Answer :
Of course! Let's break down the process of solving the given system of linear equations using matrices.
We are given the matrix equation:
[tex]$ \left[\begin{array}{cc} 3 & 10 \\ 1 & 3 \end{array}\right] \left[\begin{array}{c} x \\ y \end{array}\right]=\left[\begin{array}{c} 6 \\ 1 \end{array}\right] $[/tex]
This matrix equation corresponds to the following system of linear equations:
1. [tex]\(3x + 10y = 6\)[/tex]
2. [tex]\(1x + 3y = 1\)[/tex]
To solve for [tex]\(x\)[/tex] and [tex]\(y\)[/tex] using matrices, we represent the equation as [tex]\(AX = B\)[/tex], where:
- [tex]\(A\)[/tex] is the coefficient matrix: [tex]\(A = \left[\begin{array}{cc} 3 & 10 \\ 1 & 3 \end{array}\right]\)[/tex]
- [tex]\(X\)[/tex] is the column matrix of the variables: [tex]\(X = \left[\begin{array}{c} x \\ y \end{array}\right]\)[/tex]
- [tex]\(B\)[/tex] is the column matrix of constants: [tex]\(B = \left[\begin{array}{c} 6 \\ 1 \end{array}\right]\)[/tex]
To find [tex]\(X\)[/tex], we use the relationship [tex]\(X = A^{-1}B\)[/tex], where [tex]\(A^{-1}\)[/tex] is the inverse of matrix [tex]\(A\)[/tex].
### Step 1: Find the determinant of [tex]\(A\)[/tex]
The determinant of matrix [tex]\(A = \left[\begin{array}{cc} 3 & 10 \\ 1 & 3 \end{array}\right]\)[/tex] is calculated as follows:
[tex]$ \text{det}(A) = (3 \cdot 3) - (10 \cdot 1) = 9 - 10 = -1 $[/tex]
### Step 2: Find the inverse of [tex]\(A\)[/tex]
The inverse of a 2x2 matrix [tex]\(A = \left[\begin{array}{cc} a & b \\ c & d \end{array}\right]\)[/tex] is given by:
[tex]$ A^{-1} = \frac{1}{\text{det}(A)} \left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right] $[/tex]
Using the values from [tex]\(A = \left[\begin{array}{cc} 3 & 10 \\ 1 & 3 \end{array}\right]\)[/tex]:
[tex]$ A^{-1} = \frac{1}{-1} \left[\begin{array}{cc} 3 & -10 \\ -1 & 3 \end{array}\right] = \left[\begin{array}{cc} -3 & 10 \\ 1 & -3 \end{array}\right] $[/tex]
### Step 3: Multiply [tex]\(A^{-1}\)[/tex] with [tex]\(B\)[/tex] to find [tex]\(X\)[/tex]
Now, we multiply [tex]\(A^{-1}\)[/tex] by [tex]\(B\)[/tex] to solve for [tex]\(X\)[/tex]:
[tex]$ X = A^{-1}B = \left[\begin{array}{cc} -3 & 10 \\ 1 & -3 \end{array}\right] \left[\begin{array}{c} 6 \\ 1 \end{array}\right] $[/tex]
Performing the matrix multiplication:
[tex]$ \left[\begin{array}{cc} -3 & 10 \\ 1 & -3 \end{array}\right] \left[\begin{array}{c} 6 \\ 1 \end{array}\right] = \left[\begin{array}{c} (-3 \cdot 6) + (10 \cdot 1) \\ (1 \cdot 6) + (-3 \cdot 1) \end{array}\right] = \left[\begin{array}{c} -18 + 10 \\ 6 - 3 \end{array}\right] = \left[\begin{array}{c} -8 \\ 3 \end{array}\right] $[/tex]
Thus, the solution to the given system of equations is:
[tex]$ x = -8, \quad y = 3 $[/tex]
We are given the matrix equation:
[tex]$ \left[\begin{array}{cc} 3 & 10 \\ 1 & 3 \end{array}\right] \left[\begin{array}{c} x \\ y \end{array}\right]=\left[\begin{array}{c} 6 \\ 1 \end{array}\right] $[/tex]
This matrix equation corresponds to the following system of linear equations:
1. [tex]\(3x + 10y = 6\)[/tex]
2. [tex]\(1x + 3y = 1\)[/tex]
To solve for [tex]\(x\)[/tex] and [tex]\(y\)[/tex] using matrices, we represent the equation as [tex]\(AX = B\)[/tex], where:
- [tex]\(A\)[/tex] is the coefficient matrix: [tex]\(A = \left[\begin{array}{cc} 3 & 10 \\ 1 & 3 \end{array}\right]\)[/tex]
- [tex]\(X\)[/tex] is the column matrix of the variables: [tex]\(X = \left[\begin{array}{c} x \\ y \end{array}\right]\)[/tex]
- [tex]\(B\)[/tex] is the column matrix of constants: [tex]\(B = \left[\begin{array}{c} 6 \\ 1 \end{array}\right]\)[/tex]
To find [tex]\(X\)[/tex], we use the relationship [tex]\(X = A^{-1}B\)[/tex], where [tex]\(A^{-1}\)[/tex] is the inverse of matrix [tex]\(A\)[/tex].
### Step 1: Find the determinant of [tex]\(A\)[/tex]
The determinant of matrix [tex]\(A = \left[\begin{array}{cc} 3 & 10 \\ 1 & 3 \end{array}\right]\)[/tex] is calculated as follows:
[tex]$ \text{det}(A) = (3 \cdot 3) - (10 \cdot 1) = 9 - 10 = -1 $[/tex]
### Step 2: Find the inverse of [tex]\(A\)[/tex]
The inverse of a 2x2 matrix [tex]\(A = \left[\begin{array}{cc} a & b \\ c & d \end{array}\right]\)[/tex] is given by:
[tex]$ A^{-1} = \frac{1}{\text{det}(A)} \left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right] $[/tex]
Using the values from [tex]\(A = \left[\begin{array}{cc} 3 & 10 \\ 1 & 3 \end{array}\right]\)[/tex]:
[tex]$ A^{-1} = \frac{1}{-1} \left[\begin{array}{cc} 3 & -10 \\ -1 & 3 \end{array}\right] = \left[\begin{array}{cc} -3 & 10 \\ 1 & -3 \end{array}\right] $[/tex]
### Step 3: Multiply [tex]\(A^{-1}\)[/tex] with [tex]\(B\)[/tex] to find [tex]\(X\)[/tex]
Now, we multiply [tex]\(A^{-1}\)[/tex] by [tex]\(B\)[/tex] to solve for [tex]\(X\)[/tex]:
[tex]$ X = A^{-1}B = \left[\begin{array}{cc} -3 & 10 \\ 1 & -3 \end{array}\right] \left[\begin{array}{c} 6 \\ 1 \end{array}\right] $[/tex]
Performing the matrix multiplication:
[tex]$ \left[\begin{array}{cc} -3 & 10 \\ 1 & -3 \end{array}\right] \left[\begin{array}{c} 6 \\ 1 \end{array}\right] = \left[\begin{array}{c} (-3 \cdot 6) + (10 \cdot 1) \\ (1 \cdot 6) + (-3 \cdot 1) \end{array}\right] = \left[\begin{array}{c} -18 + 10 \\ 6 - 3 \end{array}\right] = \left[\begin{array}{c} -8 \\ 3 \end{array}\right] $[/tex]
Thus, the solution to the given system of equations is:
[tex]$ x = -8, \quad y = 3 $[/tex]
