Cramer’s Rule

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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:

\begin{cases} a_{11}x_1 + a_{12}x_2 + \dots + a_{1n}x_n = b_1 \\[0.5em] a_{21}x_1 + a_{22}x_2 + \dots + a_{2n}x_n = b_2 \\[0.5em] \quad\vdots \\[0.5em] a_{n1}x_1 + a_{n2}x_2 + \dots + a_{nn}x_n = b_n \end{cases}

We can rewrite the system using matrix notation as $A \cdot \mathbf{X} = \mathbf{B}$. The system becomes:

A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\\\ a_{21} & a_{22} & \cdots & a_{2n} \\\\ \vdots & \vdots & \ddots & \vdots \\\\ a_{m1} & a_{n2} & \cdots & a_{nn} \end{bmatrix}

The constant terms and variables can be organized into two column vectors:

X = \begin{bmatrix} x_1 \\\\ x_2 \\\\ \vdots \\\\ x_n \end{bmatrix} \quad\quad B = \begin{bmatrix} b_1 \\\\ b_2 \\\\ \vdots \\\\ b_m \end{bmatrix}

Suppose that the coefficient matrix $A$ is invertible, which means that its determinant is non-zero ($\det(A) \ne 0$). Under this condition, the system has a unique solution, and it can be expressed using the inverse of $A$ as:

\begin{bmatrix} x_1 \\\\ x_2 \\\\ \vdots \\\\ x_n \end{bmatrix} = \frac{1}{\det(A)} \begin{bmatrix} A_{11} & A_{21} & \cdots & A_{n1} \\\\ A_{12} & A_{22} & \cdots & A_{n2} \\\\ \vdots & \vdots & \ddots & \vdots \\\\ A_{1n} & A_{2n} & \cdots & A_{nn} \\\\ \end{bmatrix} \begin{bmatrix} b_1 \\\\ b_2 \\\\ \vdots \\\\ b_n \end{bmatrix}

This is the general form of Cramer’s Rule, where each $A_{ij}$ is the cofactor of the element $a_{ji}$ in the original matrix $A$.

In this context, $A_{ij}$ refers to the cofactor of the element $a_{ji}$ 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 $\det(A) \ne 0$, 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:

\begin{cases} x + y + z = 0 \\[0.5em] 2x - y + z = 0 \\[0.5em] 3x + y + 2z = 0 \end{cases}

This system can be written in matrix form as $A \cdot \mathbf{x} = \mathbf{0}$, where the coefficient matrix is:

A = \begin{bmatrix} 1 & 1 & 1 \\[0.5em] 2 & -1 & 1 \\[0.5em] 3 & 1 & 2 \end{bmatrix}

To understand the nature of the solutions, we compute the determinant of the matrix $A$:

\begin{align*} \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) \\[0.5em] &= 1(-2 - 1) - 1(4 - 3) + 1(2 + 3) \\[0.5em] &= -3 - 1 + 5 \\[0.5em] & = 1 \end{align*}

Since the determinant is non-zero, the system admits only the trivial solution:

x = 0 \quad y = 0 \quad z = 0
This outcome aligns with Cramer’s Rule: when $\det(A) \ne 0$, 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:

\begin{cases} 2x + 3y = 8 \\[0.5em] 4x - y = 2 \end{cases}

We identify the coefficient matrix $A$, the vector of unknowns $\mathbf{x}$, and the constants vector $\mathbf{b}$:

A = \begin{bmatrix} 2 & 3 \\\\ 4 & -1 \end{bmatrix} \quad\quad \mathbf{x} = \begin{bmatrix} x \\\\ y \end{bmatrix} \quad\quad \mathbf{b} = \begin{bmatrix} 8 \\\\ 2 \end{bmatrix}

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 = \begin{bmatrix} 8 & 3 \\\\ 2 & -1 \end{bmatrix} \quad \rightarrow \quad \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 = \begin{bmatrix} 2 & 8 \\\\ 4 & 2 \end{bmatrix} \quad \rightarrow \quad \det(A_2) = 2 \cdot 2 - 8 \cdot 4 = 4 - 32 = -28

Now we compute the values of the unknowns using the formula:

\begin{align*} x &= \frac{\det(A_1)}{\det(A)} = \frac{-14}{-14} = 1 \\[0.5em] y &= \frac{\det(A_2)}{\det(A)} = \frac{-28}{-14} = 2 \end{align*}

The solution to the system is:

x = 1 \quad\quad y = 2
Remember that the solution to a system of linear equations refers to the $n$-tuple of values that satisfies all equations in the system simultaneously. In this case, the pair $(x, y) = (1, 2)$ is the only combination of values that makes both equations true at the same time.