Answer :
To solve the given problem using matrices, we need to determine the cost of a t-shirt and the cost of a notebook given the sales information from two student clubs.
Let's denote:
- [tex]\( x \)[/tex] as the cost of a t-shirt.
- [tex]\( y \)[/tex] as the cost of a notebook.
We have two equations based on the sales:
1. Club A sold 2 t-shirts and 3 notebooks for a total of [tex]$40. This can be written as: \[ 2x + 3y = 40 \] 2. Club B sold 1 t-shirt and 1 notebook for a total of $[/tex]16. This can be written as:
[tex]\[ x + y = 16 \][/tex]
We can express these equations in matrix form [tex]\( A \mathbf{x} = \mathbf{b} \)[/tex] as follows:
[tex]\[ \left[\begin{array}{ll} 2 & 3 \\ 1 & 1 \end{array}\right] \left[\begin{array}{l} x \\ y \end{array}\right] = \left[\begin{array}{l} 40 \\ 16 \end{array}\right] \][/tex]
Where:
- [tex]\( A \)[/tex] is the coefficient matrix:
[tex]\[ A = \left[\begin{array}{ll} 2 & 3 \\ 1 & 1 \end{array}\right] \][/tex]
- [tex]\( \mathbf{x} \)[/tex] is the column vector of unknowns:
[tex]\[ \mathbf{x} = \left[\begin{array}{l} x \\ y \end{array}\right] \][/tex]
- [tex]\( \mathbf{b} \)[/tex] is the constants matrix:
[tex]\[ \mathbf{b} = \left[\begin{array}{l} 40 \\ 16 \end{array}\right] \][/tex]
To find the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex], we solve the matrix equation [tex]\( A \mathbf{x} = \mathbf{b} \)[/tex].
We can use the method of matrix inversion or any other suitable matrix-solving technique. For simplicity here, let's directly find the costs using given matrix knowledge:
1. Find the determinant of [tex]\( A \)[/tex]:
[tex]\[ \text{det}(A) = (2 \times 1) - (3 \times 1) = 2 - 3 = -1 \][/tex]
2. Find the inverse of [tex]\( A \)[/tex]:
The inverse formula for a 2x2 matrix [tex]\( \left[\begin{array}{ll} a & b \\ c & d \end{array}\right] \)[/tex] is [tex]\( \frac{1}{ad - bc} \left[\begin{array}{ll} d & -b \\ -c & a \end{array}\right] \)[/tex].
Using this formula:
[tex]\[ A^{-1} = \frac{1}{-1} \left[\begin{array}{ll} 1 & -3 \\ -1 & 2 \end{array}\right] = \left[\begin{array}{ll} -1 & 3 \\ 1 & -2 \end{array}\right] \][/tex]
3. Multiply [tex]\( A^{-1} \)[/tex] by [tex]\( \mathbf{b} \)[/tex] to find [tex]\( \mathbf{x} \)[/tex]:
[tex]\[ \mathbf{x} = A^{-1} \mathbf{b} = \left[\begin{array}{ll} -1 & 3 \\ 1 & -2 \end{array}\right] \left[\begin{array}{l} 40 \\ 16 \end{array}\right] \][/tex]
4. Calculate the product:
[tex]\[ \mathbf{x} = \left[\begin{array}{l} (-1 \times 40) + (3 \times 16) \\ (1 \times 40) + (-2 \times 16) \end{array}\right] = \left[\begin{array}{l} -40 + 48 \\ 40 - 32 \end{array}\right] = \left[\begin{array}{l} 8 \\ 8 \end{array}\right] \][/tex]
So, the cost of a t-shirt [tex]\( x \)[/tex] is [tex]$8 and the cost of a notebook \( y \) is $[/tex]8.
[tex]\[ x = 8, \quad y = 8 \][/tex]
Therefore, the cost of a t-shirt is [tex]\( \$8 \)[/tex] and the cost of a notebook is [tex]\( \$8 \)[/tex].
Let's denote:
- [tex]\( x \)[/tex] as the cost of a t-shirt.
- [tex]\( y \)[/tex] as the cost of a notebook.
We have two equations based on the sales:
1. Club A sold 2 t-shirts and 3 notebooks for a total of [tex]$40. This can be written as: \[ 2x + 3y = 40 \] 2. Club B sold 1 t-shirt and 1 notebook for a total of $[/tex]16. This can be written as:
[tex]\[ x + y = 16 \][/tex]
We can express these equations in matrix form [tex]\( A \mathbf{x} = \mathbf{b} \)[/tex] as follows:
[tex]\[ \left[\begin{array}{ll} 2 & 3 \\ 1 & 1 \end{array}\right] \left[\begin{array}{l} x \\ y \end{array}\right] = \left[\begin{array}{l} 40 \\ 16 \end{array}\right] \][/tex]
Where:
- [tex]\( A \)[/tex] is the coefficient matrix:
[tex]\[ A = \left[\begin{array}{ll} 2 & 3 \\ 1 & 1 \end{array}\right] \][/tex]
- [tex]\( \mathbf{x} \)[/tex] is the column vector of unknowns:
[tex]\[ \mathbf{x} = \left[\begin{array}{l} x \\ y \end{array}\right] \][/tex]
- [tex]\( \mathbf{b} \)[/tex] is the constants matrix:
[tex]\[ \mathbf{b} = \left[\begin{array}{l} 40 \\ 16 \end{array}\right] \][/tex]
To find the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex], we solve the matrix equation [tex]\( A \mathbf{x} = \mathbf{b} \)[/tex].
We can use the method of matrix inversion or any other suitable matrix-solving technique. For simplicity here, let's directly find the costs using given matrix knowledge:
1. Find the determinant of [tex]\( A \)[/tex]:
[tex]\[ \text{det}(A) = (2 \times 1) - (3 \times 1) = 2 - 3 = -1 \][/tex]
2. Find the inverse of [tex]\( A \)[/tex]:
The inverse formula for a 2x2 matrix [tex]\( \left[\begin{array}{ll} a & b \\ c & d \end{array}\right] \)[/tex] is [tex]\( \frac{1}{ad - bc} \left[\begin{array}{ll} d & -b \\ -c & a \end{array}\right] \)[/tex].
Using this formula:
[tex]\[ A^{-1} = \frac{1}{-1} \left[\begin{array}{ll} 1 & -3 \\ -1 & 2 \end{array}\right] = \left[\begin{array}{ll} -1 & 3 \\ 1 & -2 \end{array}\right] \][/tex]
3. Multiply [tex]\( A^{-1} \)[/tex] by [tex]\( \mathbf{b} \)[/tex] to find [tex]\( \mathbf{x} \)[/tex]:
[tex]\[ \mathbf{x} = A^{-1} \mathbf{b} = \left[\begin{array}{ll} -1 & 3 \\ 1 & -2 \end{array}\right] \left[\begin{array}{l} 40 \\ 16 \end{array}\right] \][/tex]
4. Calculate the product:
[tex]\[ \mathbf{x} = \left[\begin{array}{l} (-1 \times 40) + (3 \times 16) \\ (1 \times 40) + (-2 \times 16) \end{array}\right] = \left[\begin{array}{l} -40 + 48 \\ 40 - 32 \end{array}\right] = \left[\begin{array}{l} 8 \\ 8 \end{array}\right] \][/tex]
So, the cost of a t-shirt [tex]\( x \)[/tex] is [tex]$8 and the cost of a notebook \( y \) is $[/tex]8.
[tex]\[ x = 8, \quad y = 8 \][/tex]
Therefore, the cost of a t-shirt is [tex]\( \$8 \)[/tex] and the cost of a notebook is [tex]\( \$8 \)[/tex].
