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Cramers Rule

Cramer's Rule is a method for solving systems of linear equations using determinants. It applies to a system of n\(n\) linear equations with n\(n\) unknowns, provided that the determinant of the coefficient matrix has ais non-zero determinant.zero. This rule provides a straightforward 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:

[ $AX = BB$ ]

Where:

  • A\(A\) is the coefficient matrix,
  • X\(X\) is the column matrix of unknowns,
  • B\(B\) 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 \(x_i)x_i\) as:

[ $x_i = \frac{\text{det}(A_i)}{\text{det}(A)}$ ]

Where:

  • \(\text{det}(A)\) is the determinant of the coefficient matrix A\(A\).
  • \(A_i)A_i\) is the matrix obtained by replacing the \(i)i\)-th column of A\(A\) with the constants matrix B\(B\).

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\(A\) and the constants matrix B\(B\) 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\(A\)

The determinant of matrix A\(A\) is:

[ $\text{det}(A) = (2 \times 1) - (3 \times 4) = 2 - 12 = -1010$ ]

Step 3: Find the determinant for each variable

    1.
  1. Determinant for \(x)x\) (\(\text{det}(A_x)\)):

    2. Replace the first column of A\(A\) with B\(B\):

[ $A_x = \begin{bmatrix} 5 & 3 \\ 6 & 1 \end{bmatrix}$ ]

The determinant of \(A_x)A_x\) is:

[ $\text{det}(A_x) = (5 \times 1) - (3 \times 6) = 5 - 18 = -1313$ ]

    2.
  1. Determinant for \(y)y\) (\(\text{det}(A_y)\)):

    2. Replace the second column of A\(A\) with B\(B\):

[ $A_y = \begin{bmatrix} 2 & 5 \\ 4 & 6 \end{bmatrix}$ ]

The determinant of \(A_y)A_y\) is:

[ $\text{det}(A_y) = (2 \times 6) - (5 \times 4) = 12 - 20 = -88$ ]

Step 4: Solve for \(x)x\) and \(y)y\)

Using Cramer’s Rule:

[ $x = \frac{\text{det}(A_x)}{\text{det}(A)} = \frac{-13}{-10} = 1.33$ ]

[ $y = \frac{\text{det}(A_y)}{\text{det}(A)} = \frac{-8}{-10} = 0.88$ ]

Thus, the solution is \(x = 1.3)3\) and \(y = 0.8)8\).

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\(A\)

[ $\text{det}(A) = 1 \times \begin{vmatrix} 3 & 5 \\ 2 & 1 \end{vmatrix} - 2 \times \begin{vmatrix} 2 & 5 \\ 3 & 1 \end{vmatrix} + 3 \times \begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix}$ ]

[ $= 1 \times (3 \times 1 - 5 \times 2) - 2 \times (2 \times 1 - 5 \times 3) + 3 \times (2 \times 2 - 3 \times 3)$ ]

[

$= 1 \times (3 - 10) - 2 \times (2 - 15) + 3 \times (4 - 9)$ ]

[

$= 1 \times -7 - 2 \times -13 + 3 \times -5 = -7 + 26 - 15 = 44$ ]

Step 3: Find the determinant for each variable

    1.
  1. Determinant for \(x)x\) (\(\text{det}(A_x)\)):

    2. Replace the first column of A\(A\) with B\(B\):

[ $A_x = \begin{bmatrix} 9 & 2 & 3 \\ 19 & 3 & 5 \\ 10 & 2 & 1 \end{bmatrix}$ ]

[ $\text{det}(A_x) = 9 \times \begin{vmatrix} 3 & 5 \\ 2 & 1 \end{vmatrix} - 2 \times \begin{vmatrix} 19 & 5 \\ 10 & 1 \end{vmatrix} + 3 \times \begin{vmatrix} 19 & 3 \\ 10 & 2 \end{vmatrix}$ ]

[

$= 9 \times (3 - 10) - 2 \times (19 - 50) + 3 \times (38 - 30)$ ]

[

$= 9 \times -7 - 2 \times -31 + 3 \times 8 = -63 + 62 + 24 = 2323$ ]

    2.
  1. Determinant for \(y)y\) (\(\text{det}(A_y)\)):

    2. Replace the second column of A\(A\) with B\(B\):

[ $A_y = \begin{bmatrix} 1 & 9 & 3 \\ 2 & 19 & 5 \\ 3 & 10 & 1 \end{bmatrix}$ ]

After calculating, ($\text{det}(A_y) = 2)2$.

    3.
  1. Determinant for \(z)z\) (\(\text{det}(A_z)\)):

    2. Replace the third column of A\(A\) with B\(B\):

[ $A_z = \begin{bmatrix} 1 & 2 & 9 \\ 2 & 3 & 19 \\ 3 & 2 & 10 \end{bmatrix}$ ]

After calculating, ($\text{det}(A_z) = 10)10$.

Step 4: Solve for \(x)x\), \(y)y\), and \(z)z\)

Using Cramer’s Rule:

[ $x = \frac{\text{det}(A_x)}{\text{det}(A)} = \frac{23}{4} = 5.7575$ ]

[

$y = \frac{\text{det}(A_y)}{\text{det}(A)} = \frac{2}{4} = 0.55$ ]

[

$z = \frac{\text{det}(A_z)}{\text{det}(A)} = \frac{10}{4} = 2.55$ ]

Thus, the solution is \(x = 5.75)75\), \(y = 0.5)5\), and \(z = 2.5)5\).

Conclusion

Cramer's Rule is a straightforward and effectivestructured method for solving small systems of linear equations using determinants. WhileIt itis becomesclear inefficientto calculate for largesmaller systems,systems itand provides a clear andstructure structuredfor waysolving toequations solve systems where the determinant ofwhen the coefficient matrixdeterminant is non-zero,not offeringzero. insight into the relationships between variables.

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