My Math Forum Help Solving this Matrix Word Problem???

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 February 19th, 2017, 05:06 PM #1 Newbie   Joined: Feb 2017 From: nowhere Posts: 2 Thanks: 0 Help Solving this Matrix Word Problem??? I need help with this and have no idea where to start. Can someone walk me through how to do this? A nutritionist planning a diet for a rugby player wants him to consume 3,650 Calories and 650 grams of food daily. Calories from carbohydrates and fat will be 70% of the total Calories. There are 4, 4, and 9 Calories per gram for protein, carbohydrates, and fat, respectively. Find the number of daily grams of fat the diet will need to include by writing a system of equations, and then solving it using Cramer's Rule.
 February 19th, 2017, 05:33 PM #2 Member   Joined: Oct 2016 From: Melbourne Posts: 77 Thanks: 35 The first thing you need to do is write down a system of equations. As there will be three unknowns (the amount of protien, carbohydrates and fat) you will need to generate three equations. What information can you use that is given in the question to help come up with these equations?
 February 19th, 2017, 09:38 PM #3 Senior Member     Joined: Sep 2015 From: USA Posts: 1,944 Thanks: 1011 the unknowns are the grams of protein, carbs, and fat included in the diet. let these be denoted by $p,~c,~f$ respectively From the first sentence of the problem we have $p + c + f = 650$ $4 p + 4 c + 9 f = 3650$ From the 2nd sentence $4 c + 9 f = (0.7) 3650$ we write this as a matrix equation as $\begin{pmatrix}1 &1 &1 \\ 4 &4 &9 \\ 0 &4 &9 \end{pmatrix} \begin{pmatrix}p \\ c \\ f \end{pmatrix} = \begin{pmatrix} 650 \\ 3650 \\ 2555 \end{pmatrix}$ To use Cramer's rule first we need the determinant of the the matrix on the left hand side of this equation. $D =\left | \begin{pmatrix}1 &1 &1 \\ 4 &4 &9 \\ 0 &4 &9 \end{pmatrix} \right|$ then for example to find the value of $p$ we substitute the right hand side vector for the first column of the matrix on the left, and find the determinant of the resulting matrix. $D_x =\left | \begin{pmatrix}650 &1 &1 \\ 3650 &4 &9 \\ 2555 &4 &9 \end{pmatrix} \right |$ $x = \dfrac{D_x}{D}$ Similarly $D_y = \left | \begin{pmatrix}1 &650 &1 \\ 4 &3650 &9 \\ 0 &2555 &9 \end{pmatrix} \right |$ $y = \dfrac{D_y}{D}$ $D_z = \left |\begin{pmatrix}1 &1 &650 \\ 4 &4 &3650 \\ 0 &4 &2555 \end{pmatrix} \right |$ $z = \dfrac{D_z}{D}$ I leave it to you to calculate all these determinants. Thanks from deesuwalka and Trajamo
 February 19th, 2017, 10:30 PM #4 Member   Joined: Sep 2016 From: India Posts: 88 Thanks: 30 $D=1[36-36]-1[36-0]+1[16-0]$ $=-36+16=-20$ $D_x=650[36-36]-1[3650×9-2555×9]+1[3650×4-2555×4]$ $=0-9[3650-2555]+4[3650-2555]$ $=-9×1095+4×1095$ $=1095[-9+4]=1095×(-5)=-5475$ $x=\dfrac{D_x}{D}=\dfrac{-5475}{-20}=\dfrac{5475}{20}=273.75$ $D_y=1[3650×9-2555×9]-650[36-0]+1[4×2555-0]$ $=9×1095-650×36+2555×4$ $=-3325$ $y=\dfrac{D_y}{D}=\dfrac{-3325}{-20}=\dfrac{3325}{20}=166.25$ $D_z=1[4×2555-4×3650]-1[4×2555-0]+650[16-0]$ $=-4200$ $z=\dfrac{D_z}{D}=\dfrac{-4200}{-20}=210$ Thanks from Trajamo
 February 20th, 2017, 02:24 PM #5 Newbie   Joined: Feb 2017 From: nowhere Posts: 2 Thanks: 0 Thanks you so much romsek! I was wayy to overwhelmed cause I wasn't walked through a word problem like this in class. This really helped! Last edited by Trajamo; February 20th, 2017 at 02:54 PM.

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