Problem 3
Question
Solve the given system of equations by Cramer's rule. $$ \begin{aligned} 0.1 x_{1}-0.4 x_{2} &=0.13 \\ x_{1}-\quad x_{2} &=0.4 \end{aligned} $$
Step-by-Step Solution
Verified Answer
The solution is \( x_1 = 0.1 \) and \( x_2 = -0.3 \).
1Step 1: Write the System in Matrix Form
Re-write the system of linear equations as a matrix equation of the form \( A\mathbf{x} = \mathbf{b} \), where \( A \) is the coefficient matrix, \( \mathbf{x} \) is the vector of variables, and \( \mathbf{b} \) is the constant vector. Matrix form:\[A = \begin{bmatrix} 0.1 & -0.4 \ 1 & -1 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x_1 \ x_2 \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 0.13 \ 0.4 \end{bmatrix}\]
2Step 2: Calculate the Determinant of Matrix A
Find the determinant of the coefficient matrix \( A \). The determinant \( \text{det}(A) \) for a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is calculated as \( ad - bc \).For our matrix:\[\text{det}(A) = (0.1)(-1) - (1)(-0.4) = -0.1 + 0.4 = 0.3\]
3Step 3: Calculate Determinant for x1 (det(A1))
Replace the first column of \( A \) with \( \mathbf{b} \) to form matrix \( A_1 \), then find its determinant.Matrix \( A_1 \):\[A_1 = \begin{bmatrix} 0.13 & -0.4 \ 0.4 & -1 \end{bmatrix}\]Determinant of \( A_1 \):\[\text{det}(A_1) = (0.13)(-1) - (-0.4)(0.4) = -0.13 + 0.16 = 0.03\]
4Step 4: Calculate Determinant for x2 (det(A2))
Replace the second column of \( A \) with \( \mathbf{b} \) to form matrix \( A_2 \), then find its determinant.Matrix \( A_2 \):\[A_2 = \begin{bmatrix} 0.1 & 0.13 \ 1 & 0.4 \end{bmatrix}\]Determinant of \( A_2 \):\[\text{det}(A_2) = (0.1)(0.4) - (1)(0.13) = 0.04 - 0.13 = -0.09\]
5Step 5: Calculate Solutions for x1 and x2 Using Cramer's Rule
Apply Cramer's Rule to solve for the variables. For \( x_1 \), it's calculated as \( \frac{\text{det}(A_1)}{\text{det}(A)} \), and for \( x_2 \) as \( \frac{\text{det}(A_2)}{\text{det}(A)} \).Thus,\[x_1 = \frac{0.03}{0.3} = 0.1\]\[x_2 = \frac{-0.09}{0.3} = -0.3\]
Key Concepts
Determinant CalculationMatrix FormLinear System of Equations
Determinant Calculation
A determinant is a special number that can be calculated from a square matrix. For a 2x2 matrix, it is relatively simple to find the determinant. Consider a matrix of the form:
- \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \)
- \( \text{det}(A) = ad - bc \)
Matrix Form
Matrix form is a way of representing a system of linear equations as a matrix equation of the form \( A\mathbf{x} = \mathbf{b} \). Here:
- \( A \) is the coefficient matrix containing the coefficients of the variables in the equations.
- \( \mathbf{x} \) is the vector of variables we want to solve for.
- \( \mathbf{b} \) is the constant vector that contains the values on the right side of the equations.
- \[ A = \begin{bmatrix} 0.1 & -0.4 \ 1 & -1 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x_1 \ x_2 \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 0.13 \ 0.4 \end{bmatrix} \]
Linear System of Equations
A linear system of equations is a collection of one or more linear equations involving the same set of variables. For example:
Such systems can be solved using various methods, including substitution, elimination, and Cramer's Rule. Cramer's Rule offers a systematic way to solve for the variables in a linear system when the number of equations matches the number of unknowns and the determinant of the coefficient matrix is non-zero. Understanding how to represent these equations in matrix form, calculate determinants, and apply rules like Cramer's helps in solving and analyzing these systems efficiently.
- \( 0.1x_1 - 0.4x_2 = 0.13 \)
- \( x_1 - x_2 = 0.4 \)
Such systems can be solved using various methods, including substitution, elimination, and Cramer's Rule. Cramer's Rule offers a systematic way to solve for the variables in a linear system when the number of equations matches the number of unknowns and the determinant of the coefficient matrix is non-zero. Understanding how to represent these equations in matrix form, calculate determinants, and apply rules like Cramer's helps in solving and analyzing these systems efficiently.
Other exercises in this chapter
Problem 3
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