Answer :
Sure, let's solve this system of equations using matrices to determine the cost of a t-shirt and the cost of a notebook.
We have the following equations from the problem:
1. [tex]\( 2x + 3y = 40 \)[/tex]
2. [tex]\( x + y = 16 \)[/tex]
Here, [tex]\( x \)[/tex] represents the cost of a t-shirt and [tex]\( y \)[/tex] represents the cost of a notebook.
We can write these equations in matrix form as:
[tex]\[ \begin{pmatrix} 2 & 3 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 40 \\ 16 \end{pmatrix} \][/tex]
Let's denote:
- [tex]\( A \)[/tex] as the coefficient matrix [tex]\(\begin{pmatrix} 2 & 3 \\ 1 & 1 \end{pmatrix}\)[/tex]
- [tex]\( \mathbf{x} \)[/tex] as the variable matrix [tex]\(\begin{pmatrix} x \\ y \end{pmatrix}\)[/tex]
- [tex]\( \mathbf{b} \)[/tex] as the constant matrix [tex]\(\begin{pmatrix} 40 \\ 16 \end{pmatrix}\)[/tex]
Our matrix equation is:
[tex]\[ A \mathbf{x} = \mathbf{b} \][/tex]
To solve for [tex]\(\mathbf{x}\)[/tex], we can use the inverse of [tex]\( A \)[/tex] (denoted [tex]\( A^{-1} \)[/tex]) and multiply both sides of the equation by [tex]\( A^{-1} \)[/tex]:
[tex]\[ A^{-1} A \mathbf{x} = A^{-1} \mathbf{b} \][/tex]
Since [tex]\( A^{-1} A \)[/tex] is the identity matrix [tex]\( I \)[/tex], we have:
[tex]\[ I \mathbf{x} = A^{-1} \mathbf{b} \][/tex]
Which simplifies to:
[tex]\[ \mathbf{x} = A^{-1} \mathbf{b} \][/tex]
Now we need to compute [tex]\( A^{-1} \)[/tex]. The inverse of a 2x2 matrix [tex]\( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/tex] is given by:
[tex]\[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \][/tex]
For our matrix [tex]\( A = \begin{pmatrix} 2 & 3 \\ 1 & 1 \end{pmatrix} \)[/tex]:
- [tex]\( a = 2 \)[/tex]
- [tex]\( b = 3 \)[/tex]
- [tex]\( c = 1 \)[/tex]
- [tex]\( d = 1 \)[/tex]
The determinant of [tex]\( A \)[/tex] is:
[tex]\[ \det(A) = ad - bc = (2)(1) - (3)(1) = 2 - 3 = -1 \][/tex]
Thus, the inverse of [tex]\( A \)[/tex] is:
[tex]\[ A^{-1} = \frac{1}{-1} \begin{pmatrix} 1 & -3 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} -1 & 3 \\ 1 & -2 \end{pmatrix} \][/tex]
Next, we multiply [tex]\( A^{-1} \)[/tex] by [tex]\( \mathbf{b} \)[/tex] to find [tex]\( \mathbf{x} \)[/tex]:
[tex]\[ \mathbf{x} = A^{-1} \mathbf{b} = \begin{pmatrix} -1 & 3 \\ 1 & -2 \end{pmatrix} \begin{pmatrix} 40 \\ 16 \end{pmatrix} \][/tex]
Carrying out the matrix multiplication:
[tex]\[ \begin{pmatrix} -1 & 3 \\ 1 & -2 \end{pmatrix} \begin{pmatrix} 40 \\ 16 \end{pmatrix} = \begin{pmatrix} (-1 \cdot 40) + (3 \cdot 16) \\ (1 \cdot 40) + (-2 \cdot 16) \end{pmatrix} = \begin{pmatrix} -40 + 48 \\ 40 - 32 \end{pmatrix} = \begin{pmatrix} 8 \\ 8 \end{pmatrix} \][/tex]
Therefore, the cost of a t-shirt [tex]\( x \)[/tex] is [tex]$8 and the cost of a notebook \( y \) is $[/tex]8.
We have the following equations from the problem:
1. [tex]\( 2x + 3y = 40 \)[/tex]
2. [tex]\( x + y = 16 \)[/tex]
Here, [tex]\( x \)[/tex] represents the cost of a t-shirt and [tex]\( y \)[/tex] represents the cost of a notebook.
We can write these equations in matrix form as:
[tex]\[ \begin{pmatrix} 2 & 3 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 40 \\ 16 \end{pmatrix} \][/tex]
Let's denote:
- [tex]\( A \)[/tex] as the coefficient matrix [tex]\(\begin{pmatrix} 2 & 3 \\ 1 & 1 \end{pmatrix}\)[/tex]
- [tex]\( \mathbf{x} \)[/tex] as the variable matrix [tex]\(\begin{pmatrix} x \\ y \end{pmatrix}\)[/tex]
- [tex]\( \mathbf{b} \)[/tex] as the constant matrix [tex]\(\begin{pmatrix} 40 \\ 16 \end{pmatrix}\)[/tex]
Our matrix equation is:
[tex]\[ A \mathbf{x} = \mathbf{b} \][/tex]
To solve for [tex]\(\mathbf{x}\)[/tex], we can use the inverse of [tex]\( A \)[/tex] (denoted [tex]\( A^{-1} \)[/tex]) and multiply both sides of the equation by [tex]\( A^{-1} \)[/tex]:
[tex]\[ A^{-1} A \mathbf{x} = A^{-1} \mathbf{b} \][/tex]
Since [tex]\( A^{-1} A \)[/tex] is the identity matrix [tex]\( I \)[/tex], we have:
[tex]\[ I \mathbf{x} = A^{-1} \mathbf{b} \][/tex]
Which simplifies to:
[tex]\[ \mathbf{x} = A^{-1} \mathbf{b} \][/tex]
Now we need to compute [tex]\( A^{-1} \)[/tex]. The inverse of a 2x2 matrix [tex]\( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/tex] is given by:
[tex]\[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \][/tex]
For our matrix [tex]\( A = \begin{pmatrix} 2 & 3 \\ 1 & 1 \end{pmatrix} \)[/tex]:
- [tex]\( a = 2 \)[/tex]
- [tex]\( b = 3 \)[/tex]
- [tex]\( c = 1 \)[/tex]
- [tex]\( d = 1 \)[/tex]
The determinant of [tex]\( A \)[/tex] is:
[tex]\[ \det(A) = ad - bc = (2)(1) - (3)(1) = 2 - 3 = -1 \][/tex]
Thus, the inverse of [tex]\( A \)[/tex] is:
[tex]\[ A^{-1} = \frac{1}{-1} \begin{pmatrix} 1 & -3 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} -1 & 3 \\ 1 & -2 \end{pmatrix} \][/tex]
Next, we multiply [tex]\( A^{-1} \)[/tex] by [tex]\( \mathbf{b} \)[/tex] to find [tex]\( \mathbf{x} \)[/tex]:
[tex]\[ \mathbf{x} = A^{-1} \mathbf{b} = \begin{pmatrix} -1 & 3 \\ 1 & -2 \end{pmatrix} \begin{pmatrix} 40 \\ 16 \end{pmatrix} \][/tex]
Carrying out the matrix multiplication:
[tex]\[ \begin{pmatrix} -1 & 3 \\ 1 & -2 \end{pmatrix} \begin{pmatrix} 40 \\ 16 \end{pmatrix} = \begin{pmatrix} (-1 \cdot 40) + (3 \cdot 16) \\ (1 \cdot 40) + (-2 \cdot 16) \end{pmatrix} = \begin{pmatrix} -40 + 48 \\ 40 - 32 \end{pmatrix} = \begin{pmatrix} 8 \\ 8 \end{pmatrix} \][/tex]
Therefore, the cost of a t-shirt [tex]\( x \)[/tex] is [tex]$8 and the cost of a notebook \( y \) is $[/tex]8.
