Cramer’s Rule is a mathematical theorem used in linear algebra to solve systems of linear equations with as many equations as unknowns. Named after the Swiss mathematician Gabriel Cramer, this rule provides an explicit formula for the solution of a system of linear equations with a unique solution.
System of Linear Equations
Consider a system of linear equations:
$begin{align*}
a_1x + b_1y + c_1z &= d_1 \
a_2x + b_2y + c_2z &= d_2 \
a_3x + b_3y + c_3z &= d_3
end{align*}$
This can be written in matrix form as $AX = B$, where $A$ is the coefficient matrix, $X$ is the column vector of variables, and $B$ is the column vector of constants.
Determinants and Cramer’s Rule
Cramer’s Rule states that if the determinant of the coefficient matrix $A$ (denoted as $det(A)$) is non-zero, the system has a unique solution. The solutions for the variables can be found using the formula:
$X_i = frac{det(A_i)}{det(A)}$
where $det(A_i)$ is the determinant of the matrix obtained by replacing the $i$-th column of $A$ with the column vector $B$
Example
Let’s solve the following system using Cramer’s Rule:
$begin{align*}
2x + 3y – z &= 5 \
4x – y + 2z &= 6 \
-3x + 2y + 4z &= 8
end{align*}$
First, we write the coefficient matrix $A$ and the column vector $B$:
$A = begin{pmatrix} 2 & 3 & -1 \ 4 & -1 & 2 \ -3 & 2 & 4 end{pmatrix}, quad B = begin{pmatrix} 5 \ 6 \ 8 end{pmatrix}$
Calculate $det(A)$:
$det(A) = 2 begin{vmatrix} -1 & 2 \ 2 & 4 end{vmatrix} – 3 begin{vmatrix} 4 & 2 \ -3 & 4 end{vmatrix} – 1 begin{vmatrix} 4 & -1 \ -3 & 2 end{vmatrix}$
$= 2(-1 cdot 4 – 2 cdot 2) – 3(4 cdot 4 – (-3) cdot 2) – 1(4 cdot 2 – (-3) cdot -1)$
$= 2(-4 – 4) – 3(16 + 6) – 1(8 + 3)$
$= 2(-8) – 3(22) – 1(11)$
$= -16 – 66 – 11 = -93$
Since $det(A)
eq 0$, we can use Cramer’s Rule.
Solution for $x$
Replace the first column of $A$ with $B$ to get $A_1$:
$A_1 = begin{pmatrix} 5 & 3 & -1 \ 6 & -1 & 2 \ 8 & 2 & 4 end{pmatrix}$
Calculate $det(A_1)$:
$det(A_1) = 5 begin{vmatrix} -1 & 2 \ 2 & 4 end{vmatrix} – 3 begin{vmatrix} 6 & 2 \ 8 & 4 end{vmatrix} – (-1) begin{vmatrix} 6 & -1 \ 8 & 2 end{vmatrix}$
$= 5(-1 cdot 4 – 2 cdot 2) – 3(6 cdot 4 – 8 cdot 2) + 1(6 cdot 2 – (-1) cdot 8)$
$= 5(-4 – 4) – 3(24 – 16) + 1(12 + 8)$
$= 5(-8) – 3(8) + 1(20)$
$= -40 – 24 + 20 = -44$
Thus, $x = frac{det(A_1)}{det(A)} = frac{-44}{-93} = frac{44}{93}$
Solution for $y$ and $z$
Following similar steps, we can find $y$ and $z$. Replace the second and third columns of $A$ with $B$ to get $A_2$ and $A_3$ and calculate their determinants. Finally, use the formula $y = frac{det(A_2)}{det(A)}$ and $z = frac{det(A_3)}{det(A)}$
Conclusion
Cramer’s Rule is a powerful tool for solving systems of linear equations, especially when the number of equations matches the number of unknowns. While it may be computationally intensive for large systems, it provides an elegant solution for smaller ones.