Answer :
Certainly! To solve the given system of linear equations using inverse matrices, let's follow the detailed step-by-step solution:
### Step 1: Write the system of equations in matrix form
The system of linear equations given is:
[tex]\[ \begin{cases} 5x - 2y = 30 \\ x + 6y = -26 \end{cases} \][/tex]
We can represent this system in matrix form as [tex]\(Ax = B\)[/tex], where:
[tex]\[ A = \begin{pmatrix} 5 & -2 \\ 1 & 6 \end{pmatrix}, \quad x = \begin{pmatrix} x \\ y \end{pmatrix}, \quad B = \begin{pmatrix} 30 \\ -26 \end{pmatrix} \][/tex]
Here [tex]\(A\)[/tex] is the coefficient matrix, [tex]\(x\)[/tex] is the vector of variables, and [tex]\(B\)[/tex] is the constant matrix.
### Step 2: Find the inverse of the coefficient matrix [tex]\(A\)[/tex]
The inverse of matrix [tex]\(A\)[/tex], denoted as [tex]\(A^{-1}\)[/tex], is needed to solve for [tex]\(x\)[/tex]. Given:
[tex]\[ A = \begin{pmatrix} 5 & -2 \\ 1 & 6 \end{pmatrix} \][/tex]
The inverse matrix [tex]\(A^{-1}\)[/tex] can be represented as:
[tex]\[ A^{-1} = \begin{pmatrix} 0.1875 & 0.0625 \\ -0.03125 & 0.15625 \end{pmatrix} \][/tex]
### Step 3: Multiply [tex]\(A^{-1}\)[/tex] by [tex]\(B\)[/tex] to find [tex]\(x\)[/tex]
To find the solution vector [tex]\(x\)[/tex], we use the relationship:
[tex]\[ x = A^{-1}B \][/tex]
Now, let's perform the matrix multiplication:
[tex]\[ x = \begin{pmatrix} 0.1875 & 0.0625 \\ -0.03125 & 0.15625 \end{pmatrix} \begin{pmatrix} 30 \\ -26 \end{pmatrix} \][/tex]
### Step 4: Calculate the resulting vector
Perform the multiplication:
[tex]\[ \begin{pmatrix} 0.1875 & 0.0625 \\ -0.03125 & 0.15625 \end{pmatrix} \begin{pmatrix} 30 \\ -26 \end{pmatrix} = \begin{pmatrix} (0.1875 \cdot 30) + (0.0625 \cdot -26) \\ (-0.03125 \cdot 30) + (0.15625 \cdot -26) \end{pmatrix} \][/tex]
Calculate each component:
[tex]\[ \begin{pmatrix} (0.1875 \cdot 30) + (0.0625 \cdot -26) \\ (-0.03125 \cdot 30) + (0.15625 \cdot -26) \end{pmatrix} = \begin{pmatrix} 5.625 - 1.625 \\ -0.9375 - 4.0625 \end{pmatrix} = \begin{pmatrix} 4 \\ -5 \end{pmatrix} \][/tex]
### Step 5: Interpret the results
Therefore, the solution to the system of equations is:
[tex]\[ \begin{cases} x = 4 \\ y = -5 \end{cases} \][/tex]
So, the solution is:
[tex]\[ \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 4 \\ -5 \end{pmatrix} \][/tex]
This represents the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] that satisfy both equations in the given system.
### Step 1: Write the system of equations in matrix form
The system of linear equations given is:
[tex]\[ \begin{cases} 5x - 2y = 30 \\ x + 6y = -26 \end{cases} \][/tex]
We can represent this system in matrix form as [tex]\(Ax = B\)[/tex], where:
[tex]\[ A = \begin{pmatrix} 5 & -2 \\ 1 & 6 \end{pmatrix}, \quad x = \begin{pmatrix} x \\ y \end{pmatrix}, \quad B = \begin{pmatrix} 30 \\ -26 \end{pmatrix} \][/tex]
Here [tex]\(A\)[/tex] is the coefficient matrix, [tex]\(x\)[/tex] is the vector of variables, and [tex]\(B\)[/tex] is the constant matrix.
### Step 2: Find the inverse of the coefficient matrix [tex]\(A\)[/tex]
The inverse of matrix [tex]\(A\)[/tex], denoted as [tex]\(A^{-1}\)[/tex], is needed to solve for [tex]\(x\)[/tex]. Given:
[tex]\[ A = \begin{pmatrix} 5 & -2 \\ 1 & 6 \end{pmatrix} \][/tex]
The inverse matrix [tex]\(A^{-1}\)[/tex] can be represented as:
[tex]\[ A^{-1} = \begin{pmatrix} 0.1875 & 0.0625 \\ -0.03125 & 0.15625 \end{pmatrix} \][/tex]
### Step 3: Multiply [tex]\(A^{-1}\)[/tex] by [tex]\(B\)[/tex] to find [tex]\(x\)[/tex]
To find the solution vector [tex]\(x\)[/tex], we use the relationship:
[tex]\[ x = A^{-1}B \][/tex]
Now, let's perform the matrix multiplication:
[tex]\[ x = \begin{pmatrix} 0.1875 & 0.0625 \\ -0.03125 & 0.15625 \end{pmatrix} \begin{pmatrix} 30 \\ -26 \end{pmatrix} \][/tex]
### Step 4: Calculate the resulting vector
Perform the multiplication:
[tex]\[ \begin{pmatrix} 0.1875 & 0.0625 \\ -0.03125 & 0.15625 \end{pmatrix} \begin{pmatrix} 30 \\ -26 \end{pmatrix} = \begin{pmatrix} (0.1875 \cdot 30) + (0.0625 \cdot -26) \\ (-0.03125 \cdot 30) + (0.15625 \cdot -26) \end{pmatrix} \][/tex]
Calculate each component:
[tex]\[ \begin{pmatrix} (0.1875 \cdot 30) + (0.0625 \cdot -26) \\ (-0.03125 \cdot 30) + (0.15625 \cdot -26) \end{pmatrix} = \begin{pmatrix} 5.625 - 1.625 \\ -0.9375 - 4.0625 \end{pmatrix} = \begin{pmatrix} 4 \\ -5 \end{pmatrix} \][/tex]
### Step 5: Interpret the results
Therefore, the solution to the system of equations is:
[tex]\[ \begin{cases} x = 4 \\ y = -5 \end{cases} \][/tex]
So, the solution is:
[tex]\[ \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 4 \\ -5 \end{pmatrix} \][/tex]
This represents the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] that satisfy both equations in the given system.
