At Westonci.ca, we connect you with experts who provide detailed answers to your most pressing questions. Start exploring now! Discover reliable solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
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.
We appreciate your time. Please come back anytime for the latest information and answers to your questions. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.