Solve $\begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 4 \\ 11 \end{pmatrix}$.

$\det(A) = 6 - 5 = 1$.

$x = \frac{\det\begin{pmatrix} 4 & 1 \\ 11 & 3 \end{pmatrix}}{\det(A)} = \frac{12 - 11}{1} = 1, \quad y = \frac{\det\begin{pmatrix} 2 & 4 \\ 5 & 11 \end{pmatrix}}{\det(A)} = \frac{22 - 20}{1} = 2.$

Solution: $(x, y) = (1, 2)$. Verify: $2(1) + 1(2) = 4$ and $5(1) + 3(2) = 11$.

Solve: $x + 2z = 6$, $-3x + 4y + 6z = 30$, $-x - 2y + 3z = 8$.

$A = \begin{pmatrix} 1 & 0 & 2 \\ -3 & 4 & 6 \\ -1 & -2 & 3 \end{pmatrix}, \quad b = \begin{pmatrix} 6 \\ 30 \\ 8 \end{pmatrix}.$

$\det(A) = 1(12+12) - 0 + 2(6+4) = 24 + 20 = 44$.

$x = \frac{\det\begin{pmatrix} 6 & 0 & 2 \\ 30 & 4 & 6 \\ 8 & -2 & 3 \end{pmatrix}}{44} = \frac{6(12+12) - 0 + 2(-60-32)}{44} = \frac{144 - 184}{44} = \frac{-40}{44} = -\frac{10}{11}.$

Similarly one computes $y$ and $z$.

Solve $ax + by = e$, $cx + dy = f$ with $ad - bc \neq 0$:

$x = \frac{ed - bf}{ad - bc}, \quad y = \frac{af - ce}{ad - bc}.$

Cramer's Rule gives a closed-form solution that makes dependence on parameters transparent.

If $\det(A) = 0$, Cramer's Rule does not apply. The system $Ax = b$ either has no solution or infinitely many solutions, depending on whether $b \in C(A)$.

$\begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3 \\ 6 \end{pmatrix}$: $\det = 0$. Infinitely many solutions: $x = 3 - 2t$, $y = t$.

$\begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3 \\ 7 \end{pmatrix}$: $\det = 0$ and $b \notin C(A)$. No solution.

A key advantage of Cramer's Rule: to find just $x_3$ in a $5 \times 5$ system, compute only two determinants ($\det(A)$ and $\det(A_3(b))$), rather than solving the full system. This can be more efficient than row reduction if only one variable is needed.

Cramer's Rule is rarely used for numerical computation (too expensive for large $n$), but it is invaluable in theory:

• It shows the solution of $Ax = b$ depends rationally on the entries of $A$ and $b$.
• It proves that solutions vary smoothly (or algebraically) with parameters.
• It is used in algebraic geometry to study families of linear systems.

Setting $b = e_j$ in $Ax = e_j$ gives the $j$-th column of $A^{-1}$:

$(A^{-1})_{ij} = \frac{\det(A_i(e_j))}{\det(A)} = \frac{C_{ji}}{\det(A)} = \frac{(\text{adj}(A))_{ij}}{\det(A)}.$

This recovers the adjugate formula $A^{-1} = \frac{1}{\det(A)}\text{adj}(A)$.

Solve $(1+i)z_1 + 2z_2 = 3$ and $iz_1 + (1-i)z_2 = 1$ over $\mathbb{C}$:

$\det(A) = (1+i)(1-i) - 2i = 2 - 2i$.

$z_1 = \frac{3(1-i) - 2}{2 - 2i} = \frac{1 - 3i}{2 - 2i}$. Rationalize by multiplying by $\overline{2-2i} = 2 + 2i$:

$z_1 = \frac{(1-3i)(2+2i)}{(2-2i)(2+2i)} = \frac{2 + 2i - 6i - 6i^2}{8} = \frac{8 - 4i}{8} = 1 - \frac{i}{2}$.

For $Ax = b$ in $\mathbb{R}^2$: $x_1 = \frac{\det(b, a_2)}{\det(a_1, a_2)}$ where $a_1, a_2$ are the columns of $A$.

$\det(b, a_2)$ is the signed area of the parallelogram spanned by $b$ and $a_2$, and $\det(a_1, a_2)$ is the signed area spanned by $a_1$ and $a_2$. The ratio gives the "proportion" of $a_1$ needed to reach $b$ from the $a_2$ direction.

Cramer's Rule is a consequence of the multilinear and alternating nature of the determinant. Replacing column $i$ by $b = \sum x_j a_j$ and using multilinearity:

$\det(a_1, \ldots, b, \ldots, a_n) = \sum_{j} x_j \det(a_1, \ldots, a_j, \ldots, a_n) = x_i \det(A)$

since $\det$ vanishes when two columns are equal. This is the deepest way to understand Cramer's Rule.

Cramer's Rule requires computing $n + 1$ determinants of $n \times n$ matrices. Using cofactor expansion, this is $O(n \cdot n!) = O((n+1)!)$. Even using LU decomposition for each determinant, it is $O(n \cdot n^3) = O(n^4)$, compared to $O(n^3)$ for Gaussian elimination on the augmented matrix.

Over a commutative ring $R$ (not necessarily a field), Cramer's Rule still holds in the form: $\det(A) \cdot x_i = \det(A_i(b))$. If $\det(A)$ is a unit in $R$, we can divide. If not, the formula still gives useful information (e.g., $\det(A)$ annihilates the cokernel of the map defined by $A$).