Westonci.ca makes finding answers easy, with a community of experts ready to provide you with the information you seek. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
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.
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.
