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# High School Inverse Matrices help.

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Almonds cost $8/pound and cashews cost$5/pound. Robin wants to make a 60 pounds of a mixture that will cost \$7/pound.

Find out how many pounds each of almonds and cashews are needed to create this mixture.

You have to use an inverse matrix to do this problem. How do I construct the first matrix?

Jun 19, 2019

#3
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Set up systems of equations:

We have a+c=60 and 8a+5c=420.

$$\begin{bmatrix} 1 & 1 \\ 8 & 5 \end{bmatrix}$$$$\begin{bmatrix} 60\\420 \end{bmatrix}$$, and we take the determinant of the first matrix, which is $$D$$, which can be found by solving $$ad-bc$$, and that gives $$(1)(5)-(1)(8)=5-8=-3$$.

Now, we have to find the inverse of this matrix, $$\begin{bmatrix} 1 & 1 \\ 8 & 5 \end{bmatrix}$$, where you have to switch the $$a$$ and $$d$$, and find the negative values of $$b$$ and $$c$$.

Thus, the new matrix is: $$\begin{bmatrix} 5 & -1 \\ -8 & 1 \end{bmatrix}$$. And, now we have this: $$\frac{1}{-3}\begin{bmatrix} 5 & -1 \\ -8 & 1 \end{bmatrix}$$, and "distributing" them $$-3$$ to all the terms in the matrix, we get(have) $$\begin{bmatrix} \frac{5}{-3} & \frac{-1}{-3} \\ \frac{-8}{-3} & \frac{1}{-3} \end{bmatrix}$$. and this can be re-written as $$\begin{bmatrix} \frac{-5}{3} & \frac{1}{3} \\ \frac{8}{3} & \frac{-1}{3} \end{bmatrix}$$. Now, we have to multiply $$\begin{bmatrix} \frac{-5}{3} & \frac{1}{3} \\ \frac{8}{3} & \frac{-1}{3} \end{bmatrix} * \begin{bmatrix} 60\\420 \end{bmatrix}$$, and multiplying the first column by the first row, and the first column by the second row! (The second matrix has only one column), we get $$\begin{bmatrix} 40\\20 \end{bmatrix}$$. Thus, the answer is forty$$(40)$$ almonds and twenty$$(20)$$ cashews.

-tertre

Jun 20, 2019
edited by tertre  Jun 20, 2019

#1
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#almonds =x    cost of almonds   8 x

#cashews = 60-x   cost of cashews = 5 (60-x)

Added together they equal  7 * 60

8x  +  5(60-x) = 7*60

8x + 300-5x = 420

3x = 120

x = #almonds = 40

60-x = cashews = 60-40 = 20#

Jun 19, 2019
#2
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How would you use an inverse matrix?

Bxtterman  Jun 19, 2019
#3
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Set up systems of equations:

We have a+c=60 and 8a+5c=420.

$$\begin{bmatrix} 1 & 1 \\ 8 & 5 \end{bmatrix}$$$$\begin{bmatrix} 60\\420 \end{bmatrix}$$, and we take the determinant of the first matrix, which is $$D$$, which can be found by solving $$ad-bc$$, and that gives $$(1)(5)-(1)(8)=5-8=-3$$.

Now, we have to find the inverse of this matrix, $$\begin{bmatrix} 1 & 1 \\ 8 & 5 \end{bmatrix}$$, where you have to switch the $$a$$ and $$d$$, and find the negative values of $$b$$ and $$c$$.

Thus, the new matrix is: $$\begin{bmatrix} 5 & -1 \\ -8 & 1 \end{bmatrix}$$. And, now we have this: $$\frac{1}{-3}\begin{bmatrix} 5 & -1 \\ -8 & 1 \end{bmatrix}$$, and "distributing" them $$-3$$ to all the terms in the matrix, we get(have) $$\begin{bmatrix} \frac{5}{-3} & \frac{-1}{-3} \\ \frac{-8}{-3} & \frac{1}{-3} \end{bmatrix}$$. and this can be re-written as $$\begin{bmatrix} \frac{-5}{3} & \frac{1}{3} \\ \frac{8}{3} & \frac{-1}{3} \end{bmatrix}$$. Now, we have to multiply $$\begin{bmatrix} \frac{-5}{3} & \frac{1}{3} \\ \frac{8}{3} & \frac{-1}{3} \end{bmatrix} * \begin{bmatrix} 60\\420 \end{bmatrix}$$, and multiplying the first column by the first row, and the first column by the second row! (The second matrix has only one column), we get $$\begin{bmatrix} 40\\20 \end{bmatrix}$$. Thus, the answer is forty$$(40)$$ almonds and twenty$$(20)$$ cashews.

-tertre

tertre Jun 20, 2019
edited by tertre  Jun 20, 2019
#4
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Very nice, tertre......EXACTLY what was wanted!!!   CPhill  Jun 20, 2019