Claire bought 5 kg of apples and 2 kg of bananas and paid altogether $22 Dale bought 4 kg of apples and 6 kg of bananas and paid altogether $33 Use matrices to find the cost of 1 kg of bananas

Respuesta :

Answer:

$3.5

Explanation:

Let the cost of 1 kg of apples = x

Let the cost of 1 kg of bananas​ =y

Claire bought 5 kg of apples and 2 kg of bananas and paid altogether $22.

[tex]5x+2y=22[/tex]

Dale bought 4 kg of apples and 6 kg of bananas and paid altogether $33.

[tex]4x+6y=33[/tex]

We set up the system of linear equations as a matrix below:

[tex]\begin{bmatrix}{5} & {2} & \\ {4} & {6} & {}{}\end{bmatrix}\begin{bmatrix}{x} & \\ {y} & {}{}\end{bmatrix}=\begin{bmatrix}{22} & \\ {33} & {}{}\end{bmatrix}[/tex]

We then solve for the variables x and y as follows.

[tex]\begin{gathered} \begin{bmatrix}{x} & \\ {y} & {}{}\end{bmatrix}=\begin{bmatrix}{5} & {2} & \\ {4} & {6} & {}{}\end{bmatrix}^{-1}\begin{bmatrix}{22} & \\ {33} & {}{}\end{bmatrix} \\ =\frac{1}{30-8}\begin{bmatrix}{6} & {-2} & \\ {-4} & {5} & {}{}\end{bmatrix}\begin{bmatrix}{22} & \\ {33} & {}{}\end{bmatrix} \\ =\frac{1}{22}\begin{bmatrix}{6} & {-2} & \\ {-4} & {5} & {}{}\end{bmatrix}\begin{bmatrix}{22} & \\ {33} & {}{}\end{bmatrix} \end{gathered}[/tex]

We proceed to simplify further.

[tex]\begin{gathered} =\begin{bmatrix}{\frac{6}{22}} & {-\frac{2}{22}} & \\ {-\frac{4}{22}} & {\frac{5}{22}} & {}{}\end{bmatrix}\begin{bmatrix}{22} & \\ {33} & {}{}\end{bmatrix} \\ =\begin{bmatrix}{\frac{6}{22}\times22-\frac{2}{22}\times33} & {} & \\ {\frac{-4}{22}\times22+\frac{5}{22}\times33} & & {}{}\end{bmatrix} \\ =\begin{bmatrix}{6-3} & {} & \\ {-4+7.5} & & {}{}\end{bmatrix} \\ =\begin{bmatrix}{3} & {} & \\ {3.5} & & {}{}\end{bmatrix} \end{gathered}[/tex]

Therefore:

x=3 and y=3.5.

The cost of 1 kg of bananas​ is $3.5.