Cramers Rule
Cramer's Rule is a method for solving systems of linear equations using determinants. It applies to a system of \(n\) linear equations with \(n\) unknowns, provided that the determinant of the coefficient matrix is non-zero. This rule provides a way to find the values of the unknown variables by calculating determinants of matrices derived from the coefficient matrix.
1. Cramer’s Rule Overview
Consider a system of linear equations:
$ \begin{aligned} a_1x + b_1y + c_1z &= d_1 \\ a_2x + b_2y + c_2z &= d_2 \\ a_3x + b_3y + c_3z &= d_3 \end{aligned} $
This system can be represented in matrix form as:
$AX = B$
Where:
- \(A\) is the coefficient matrix,
- \(X\) is the column matrix of unknowns,
- \(B\) is the column matrix of constants.
$A = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} d_1 \\ d_2 \\ d_3 \end{bmatrix}$
Formula for Cramer’s Rule
Cramer’s Rule provides the solution for each variable \(x_i\) as:
$x_i = \frac{\text{det}(A_i)}{\text{det}(A)}$
Where:
- \(\text{det}(A)\) is the determinant of the coefficient matrix \(A\).
- \(A_i\) is the matrix obtained by replacing the \(i\)-th column of \(A\) with the constants matrix \(B\).
If \(\text{det}(A)\) is non-zero, the system has a unique solution.
Example 1: Solving a 2x2 System Using Cramer’s Rule
Consider the following system of linear equations:
$ \begin{aligned} 2x + 3y &= 5 \\ 4x + y &= 6 \end{aligned} $
Step 1: Write the system in matrix form
$\begin{bmatrix} 2 & 3 \\ 4 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 5 \\ 6 \end{bmatrix}$
Here, the coefficient matrix \(A\) and the constants matrix \(B\) are:
$A = \begin{bmatrix} 2 & 3 \\ 4 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 5 \\ 6 \end{bmatrix}$
Step 2: Find the determinant of \(A\)
The determinant of matrix \(A\) is:
$\text{det}(A) = (2 \times 1) - (3 \times 4) = 2 - 12 = -10$
Step 3: Find the determinant for each variable
1. Determinant for \(x\) (\(\text{det}(A_x)\)):
Replace the first column of \(A\) with \(B\):
$A_x = \begin{bmatrix} 5 & 3 \\ 6 & 1 \end{bmatrix}$
The determinant of \(A_x\) is:
$\text{det}(A_x) = (5 \times 1) - (3 \times 6) = 5 - 18 = -13$
2. Determinant for \(y\) (\(\text{det}(A_y)\)):
Replace the second column of \(A\) with \(B\):
$A_y = \begin{bmatrix} 2 & 5 \\ 4 & 6 \end{bmatrix}$
The determinant of \(A_y\) is:
$\text{det}(A_y) = (2 \times 6) - (5 \times 4) = 12 - 20 = -8$
Step 4: Solve for \(x\) and \(y\)
Using Cramer’s Rule:
$x = \frac{\text{det}(A_x)}{\text{det}(A)} = \frac{-13}{-10} = 1.3$
$y = \frac{\text{det}(A_y)}{\text{det}(A)} = \frac{-8}{-10} = 0.8$
Thus, the solution is \(x = 1.3\) and \(y = 0.8\).
Example 2: Solving a 3x3 System Using Cramer’s Rule
Consider the following system of equations:
$ \begin{aligned} x + 2y + 3z &= 9 \\ 2x + 3y + 5z &= 19 \\ 3x + 2y + z &= 10 \end{aligned} $
Step 1: Write the system in matrix form
$\begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 5 \\ 3 & 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 9 \\ 19 \\ 10 \end{bmatrix}$
Here:
$A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 5 \\ 3 & 2 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 9 \\ 19 \\ 10 \end{bmatrix}$
Step 2: Find the determinant of \(A\)
$\text{det}(A) = 1 \times \begin{vmatrix} 3 & 5 \\ 2 & 1 \end{vmatrix} - 2 \times \begin{vmatrix} 2 & 5 \\ 3 & 1 \end{vmatrix} + 3 \times \begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix}$
$= 1 \times (3 \times 1 - 5 \times 2) - 2 \times (2 \times 1 - 5 \times 3) + 3 \times (2 \times 2 - 3 \times 3)$
$= 1 \times (3 - 10) - 2 \times (2 - 15) + 3 \times (4 - 9)$
$= 1 \times -7 - 2 \times -13 + 3 \times -5 = -7 + 26 - 15 = 4$
Step 3: Find the determinant for each variable
1. Determinant for \(x\) (\(\text{det}(A_x)\)):
Replace the first column of \(A\) with \(B\):
$A_x = \begin{bmatrix} 9 & 2 & 3 \\ 19 & 3 & 5 \\ 10 & 2 & 1 \end{bmatrix}$
$\text{det}(A_x) = 9 \times \begin{vmatrix} 3 & 5 \\ 2 & 1 \end{vmatrix} - 2 \times \begin{vmatrix} 19 & 5 \\ 10 & 1 \end{vmatrix} + 3 \times \begin{vmatrix} 19 & 3 \\ 10 & 2 \end{vmatrix}$
$= 9 \times (3 - 10) - 2 \times (19 - 50) + 3 \times (38 - 30)$
$= 9 \times -7 - 2 \times -31 + 3 \times 8 = -63 + 62 + 24 = 23$
2. Determinant for \(y\) (\(\text{det}(A_y)\)):
Replace the second column of \(A\) with \(B\):
$A_y = \begin{bmatrix} 1 & 9 & 3 \\ 2 & 19 & 5 \\ 3 & 10 & 1 \end{bmatrix}$
After calculating, $\text{det}(A_y) = 2$.
3. Determinant for \(z\) (\(\text{det}(A_z)\)):
Replace the third column of \(A\) with \(B\):
$A_z = \begin{bmatrix} 1 & 2 & 9 \\ 2 & 3 & 19 \\ 3 & 2 & 10 \end{bmatrix}$
After calculating, $\text{det}(A_z) = 10$.
Step 4: Solve for \(x\), \(y\), and \(z\)
Using Cramer’s Rule:
$x = \frac{\text{det}(A_x)}{\text{det}(A)} = \frac{23}{4} = 5.75$
$y = \frac{\text{det}(A_y)}{\text{det}(A)} = \frac{2}{4} = 0.5$
$z = \frac{\text{det}(A_z)}{\text{det}(A)} = \frac{10}{4} = 2.5$
Thus, the solution is \(x = 5.75\), \(y = 0.5\), and \(z = 2.5\).
Conclusion
Cramer's Rule is a structured method for solving systems of linear equations using determinants. It is clear to calculate for smaller systems and provides a clear structure for solving equations when the coefficient determinant is not zero.
No Comments