Answer :
To solve the system of linear equations and determine the cost of a t-shirt and the cost of a notebook, we can use matrices.
The information provided can be formulated into a system of linear equations. Let [tex]\( x \)[/tex] represent the cost of a t-shirt in dollars, and [tex]\( y \)[/tex] represent the cost of a notebook in dollars.
From the given data, we can write the following equations:
1. For club A: [tex]\( 2x + 3y = 20 \)[/tex]
2. For club B: [tex]\( 2x + y = 8 \)[/tex]
These equations can be represented in matrix form as:
[tex]\[ \begin{pmatrix} 2 & 3 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 20 \\ 8 \end{pmatrix} \][/tex]
Let's denote:
[tex]\[ A = \begin{pmatrix} 2 & 3 \\ 2 & 1 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 20 \\ 8 \end{pmatrix} \][/tex]
We need to find the vector [tex]\( \mathbf{x} \)[/tex] by solving the matrix equation [tex]\( A\mathbf{x} = \mathbf{b} \)[/tex].
To solve this system, we can find the inverse of matrix [tex]\( A \)[/tex] and then multiply it by vector [tex]\( \mathbf{b} \)[/tex]:
[tex]\[ \mathbf{x} = A^{-1}\mathbf{b} \][/tex]
First, we need to find the inverse of [tex]\( A \)[/tex].
The formula for the inverse of a 2x2 matrix [tex]\( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/tex] is:
[tex]\[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \][/tex]
For our matrix [tex]\( A = \begin{pmatrix} 2 & 3 \\ 2 & 1 \end{pmatrix} \)[/tex]:
[tex]\[ a = 2, \; b = 3, \; c = 2, \; d = 1 \][/tex]
Calculate the determinant ([tex]\( ad - bc \)[/tex]):
[tex]\[ \text{det}(A) = (2 \cdot 1) - (3 \cdot 2) = 2 - 6 = -4 \][/tex]
The inverse of [tex]\( A \)[/tex] is:
[tex]\[ A^{-1} = \frac{1}{-4} \begin{pmatrix} 1 & -3 \\ -2 & 2 \end{pmatrix} = \begin{pmatrix} \frac{1}{-4} & \frac{-3}{-4} \\ \frac{-2}{-4} & \frac{2}{-4} \end{pmatrix} = \begin{pmatrix} -\frac{1}{4} & \frac{3}{4} \\ -\frac{1}{2} & \frac{1}{2} \end{pmatrix} \][/tex]
Now, multiply [tex]\( A^{-1} \)[/tex] by [tex]\( \mathbf{b} \)[/tex]:
[tex]\[ \mathbf{x} = A^{-1} \mathbf{b} = \begin{pmatrix} -\frac{1}{4} & \frac{3}{4} \\ -\frac{1}{2} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} 20 \\ 8 \end{pmatrix} \][/tex]
Perform the matrix multiplication:
[tex]\[ \begin{pmatrix} -\frac{1}{4} & \frac{3}{4} \\ -\frac{1}{2} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} 20 \\ 8 \end{pmatrix} = \begin{pmatrix} (-\frac{1}{4} \cdot 20) + (\frac{3}{4} \cdot 8) \\ (-\frac{1}{2} \cdot 20) + (\frac{1}{2} \cdot 8) \end{pmatrix} = \begin{pmatrix} -5 + 6 \\ -10 + 4 \end{pmatrix} = \begin{pmatrix} 1 \\ -6 \end{pmatrix} \][/tex]
Therefore, the solution to the system of equations is:
[tex]\[ x = 1, \quad y = 6 \][/tex]
Hence, the cost of one t-shirt is [tex]\(\$1.00\)[/tex] and the cost of one notebook is [tex]\(\$6.00\)[/tex].
The information provided can be formulated into a system of linear equations. Let [tex]\( x \)[/tex] represent the cost of a t-shirt in dollars, and [tex]\( y \)[/tex] represent the cost of a notebook in dollars.
From the given data, we can write the following equations:
1. For club A: [tex]\( 2x + 3y = 20 \)[/tex]
2. For club B: [tex]\( 2x + y = 8 \)[/tex]
These equations can be represented in matrix form as:
[tex]\[ \begin{pmatrix} 2 & 3 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 20 \\ 8 \end{pmatrix} \][/tex]
Let's denote:
[tex]\[ A = \begin{pmatrix} 2 & 3 \\ 2 & 1 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 20 \\ 8 \end{pmatrix} \][/tex]
We need to find the vector [tex]\( \mathbf{x} \)[/tex] by solving the matrix equation [tex]\( A\mathbf{x} = \mathbf{b} \)[/tex].
To solve this system, we can find the inverse of matrix [tex]\( A \)[/tex] and then multiply it by vector [tex]\( \mathbf{b} \)[/tex]:
[tex]\[ \mathbf{x} = A^{-1}\mathbf{b} \][/tex]
First, we need to find the inverse of [tex]\( A \)[/tex].
The formula for the inverse of a 2x2 matrix [tex]\( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/tex] is:
[tex]\[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \][/tex]
For our matrix [tex]\( A = \begin{pmatrix} 2 & 3 \\ 2 & 1 \end{pmatrix} \)[/tex]:
[tex]\[ a = 2, \; b = 3, \; c = 2, \; d = 1 \][/tex]
Calculate the determinant ([tex]\( ad - bc \)[/tex]):
[tex]\[ \text{det}(A) = (2 \cdot 1) - (3 \cdot 2) = 2 - 6 = -4 \][/tex]
The inverse of [tex]\( A \)[/tex] is:
[tex]\[ A^{-1} = \frac{1}{-4} \begin{pmatrix} 1 & -3 \\ -2 & 2 \end{pmatrix} = \begin{pmatrix} \frac{1}{-4} & \frac{-3}{-4} \\ \frac{-2}{-4} & \frac{2}{-4} \end{pmatrix} = \begin{pmatrix} -\frac{1}{4} & \frac{3}{4} \\ -\frac{1}{2} & \frac{1}{2} \end{pmatrix} \][/tex]
Now, multiply [tex]\( A^{-1} \)[/tex] by [tex]\( \mathbf{b} \)[/tex]:
[tex]\[ \mathbf{x} = A^{-1} \mathbf{b} = \begin{pmatrix} -\frac{1}{4} & \frac{3}{4} \\ -\frac{1}{2} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} 20 \\ 8 \end{pmatrix} \][/tex]
Perform the matrix multiplication:
[tex]\[ \begin{pmatrix} -\frac{1}{4} & \frac{3}{4} \\ -\frac{1}{2} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} 20 \\ 8 \end{pmatrix} = \begin{pmatrix} (-\frac{1}{4} \cdot 20) + (\frac{3}{4} \cdot 8) \\ (-\frac{1}{2} \cdot 20) + (\frac{1}{2} \cdot 8) \end{pmatrix} = \begin{pmatrix} -5 + 6 \\ -10 + 4 \end{pmatrix} = \begin{pmatrix} 1 \\ -6 \end{pmatrix} \][/tex]
Therefore, the solution to the system of equations is:
[tex]\[ x = 1, \quad y = 6 \][/tex]
Hence, the cost of one t-shirt is [tex]\(\$1.00\)[/tex] and the cost of one notebook is [tex]\(\$6.00\)[/tex].
