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To determine the cost of a t-shirt and the cost of a notebook using matrices, we can set up and solve a system of linear equations. Let's denote the cost of a t-shirt by [tex]\( x \)[/tex] and the cost of a notebook by [tex]\( y \)[/tex].
### Step 1: Translate the word problem into a system of equations
From the problem, we have the following information:
1. Club A sold 3 t-shirts and 2 notebooks and made \[tex]$19. 2. Club B sold 1 t-shirt and 1 notebook and made \$[/tex]8.
This translates to the following system of linear equations:
[tex]\[ \begin{cases} 3x + 2y = 19 \\ x + y = 8 \end{cases} \][/tex]
### Step 2: Represent the system of equations in matrix form
The given system of equations can be represented in matrix form as follows:
[tex]\[ \begin{pmatrix} 3 & 2 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 19 \\ 8 \end{pmatrix} \][/tex]
Here, the coefficient matrix [tex]\( A \)[/tex], the variable matrix [tex]\( X \)[/tex], and the constant matrix [tex]\( B \)[/tex] are:
[tex]\[ A = \begin{pmatrix} 3 & 2 \\ 1 & 1 \end{pmatrix} , \quad X = \begin{pmatrix} x \\ y \end{pmatrix} , \quad B = \begin{pmatrix} 19 \\ 8 \end{pmatrix} \][/tex]
### Step 3: Solve the matrix equation [tex]\( AX = B \)[/tex]
To find [tex]\( X \)[/tex], we need to solve the equation [tex]\( AX = B \)[/tex]. One way to do this is by finding the inverse of matrix [tex]\( A \)[/tex] and then multiplying it by matrix [tex]\( B \)[/tex]. This gives us:
[tex]\[ X = A^{-1} B \][/tex]
First, we need to determine the inverse of matrix [tex]\( A \)[/tex]. The formula for the inverse of a [tex]\( 2 \times 2 \)[/tex] matrix [tex]\( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/tex] is:
[tex]\[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \][/tex]
Given [tex]\( A = \begin{pmatrix} 3 & 2 \\ 1 & 1 \end{pmatrix} \)[/tex], we compute the determinant [tex]\( \text{det}(A) \)[/tex]:
[tex]\[ \text{det}(A) = (3)(1) - (2)(1) = 3 - 2 = 1 \][/tex]
Since the determinant is 1 (non-zero), [tex]\( A \)[/tex] is invertible, and its inverse is:
[tex]\[ A^{-1} = \begin{pmatrix} 1 & -2 \\ -1 & 3 \end{pmatrix} \][/tex]
### Step 4: Multiply [tex]\( A^{-1} \)[/tex] by [tex]\( B \)[/tex] to find [tex]\( X \)[/tex]
Now, we multiply the inverse matrix [tex]\( A^{-1} \)[/tex] by the constant matrix [tex]\( B \)[/tex]:
[tex]\[ X = A^{-1} B = \begin{pmatrix} 1 & -2 \\ -1 & 3 \end{pmatrix} \begin{pmatrix} 19 \\ 8 \end{pmatrix} \][/tex]
Perform the matrix multiplication:
[tex]\[ X = \begin{pmatrix} 1 \cdot 19 + (-2) \cdot 8 \\ -1 \cdot 19 + 3 \cdot 8 \end{pmatrix} = \begin{pmatrix} 19 - 16 \\ -19 + 24 \end{pmatrix} = \begin{pmatrix} 3 \\ 5 \end{pmatrix} \][/tex]
Thus, the solution [tex]\( X \)[/tex] is:
[tex]\[ X = \begin{pmatrix} 3 \\ 5 \end{pmatrix} \][/tex]
### Conclusion
The cost of a t-shirt is [tex]\( x = \$3 \)[/tex], and the cost of a notebook is [tex]\( y = \$5 \)[/tex].
### Step 1: Translate the word problem into a system of equations
From the problem, we have the following information:
1. Club A sold 3 t-shirts and 2 notebooks and made \[tex]$19. 2. Club B sold 1 t-shirt and 1 notebook and made \$[/tex]8.
This translates to the following system of linear equations:
[tex]\[ \begin{cases} 3x + 2y = 19 \\ x + y = 8 \end{cases} \][/tex]
### Step 2: Represent the system of equations in matrix form
The given system of equations can be represented in matrix form as follows:
[tex]\[ \begin{pmatrix} 3 & 2 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 19 \\ 8 \end{pmatrix} \][/tex]
Here, the coefficient matrix [tex]\( A \)[/tex], the variable matrix [tex]\( X \)[/tex], and the constant matrix [tex]\( B \)[/tex] are:
[tex]\[ A = \begin{pmatrix} 3 & 2 \\ 1 & 1 \end{pmatrix} , \quad X = \begin{pmatrix} x \\ y \end{pmatrix} , \quad B = \begin{pmatrix} 19 \\ 8 \end{pmatrix} \][/tex]
### Step 3: Solve the matrix equation [tex]\( AX = B \)[/tex]
To find [tex]\( X \)[/tex], we need to solve the equation [tex]\( AX = B \)[/tex]. One way to do this is by finding the inverse of matrix [tex]\( A \)[/tex] and then multiplying it by matrix [tex]\( B \)[/tex]. This gives us:
[tex]\[ X = A^{-1} B \][/tex]
First, we need to determine the inverse of matrix [tex]\( A \)[/tex]. The formula for the inverse of a [tex]\( 2 \times 2 \)[/tex] matrix [tex]\( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/tex] is:
[tex]\[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \][/tex]
Given [tex]\( A = \begin{pmatrix} 3 & 2 \\ 1 & 1 \end{pmatrix} \)[/tex], we compute the determinant [tex]\( \text{det}(A) \)[/tex]:
[tex]\[ \text{det}(A) = (3)(1) - (2)(1) = 3 - 2 = 1 \][/tex]
Since the determinant is 1 (non-zero), [tex]\( A \)[/tex] is invertible, and its inverse is:
[tex]\[ A^{-1} = \begin{pmatrix} 1 & -2 \\ -1 & 3 \end{pmatrix} \][/tex]
### Step 4: Multiply [tex]\( A^{-1} \)[/tex] by [tex]\( B \)[/tex] to find [tex]\( X \)[/tex]
Now, we multiply the inverse matrix [tex]\( A^{-1} \)[/tex] by the constant matrix [tex]\( B \)[/tex]:
[tex]\[ X = A^{-1} B = \begin{pmatrix} 1 & -2 \\ -1 & 3 \end{pmatrix} \begin{pmatrix} 19 \\ 8 \end{pmatrix} \][/tex]
Perform the matrix multiplication:
[tex]\[ X = \begin{pmatrix} 1 \cdot 19 + (-2) \cdot 8 \\ -1 \cdot 19 + 3 \cdot 8 \end{pmatrix} = \begin{pmatrix} 19 - 16 \\ -19 + 24 \end{pmatrix} = \begin{pmatrix} 3 \\ 5 \end{pmatrix} \][/tex]
Thus, the solution [tex]\( X \)[/tex] is:
[tex]\[ X = \begin{pmatrix} 3 \\ 5 \end{pmatrix} \][/tex]
### Conclusion
The cost of a t-shirt is [tex]\( x = \$3 \)[/tex], and the cost of a notebook is [tex]\( y = \$5 \)[/tex].
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