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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

Best Answer 

 #3
avatar+4296 
+1

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
avatar
+2

#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
avatar+70 
+1

How would you use an inverse matrix?

Bxtterman  Jun 19, 2019
 #3
avatar+4296 
+1
Best Answer

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
avatar+101856 
+1

Very nice, tertre......EXACTLY what was wanted!!!

 

 

cool cool cool

CPhill  Jun 20, 2019

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