Question

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

Answer 1

X = $30, Y = $24, Z = $28

Explanation:

PLATO <3

Answer 2

The answer is explained below

Explanation:

Let x be the price of fertilizer A, y be the price of fertilizer B, and z be the price of fertilizer 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 first combination is given by the equation:

6X + 5Y + 3Z = 384

The second combination consists of 10 liters of A, 2 liters of B, and 6 liters of C, and it costs $516. The second combination is given by the equation:

10X + 2Y + 6Z = 516

The last combination consists of 4 liters of A, 8 liters of B, and 2 liters of C, with a cost of $368. The last combination is given by the equation:

4X + 8Y + 2Z = 368

In Matrix form it can be represented as:

`\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]`

`\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]`

Therefore:

X = $30, Y = $24, Z = $28

The price of fertilizer A = $30 per liter, The price of fertilizer B = $24 per liter and The price of fertilizer c = $28 per liter