Problem 22
Question
Solve using Cramer's rule. $$ \begin{aligned} x-y+2 z &=-3 \\ x+2 y+3 z &=4 \\ 2 x+y+z &=-3 \end{aligned} $$
Step-by-Step Solution
Verified Answer
The solutions are \( x = -\frac{5}{6} \), \( y = 2 \), \( z = 1 \).
1Step 1: Write the system of linear equations in matrix form
Express the system of equations as a matrix equation:\[\begin{pmatrix} 1 & -1 & 2 \ 1 & 2 & 3 \ 2 & 1 & 1 \end{pmatrix}\begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} -3 \ 4 \ -3 \end{pmatrix}\]
2Step 2: Find the determinant of the coefficient matrix
Calculate the determinant of the coefficient matrix:\[\Delta = \begin{vmatrix} 1 & -1 & 2 \ 1 & 2 & 3 \ 2 & 1 & 1 \end{vmatrix}\].Use the formula for determinants:\[\Delta = 1(2 \cdot 1 - 3 \cdot 1) - (-1)(1 \cdot 1 - 3 \cdot 2) + 2(1 \cdot 1 - 2 \cdot 2) = 1(-1) - (-1)(-5) + 2(-3) = -1 - 5 - 6 = -12\]
3Step 3: Find the determinant of the matrix for x
Replace the first column of the coefficient matrix with the constants vector and find the determinant:\[\Delta_x = \begin{vmatrix} -3 & -1 & 2 \ 4 & 2 & 3 \ -3 & 1 & 1 \end{vmatrix}\].Calculate the determinant:\[\Delta_x = (-3)(2 \cdot 1 - 1 \cdot 3) - (-1)(4 \cdot 1 - (-3) \cdot 3) + 2(4 \cdot 1 - 2 \cdot (-3)) = -3(-1) - (-1)(13) + 2(10) = 3 - 13 + 20 = 10\]
4Step 4: Find the determinant of the matrix for y
Replace the second column of the coefficient matrix with the constants vector and find the determinant:\[\Delta_y = \begin{vmatrix} 1 & -3 & 2 \ 1 & 4 & 3 \ 2 & -3 & 1 \end{vmatrix}\].Calculate the determinant:\[\Delta_y = 1(4 \cdot 1 - 3 \cdot (-3)) - (-3)(1 \cdot 1 - 3 \cdot 2) + 2(1 \cdot (-3) - 4 \cdot 2) = 1(13) - (-3)(-5) + 2(-11) = 13 - 15 - 22 = -24\]
5Step 5: Find the determinant of the matrix for z
Replace the third column of the coefficient matrix with the constants vector and find the determinant:\[\Delta_z = \begin{vmatrix} 1 & -1 & -3 \ 1 & 2 & 4 \ 2 & 1 & -3 \end{vmatrix}\].Calculate the determinant:\[\Delta_z = 1(2 \cdot (-3) - 4 \cdot 1) - (-1)(1 \cdot (-3) - 4 \cdot 2) + (-3)(1 \cdot 1 - 2 \cdot 2) = 1(-10) - (-1)(-11) + (-3)(-3) = -10 - 11 + 9 = -12\]
6Step 6: Solve for the variables using Cramer's Rule
Use Cramer's Rule to find values of x, y, and z:\[x = \frac{\Delta_x}{\Delta} = \frac{10}{-12} = -\frac{5}{6} \ y = \frac{\Delta_y}{\Delta} = \frac{-24}{-12} = 2 \ z = \frac{\Delta_z}{\Delta} = \frac{-12}{-12} = 1\]
Key Concepts
System of Linear EquationsMatrix DeterminantsSolving Linear Equations
System of Linear Equations
A system of linear equations consists of two or more linear equations with the same set of variables. The equations are typically written like this:
- Equation 1: \( a_1x + b_1y + c_1z = d_1 \)
- Equation 2: \( a_2x + b_2y + c_2z = d_2 \)
- Equation 3: \( a_3x + b_3y + c_3z = d_3 \)
Matrix Determinants
A determinant is a scalar value that can be computed from the elements of a square matrix. It provides essential properties about the matrix. Let's consider a 3x3 matrix, denoted by \( A \):\[ A = \begin{pmatrix} a & b & c \ d & e & f \ g & h & i \end{pmatrix} \]The determinant of matrix \( A \), denoted as \( \Delta \) or \( \text{det}(A) \), is calculated as follows:\[ \Delta = a(ei - fh) - b(di - fg) + c(dh - eg) \]Here, the calculation involves the sum and difference of products of the elements of the matrix, reflecting its geometric and algebraic properties. The determinant value can inform you if a matrix is singular (i.e., non-invertible) when it equals zero, or invertible when it doesn't. Matrix determinants are particularly useful in Cramer's rule for solving systems of linear equations.
Solving Linear Equations
To solve a system of linear equations using Cramer's rule, you need to follow these steps:
- Write the system of equations in matrix form.
- Find the determinant of the coefficient matrix (\( \Delta \)).
- Replace each column of the coefficient matrix by the constants vector to find the determinants for each variable (\( \Delta_x, \Delta_y, \Delta_z \)).
- Compute each variable using the ratios of these determinants: \( x = \frac{\Delta_x}{\Delta}, y = \frac{\Delta_y}{\Delta}, z = \frac{\Delta_z}{\Delta} \).
Other exercises in this chapter
Problem 21
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