Cramer’s rule

1.2 The method of determinants

Suppose that we wish to solve the following pair of equations for $x$ and $y$

$$ ax + by = c $$

$$ lx + my = n $$

where $a, b, c, l, m$ and $n$ are constants.

Multiplying the first equation by $m$, the second equation by $b$ and subtracting eliminates $y$ leaving us with the linear equation in $x$

$$ (am - lb)x = cm - nb . $$

Similarly, multiplying the first equation by $l$, the second by $a$ and subtracting would eliminate $x$ and give us a linear equation in $y$ only

$$ (bl - am)y = cl - na . $$

The solution to the pair of equations is then

$$ x = \frac{cm - nb}{am - lb}, \quad y = \frac{an - lc}{am - lb} $$

provided $am - lb \neq 0$. The denominator in these expressions for $x$ and $y$ is called the determinant of the coefficients of $x$ and $y$ and is defined by the notation

$$ \begin{vmatrix} a & b \\ l & m \end{vmatrix} = am - lb . $$

This is called a second order determinant because it has 2 rows and 2 columns bounded by two vertical lines, not brackets. A third order determinant would have 3 rows and 3 columns. A determinant has a single value, for example

$$ \begin{vmatrix} 2 & 4 \\ 1 & 3 \end{vmatrix} = 2 \times 3 - 4 \times 1 = 2 . $$

Notice now that $x$ and $y$ can be written as fractions involving two determinants

$$ x = \frac{\begin{vmatrix} c & b \\ n & m \end{vmatrix}}{\begin{vmatrix} a & b \\ l & m \end{vmatrix}},  \quad y = \frac{\begin{vmatrix} a & c \\ l & n \end{vmatrix}}{\begin{vmatrix} a & b \\ l & m \end{vmatrix}} . $$

The determinants in the numerators are very similar to the denominator. We write the solution in the form

$$ x = \frac{\Delta_x}{\Delta}, \quad y = \frac{\Delta_y}{\Delta} $$

where

$$ \Delta = \begin{vmatrix} a & b \\ l & m \end{vmatrix},  \quad \Delta_x = \begin{vmatrix} c & b \\ n & m \end{vmatrix},  \quad \Delta_y = \begin{vmatrix} a & c \\ l & n \end{vmatrix} $$

The determinants $\Delta_x$ and $\Delta_y$ are obtained from $\Delta$ by replacing the first and second columns respectively by the right-hand sides of the original linear equations.

Remember, the solution is defined only if $\Delta \neq 0$. This approach is called the method of determinants, or Cramer’s rule.

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Please tell me more about Cramer’s rule.

Here’s the big picture and the useful details about Cramer’s rule.

What it is (general form)

For a square linear system $A\mathbf{x}=\mathbf{b}$ of size $n\times n$ with $\det(A)\neq 0$,

$$ x_i=\frac{\det(A_i)}{\det(A)}\quad(i=1,\dots,n), $$

where $A_i$ is the matrix $A$ with its $i$-th column replaced by $\mathbf{b}$. That’s exactly what you saw for $n=2$. It works over real or complex numbers.

When it applies

• Unique solution: only if $\det(A)\ne 0$ (i.e., $A$ is invertible).
• $\det(A)=0$: either no solution or infinitely many. (Rank test: if $\operatorname{rank}(A)=\operatorname{rank}([A|\mathbf{b}])<n$ → infinitely many; if ranks differ → no solution.)

Why it works (connection to the adjugate)

The matrix inverse has the explicit form

$$ A^{-1}=\frac{1}{\det(A)}\,\operatorname{adj}(A), $$

where $\operatorname{adj}(A)$ is built from cofactors. Multiplying $A^{-1}\mathbf{b}$ yields the same “replace-a-column” determinants in each component—those are the $\det(A_i)$.

A quick 3×3 example

Solve

$$ \begin{cases} 2x-y+3z=9\\ x+4y-z=1\\ 3x+2y+2z=12 \end{cases} \quad\Longleftrightarrow\quad A=\begin{bmatrix}2&-1&3\\[2pt]1&4&-1\\[2pt]3&2&2\end{bmatrix},\ \mathbf{b}=\begin{bmatrix}9\\1\\12\end{bmatrix}. $$

Compute determinants:

$$ \Delta=\det(A)=-5,\quad \Delta_x=\det\!\begin{bmatrix}\mathbf{b}&-1&3\\1&4&-1\\12&2&2\end{bmatrix}=-34,\quad \Delta_y=10,\quad \Delta_z=11. $$

Hence

$$ x=\frac{-34}{-5}=\tfrac{34}{5},\quad y=\frac{10}{-5}=-2,\quad z=\frac{11}{-5}=-\tfrac{11}{5}. $$

Geometric intuition (2×2 case)

$|\det(A)|$ is the area scaling of the unit square under the linear map $A$. Replacing a column by $\mathbf{b}$ forms a new parallelogram; the ratio $\det(A_i)/\det(A)$ extracts the coordinate $x_i$.

Pros & cons

Pros

• Gives closed-form expressions for each variable.
• Great for symbolic work (e.g., showing how solutions change with parameters) and small systems (2×2, occasionally 3×3).

Cons

• Not computationally efficient for large $n$: computing $n+1$ determinants is expensive (typically $O(n^4)$ if done via LU per determinant).
• Numerically less stable than Gaussian elimination/QR for floating-point data; determinants can overflow/underflow and amplify conditioning issues.

Practical tips

• For hand calculations: use it for 2×2 and 3×3 (Sarrus’ rule helps for 3×3).
• For larger systems or numerics: prefer Gaussian elimination, LU/QR factorization, or direct $A^{-1}\mathbf{b}$ via a solver.
• If $\det(A)$ is (nearly) zero, the system is ill-conditioned or singular—avoid Cramer’s rule and use a stable solver and conditioning checks.

If you want, I can walk you through Sarrus’ rule for 3×3 determinants or set up a couple of practice problems with solutions.

Yes, please tell me about Sarrus’ rule.

happy to! Sarrus’ rule is a quick way to compute only 3×3 determinants.

What it is

For

$$ \begin{vmatrix} a&b&c\\ d&e&f\\ g&h&i \end{vmatrix} $$

Sarrus’ rule says

$$ \det = aei + bfg + cdh \;-\; ceg - bdi - afh. $$

How to do it (the “copy columns” trick)

1. Rewrite the first two columns to the right:

$$ \begin{array}{ccc|cc} a&b&c& a&b\\ d&e&f& d&e\\ g&h&i& g&h \end{array} $$

2. Multiply the three down-right diagonals and add:

$$ (a\cdot e\cdot i) + (b\cdot f\cdot g) + (c\cdot d\cdot h). $$

3. Multiply the three up-right diagonals and add:

$$ (c\cdot e\cdot g) + (b\cdot d\cdot i) + (a\cdot f\cdot h). $$

4. Subtract: (down sum) − (up sum).

Worked example

Compute

$$ \det\!\begin{bmatrix} 2&-1&3\\ 1&4&-1\\ 3&2&2 \end{bmatrix}. $$

Copy first two columns:

$$ \begin{array}{ccc|cc} 2&-1&3&2&-1\\ 1&4&-1&1&4\\ 3&2&2&3&2 \end{array} $$

Down sum: $2\cdot4\cdot2=16$, $(-1)\cdot(-1)\cdot3=3$, $3\cdot1\cdot2=6$ → $16+3+6=25$. Up sum: $3\cdot4\cdot3=36$, $2\cdot(-1)\cdot2=-4$, $2\cdot1\cdot(-1)=-2$ → $36-4-2=30$. Determinant $=25-30=-5$.

Why it works (intuition)

For 3×3 matrices the determinant is a signed sum over the six permutations of $\{1,2,3\}$. Sarrus’ diagram just groups the three even permutations (the “down” diagonals) and subtracts the three odd permutations (the “up” diagonals).

Tips & pitfalls

• Only for 3×3. There’s no Sarrus shortcut for 4×4 or higher; use cofactor expansion or row-reduction.
• Row/column ops help: If you can create a zero, expansion along that row/column gets easier than Sarrus.
• Signs: The most common mistake is dropping a minus in the “up” group—write the two sums separately, then subtract.

Want a couple of quick practice determinants with answers?