Cramer’s Rule
What is Cramer’s rule?
Cramer’s Rule provides a method for solving systems of $n$ linear equations in $n$ unknowns, by using the determinant of the system’s coefficient matrix. This rule applies only when the coefficient matrix is square and its determinant is non-zero, ensuring that the system has a unique solution.
Applying Cramer’s Rule
Let us consider a general system of $n$ equations in $n$ unknowns:
\[\{ a_{11} x_{1} + a_{12} x_{2} + \hdots + a_{1 n} x_{n} = b_{1} \\ a_{21} x_{1} + a_{22} x_{2} + \hdots + a_{2 n} x_{n} = b_{2} \\ \vdots \\ a_{n 1} x_{1} + a_{n 2} x_{2} + \hdots + a_{n n} x_{n} = b_{n}\]We can rewrite the system using matrix notation as $A \cdot \mathbf{X} = \mathbf{B}$. The system becomes:
\[A = [ a_{11} & a_{12} & \hdots & a_{1 n} \\ a_{21} & a_{22} & \hdots & a_{2 n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m 1} & a_{n 2} & \hdots & a_{n n} ]\]The constant terms and variables can be organized into two column vectors:
\[X = [ x_{1} \\ x_{2} \\ \vdots \\ x_{n} ] B = [ b_{1} \\ b_{2} \\ \vdots \\ b_{m} ]\]Suppose that the coefficient matrix $A$ is invertible, which means that its determinant is non-zero ($det ( A ) \neq 0$). Under this condition, the system has a unique solution, and it can be expressed using the inverse of $A$ as:
\[[ x_{1} \\ x_{2} \\ \vdots \\ x_{n} ] = \frac{1}{det ( A )} [ A_{11} & A_{21} & \hdots & A_{n 1} \\ A_{12} & A_{22} & \hdots & A_{n 2} \\ \vdots & \vdots & \ddots & \vdots \\ A_{1 n} & A_{2 n} & \hdots & A_{n n} ] [ b_{1} \\ b_{2} \\ \vdots \\ b_{n} ]\]This is the general form of Cramer’s Rule, where each $A_{i j}$ is the cofactor of the element $a_{j i}$ in the original matrix $A$.
In this context, $A_{i j}$ refers to the cofactor of the element $a_{j i}$ from the original matrix $A$, not to the entry of the matrix itself. The matrix used here is the adjugate of $A$, denoted as $\text{adj} ( A )$.
The value of the unknown in position $k$ is given by a fraction. Its denominator is $det ( A )$ and its numerator is the determinant of the matrix obtained by replacing the $k$-th column of $A$ with the column of constants:
\[x_{k} = \frac{det ( A_{k} )}{det ( A )}\]For a clearer understanding of this step, see the detailed explanation in Example 2.
Solutions of homogeneous systems
A homogeneous system is a system of linear equations where all the constant terms are zero. These systems always have at least one solution: the trivial solution, where all the variables are zero. But something interesting happens when we look at the determinant of the coefficient matrix:
- If $det ( A ) \neq 0$, the system has only the trivial solution.
- If $det ( A ) = 0$, the system admits infinitely many solutions, including non-trivial ones, where at least one variable is not zero.
Homogeneous systems never have no solution. They’re always consistent, but the number of solutions depends entirely on the determinant.
Example 1
Let’s consider the following homogeneous system of three equations in three unknowns:
\[\{ x + y + z = 0 \\ 2 x - y + z = 0 \\ 3 x + y + 2 z = 0\]This system can be written in matrix form as $A \cdot \mathbf{x} = 0$, where the coefficient matrix is:
\[A = [ 1 & 1 & 1 \\ 2 & - 1 & 1 \\ 3 & 1 & 2 ]\]To understand the nature of the solutions, we compute the determinant of the matrix $A$:
\[det ( A ) & = 1 \cdot ( - 1 \cdot 2 - 1 \cdot 1 ) - 1 \cdot ( 2 \cdot 2 - 1 \cdot 3 ) + 1 \cdot ( 2 \cdot 1 - ( - 1 ) \cdot 3 ) \\ & = 1 ( - 2 - 1 ) - 1 ( 4 - 3 ) + 1 ( 2 + 3 ) \\ & = - 3 - 1 + 5 \\ & = 1\]Since the determinant is non-zero, the system admits only the trivial solution:
\[x = 0 y = 0 z = 0\]This outcome aligns with Cramer’s Rule: when $det ( A ) \neq 0$, the only possible solution to a homogeneous system is the trivial one, since all the determinants in the numerators of Cramer’s formula become zero.
Example 2
Let’s solve the following system of two linear equations in two unknowns:
\[\{ 2 x + 3 y = 8 \\ 4 x - y = 2\]We identify the coefficient matrix $A$, the vector of unknowns $\mathbf{x}$, and the constants vector $\mathbf{b}$:
\[A = [ 2 & 3 \\ 4 & - 1 ] \mathbf{x} = [ x \\ y ] \mathbf{b} = [ 8 \\ 2 ]\]We compute the determinant of the coefficient matrix $A$:
\[det ( A ) = 2 \cdot ( - 1 ) - 3 \cdot 4 = - 2 - 12 = - 14\]Since the determinant is non-zero, the system has exactly one solution, and we can apply Cramer’s rule.
We build the matrix $A_{1}$ by replacing the first column of $A$ with the constants:
\[A_{1} = [ 8 & 3 \\ 2 & - 1 ] \rightarrow det ( A_{1} ) = 8 \cdot ( - 1 ) - 3 \cdot 2 = - 8 - 6 = - 14\]We now build $A_{2}$ by replacing the second column of $A$ with the constants:
\[A_{2} = [ 2 & 8 \\ 4 & 2 ] \rightarrow det ( A_{2} ) = 2 \cdot 2 - 8 \cdot 4 = 4 - 32 = - 28\]Now we compute the values of the unknowns using the formula:
\[x & = \frac{det ( A_{1} )}{det ( A )} = \frac{- 14}{- 14} = 1 \\ y & = \frac{det ( A_{2} )}{det ( A )} = \frac{- 28}{- 14} = 2\]The solution to the system is:
\[x = 1 y = 2\]