Mathematics, 02.07.2020 20:01 aimeeoconnor20p6b43o
To boost the growth of a crop, a farmer decided to use different combinations of three fertilizers, A, B, and C. The first combination
costs $384 and consists of 6 liters of fertilizer A, 5 liters of fertilizer B, and 3 liters of fertilizer C. The second combination consists of 10
liters of A, 2 liters of B, and 6 liters of C, and it costs $516. The last combination consists of 4 liters of A, 8 liters of B, and 2 liters of C,
with a cost of $368. Let x be the price of fertilizer A, y be the price of fertilizer B, and z be the price of fertilizer C. Use matrices to
determine the cost of each type of fertilizer.
X=
y =
ZE
Answers: 1
![\left[\begin{array}{ccc}6&5&3\\10&2&6\\4&8&2\end{array}\right]\left[\begin{array}{c}X\\Y\\Z\end{array}\right]=\left[\begin{array}{c}384\\516\\368\end{array}\right]](/tpl/images/0700/0585/59230.png)
![\left[\begin{array}{c}X\\Y\\Z\end{array}\right]=\left[\begin{array}{ccc}6&5&3\\10&2&6\\4&8&2\end{array}\right]^{-1}\left[\begin{array}{c}384\\516\\368\end{array}\right]\\\\\left[\begin{array}{c}X\\Y\\Z\end{array}\right]=\left[\begin{array}{ccc}1.5714&-0.5&-0.857\\-0.142&0&0.2142\\-2.571&1&1.3571\end{array}\right]\left[\begin{array}{c}384\\516\\368\end{array}\right]\\\\\left[\begin{array}{c}X\\Y\\Z\end{array}\right]=\left[\begin{array}{c}30\\24\\28\end{array}\right]](/tpl/images/0700/0585/5e87d.png)
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