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. Tags inverse, matrix, problem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post whitegreen Linear Algebra 2 February 19th, 2016 04:50 AM the_doctor1112 Linear Algebra 2 October 15th, 2012 09:50 AM soft001 Linear Algebra 1 September 13th, 2012 05:08 AM question Algebra 1 April 7th, 2012 08:56 AM soft001 Algebra 0 December 31st, 1969 04:00 PM

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