+104 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?
+4622 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
#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#
+4622 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
