Question

two student clubs were selling T-shirts and school notebooks to raise money for an upcoming school event. In the first few minutes, Club A sold two T-shirts and three notebooks, and made $40. Club B sold one T-shirt and one notebook for a total of $16. Use the given matrix equation to solve for the cost of T-shirts and notebooks sold. explain the steps that you took to solve this problem( FULL PROBLEM IN PHOTO)

Answer 1

Given

$\begin{bmatrix}{2} & 3 \\ {1} & {1}\end{bmatrix}\begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix}=\begin{bmatrix}{40} & {} \\ 16 & {}\end{bmatrix}$

To solve for x and y.

Explanation:

It is given that,

$\begin{bmatrix}{2} & 3 \\ {1} & {1}\end{bmatrix}\begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix}=\begin{bmatrix}{40} & {} \\ 16 & {}\end{bmatrix}$

That implies,

$\begin{bmatrix}{2} & 3 \\ {1} & {1}\end{bmatrix}\begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix}=\begin{bmatrix}{40} & {} \\ 16 & {}\end{bmatrix}$

The solution to the given matrix is,

$X=A^{-1}B$

Therefore,

$\begin{gathered} A^{-1}=\frac{1}{|A|}\times adj(A) \\ \because adj(A)=\begin{bmatrix}{1} & {-3} \\ {-1} & {2}\end{bmatrix} \\ |A|=det\begin{bmatrix}{2} & {3} \\ {1} & {1}\end{bmatrix} \\ =2-3 \\ =-1 \\ \therefore A^{-1}=\frac{1}{-1}\begin{bmatrix}{1} & {-3} \\ {-1} & {2}\end{bmatrix} \\ =\begin{bmatrix}-{1} & {3} \\ {1} & -{2}\end{bmatrix} \end{gathered}$

Then,

$\begin{gathered} X=\begin{bmatrix}-{1} & {3} \\ {1} & -{2}\end{bmatrix}\begin{bmatrix}{40} & \\ 16 & \end{bmatrix} \\ =\begin{bmatrix}{-40+48} & {} \\ {40-32} & \end{bmatrix} \\ \begin{bmatrix}{x} & \\ {y} & {}\end{bmatrix}=\begin{bmatrix}{8} & {} \\ {8} & \end{bmatrix} \end{gathered}$

Hence, x=8 and y=8.