Cramer's Rule & Invertible Matrices

🎯 Learning Objectives

• State and apply Cramer's rule to solve $A\mathbf{x} = \mathbf{b}$ when $A$ is square and invertible
• Compute the adjugate (classical adjoint) of a matrix
• Find the inverse of a matrix using the adjugate formula $A^{-1} = \frac{\text{adj}(A)}{\det(A)}$
• Determine when Cramer's rule is (and isn't) computationally practical
• Express solutions to linear systems using determinants

📋 Prerequisites

• Determinants (Week 1) — computing $\det$, minors, cofactors
• Gaussian Elimination (this week) — the standard method for solving systems
• Matrix Multiplication (Week 1) — $A^{-1}$ is defined by $AA^{-1} = I$
• Cramer's rule provides an elegant theoretical formula, though Gaussian elimination is usually more practical

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1. Intuition: What Cramer's Rule Does

1. Take the coefficient matrix $A$
2. Replace column $j$ of $A$ with $\mathbf{b}$
3. Compute the determinant of this new matrix
4. Divide by $\det(A)$ The result is $x_j$. It's a beautiful formula, but computationally expensive for large $n$ (requires $n+1$ determinants).

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2. Cramer's Rule

2.1 Statement

Theorem (Cramer's Rule). Let $A$ be an $n \times n$ invertible matrix (so $\det(A) \neq 0$). For any $\mathbf{b} \in \mathbb{R}^n$, the unique solution to $A\mathbf{x} = \mathbf{b}$ is:

$$> x_j = \frac{\det(A_j)}{\det(A)}, \quad j = 1, 2, \dots, n >$$

where $A_j$ is the matrix formed by replacing the $j$-th column of $A$ with $\mathbf{b}$. Proof Sketch

Let $A = [\mathbf{a}_1 \; \mathbf{a}_2 \; \dots \; \mathbf{a}_n]$ where $\mathbf{a}_j$ is column $j$. Then $A\mathbf{x} = \mathbf{b}$ means:

$$> x_1\mathbf{a}_1 + x_2\mathbf{a}_2 + \cdots + x_n\mathbf{a}_n = \mathbf{b} >$$

Consider $A_j = [\mathbf{a}_1 \; \dots \; \mathbf{b} \; \dots \; \mathbf{a}_n]$ (column $j$ replaced by $\mathbf{b}$). Using multilinearity of the determinant:

$$> \det(A_j) = \det[\mathbf{a}_1 \; \dots \; \sum_{k=1}^n x_k\mathbf{a}_k \; \dots \; \mathbf{a}_n] = x_j \det(A) >$$

because all terms with $k \neq j$ have two identical columns (determinant zero). Hence $x_j = \det(A_j)/\det(A)$.

2.2 $2 \times 2$ Case

$$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$ $$\mathbf{b} = \begin{bmatrix} b_1 \\ b_2 \end{bmatrix}$$ $$x_1 = \frac{\begin{vmatrix} b_1 & b \\ b_2 & d \end{vmatrix}}{\begin{vmatrix} a & b \\ c & d \end{vmatrix}} = \frac{b_1 d - b b_2}{ad - bc}, \quad x_2 = \frac{\begin{vmatrix} a & b_1 \\ c & b_2 \end{vmatrix}}{\begin{vmatrix} a & b \\ c & d \end{vmatrix}} = \frac{a b_2 - b_1 c}{ad - bc}$$

Example 1: 2×2 Cramer's rule

Solve:

$$> \begin{cases} 2x + 3y = 7 \\ 4x - y = 1 \end{cases} >$$

$\det(A) = 2(-1) - 3(4) = -2 - 12 = -14$

$$> A_1 = \begin{bmatrix} 7 & 3 \\ 1 & -1 \end{bmatrix} >$$

, $\det(A_1) = 7(-1) - 3(1) = -7 - 3 = -10$

$$> A_2 = \begin{bmatrix} 2 & 7 \\ 4 & 1 \end{bmatrix} >$$

, $\det(A_2) = 2(1) - 7(4) = 2 - 28 = -26$

$x = \frac{-10}{-14} = \frac{5}{7}$, $y = \frac{-26}{-14} = \frac{13}{7}$

Check: $2(\frac{5}{7}) + 3(\frac{13}{7}) = \frac{10+39}{7} = 7$ ✓

2.3 $3 \times 3$ Case

Example 2: 3×3 Cramer's rule

Solve:

$$> \begin{cases} x + 2y + z = 3 \\ 2x - y + 3z = 2 \\ 3x + y - z = 6 \end{cases} >$$

$$> A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & -1 & 3 \\ 3 & 1 & -1 \end{bmatrix} >$$

Step 1: Compute $\det(A)$:

$$> \begin{aligned} \det(A) &= 1(-1)(-1) + 2(3)(3) + 1(2)(1) - 1(-1)(3) - 1(3)(1) - 2(2)(-1) \\ &= 1 + 18 + 2 - (-3) - 3 - (-4) \\ &= 1 + 18 + 2 + 3 - 3 + 4 = 25 \end{aligned} >$$

Step 2: Compute $A_1$ (replace column 1 with $\mathbf{b}$):

$$> A_1 = \begin{bmatrix} 3 & 2 & 1 \\ 2 & -1 & 3 \\ 6 & 1 & -1 \end{bmatrix} >$$

$$> \begin{aligned} \det(A_1) &= 3(-1)(-1) + 2(3)(6) + 1(2)(1) \\ &\quad - 1(-1)(6) - 3(3)(1) - 2(2)(-1) \\ &= 3 + 36 + 2 + 6 - 9 + 4 = 42 \end{aligned} >$$

$x = \frac{42}{25}$

Step 3: $A_2$ (replace column 2):

$$> A_2 = \begin{bmatrix} 1 & 3 & 1 \\ 2 & 2 & 3 \\ 3 & 6 & -1 \end{bmatrix} >$$

$$> \det(A_2) = 1(2)(-1) + 3(3)(3) + 1(2)(6) - 1(2)(3) - 1(3)(6) - 3(2)(-1) = -2 + 27 + 12 - 6 - 18 + 6 = 19 >$$

$y = \frac{19}{25}$

Step 4: $A_3$ (replace column 3):

$$> A_3 = \begin{bmatrix} 1 & 2 & 3 \\ 2 & -1 & 2 \\ 3 & 1 & 6 \end{bmatrix} >$$

$$> \det(A_3) = 1(-1)(6) + 2(2)(3) + 3(2)(1) - 3(-1)(3) - 1(2)(1) - 2(2)(6) = -6 + 12 + 6 + 9 - 2 - 24 = -5 >$$

$z = \frac{-5}{25} = -\frac{1}{5}$

Solution: $x = \frac{42}{25}$, $y = \frac{19}{25}$, $z = -\frac{1}{5}$.

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3. Cramer's Rule vs. Gaussian Elimination

Why Cramer's rule matters despite being expensive: It proves that the solution depends continuously on the entries of $A$ and $\mathbf{b}$ (since determinants are continuous functions). This is useful in theoretical arguments.

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4. The Adjugate and Matrix Inverse

4.1 Cofactor Matrix and Adjugate

Definition (Cofactor Matrix). The cofactor matrix $C$ of $A$ has entries $C_{ij} = (-1)^{i+j} M_{ij}$ (the cofactors). Definition (Adjugate). The adjugate (or classical adjoint) of $A$ is the transpose of the cofactor matrix:

$$> \text{adj}(A) = C^T >$$

4.2 Inverse via Adjugate

Theorem. If $\det(A) \neq 0$, then:

$$> A^{-1} = \frac{1}{\det(A)} \text{adj}(A) >$$

Example 3: Inverse of a 2×2 matrix using adjugate

$$> A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} >$$

Cofactor matrix:

$$> C = \begin{bmatrix} d & -c \\ -b & a \end{bmatrix} >$$

Adjugate:

$$> \text{adj}(A) = C^T = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} >$$

$$> A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} >$$

Check:

$$> AA^{-1} = \frac{1}{ad-bc} \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} = \frac{1}{ad-bc} \begin{bmatrix} ad-bc & 0 \\ 0 & ad-bc \end{bmatrix} = I >$$

✓ Example 4: Inverse of a 3×3 matrix

$$> A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{bmatrix} >$$

Step 1: Compute $\det(A)$.

$$> \det(A) = 1(1\cdot0 - 4\cdot6) - 2(0\cdot0 - 4\cdot5) + 3(0\cdot6 - 1\cdot5) = 1(-24) - 2(-20) + 3(-5) = -24 + 40 - 15 = 1 >$$

Step 2: Compute all 9 cofactors.

$$> C_{11} = +\det\begin{bmatrix} 1 & 4 \\ 6 & 0 \end{bmatrix} = 0 - 24 = -24 >$$

$$> C_{12} = -\det\begin{bmatrix} 0 & 4 \\ 5 & 0 \end{bmatrix} = -(0 - 20) = 20 >$$

$$> C_{13} = +\det\begin{bmatrix} 0 & 1 \\ 5 & 6 \end{bmatrix} = 0 - 5 = -5 >$$

$$> C_{21} = -\det\begin{bmatrix} 2 & 3 \\ 6 & 0 \end{bmatrix} = -(0 - 18) = 18 >$$

$$> C_{22} = +\det\begin{bmatrix} 1 & 3 \\ 5 & 0 \end{bmatrix} = 0 - 15 = -15 >$$

$$> C_{23} = -\det\begin{bmatrix} 1 & 2 \\ 5 & 6 \end{bmatrix} = -(6 - 10) = 4 >$$

$$> C_{31} = +\det\begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} = 8 - 3 = 5 >$$

$$> C_{32} = -\det\begin{bmatrix} 1 & 3 \\ 0 & 4 \end{bmatrix} = -(4 - 0) = -4 >$$

$$> C_{33} = +\det\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} = 1 - 0 = 1 >$$

Step 3: Form the adjugate (transpose of cofactor matrix):

$$> \text{adj}(A) = \begin{bmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{bmatrix} >$$

Step 4: Since $\det(A) = 1$:

$$> A^{-1} = \frac{1}{1} \text{adj}(A) = \begin{bmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{bmatrix} >$$

Verify: $AA^{-1} = I$ (check one entry: first row of $A$ times first column of $A^{-1}$: $1(-24) + 2(20) + 3(-5) = -24 + 40 - 15 = 1$ ✓).

4.3 Properties of the Adjugate

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5. Edge Cases & Gotchas

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6. Common Pitfalls

❌ Pitfall 1: Forgetting to divide by $\det(A)$

❌ Pitfall 2: Misplacing $\mathbf{b}$ in the wrong column

❌ Pitfall 3: Applying Cramer's rule to non-square or singular systems

❌ Pitfall 4: Confusing adjugate and inverse

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7. Formula Summary Table

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8. 📝 Practice Questions

Q1: 2×2 Cramer's rule

Solve:

$$> \begin{cases} 3x + 2y = 7 \\ x - 5y = -2 \end{cases} >$$

Strategy: Compute $\det(A)$, $\det(A_1)$, $\det(A_2)$, then $x$ and $y$.

Solution: $\det(A) = 3(-5) - 2(1) = -15 - 2 = -17$

$$> \det(A_1) = \begin{vmatrix} 7 & 2 \\ -2 & -5 \end{vmatrix} = 7(-5) - 2(-2) = -35 + 4 = -31 >$$

$$> \det(A_2) = \begin{vmatrix} 3 & 7 \\ 1 & -2 \end{vmatrix} = 3(-2) - 7(1) = -6 - 7 = -13 >$$

$x = \frac{-31}{-17} = \frac{31}{17}$, $y = \frac{-13}{-17} = \frac{13}{17}$. Q2: 3×3 Cramer's rule

Solve:

$$> \begin{cases} x + y + z = 6 \\ 2x - y + z = 3 \\ x + 2y - z = 2 \end{cases} >$$

Solution:

$$> \det(A) = \begin{vmatrix} 1 & 1 & 1 \\ 2 & -1 & 1 \\ 1 & 2 & -1 \end{vmatrix} >$$

$= 1(-1)(-1) + 1(1)(1) + 1(2)(2) - 1(-1)(1) - 1(1)(2) - 1(2)(-1)$ $= 1 + 1 + 4 + 1 - 2 + 2 = 7$

$$> A_1 = \begin{bmatrix} 6 & 1 & 1 \\ 3 & -1 & 1 \\ 2 & 2 & -1 \end{bmatrix} >$$

, $\det(A_1) = 6(-1)(-1) + 1(1)(2) + 1(3)(2) - 1(-1)(2) - 6(1)(2) - 1(3)(-1)$ $= 6 + 2 + 6 + 2 - 12 + 3 = 7$

$x = \frac{7}{7} = 1$

$$> A_2 = \begin{bmatrix} 1 & 6 & 1 \\ 2 & 3 & 1 \\ 1 & 2 & -1 \end{bmatrix} >$$

, $\det(A_2) = 1(3)(-1) + 6(1)(1) + 1(2)(2) - 1(3)(1) - 1(1)(2) - 6(2)(-1)$ $= -3 + 6 + 4 - 3 - 2 + 12 = 14$

$y = \frac{14}{7} = 2$

$$> A_3 = \begin{bmatrix} 1 & 1 & 6 \\ 2 & -1 & 3 \\ 1 & 2 & 2 \end{bmatrix} >$$

, $\det(A_3) = 1(-1)(2) + 1(3)(1) + 6(2)(2) - 6(-1)(1) - 1(3)(2) - 1(2)(2)$ $= -2 + 3 + 24 + 6 - 6 - 4 = 21$

$z = \frac{21}{7} = 3$

Solution: $x = 1$, $y = 2$, $z = 3$. Q3: 2×2 inverse

Find $A^{-1}$ for

$$> A = \begin{bmatrix} 4 & 3 \\ 2 & 1 \end{bmatrix} >$$

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Solution: $\det(A) = 4(1) - 3(2) = 4 - 6 = -2$

$$> A^{-1} = \frac{1}{-2} \begin{bmatrix} 1 & -3 \\ -2 & 4 \end{bmatrix} = \begin{bmatrix} -\frac{1}{2} & \frac{3}{2} \\ 1 & -2 \end{bmatrix} >$$

Verify:

$$> AA^{-1} = \begin{bmatrix} 4 & 3 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} -\frac{1}{2} & \frac{3}{2} \\ 1 & -2 \end{bmatrix} = \begin{bmatrix} -2+3 & 6-6 \\ -1+1 & 3-2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} >$$

✓ Q4: 3×3 inverse

Find $A^{-1}$ for

$$> A = \begin{bmatrix} 1 & 0 & 2 \\ 2 & -1 & 3 \\ 4 & 1 & 8 \end{bmatrix} >$$

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Solution: $\det(A) = 1(-1)(8) + 0(3)(4) + 2(2)(1) - 2(-1)(4) - 1(3)(1) - 0(2)(8)$ $= -8 + 0 + 4 + 8 - 3 - 0 = 1$

Cofactors:

$$> C_{11} = +\det\begin{bmatrix} -1 & 3 \\ 1 & 8 \end{bmatrix} = -8 - 3 = -11 >$$

$$> C_{12} = -\det\begin{bmatrix} 2 & 3 \\ 4 & 8 \end{bmatrix} = -(16 - 12) = -4 >$$

$$> C_{13} = +\det\begin{bmatrix} 2 & -1 \\ 4 & 1 \end{bmatrix} = 2 - (-4) = 6 >$$

$$> C_{21} = -\det\begin{bmatrix} 0 & 2 \\ 1 & 8 \end{bmatrix} = -(0 - 2) = 2 >$$

$$> C_{22} = +\det\begin{bmatrix} 1 & 2 \\ 4 & 8 \end{bmatrix} = 8 - 8 = 0 >$$

$$> C_{23} = -\det\begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix} = -(1 - 0) = -1 >$$

$$> C_{31} = +\det\begin{bmatrix} 0 & 2 \\ -1 & 3 \end{bmatrix} = 0 - (-2) = 2 >$$

$$> C_{32} = -\det\begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} = -(3 - 4) = 1 >$$

$$> C_{33} = +\det\begin{bmatrix} 1 & 0 \\ 2 & -1 \end{bmatrix} = -1 - 0 = -1 >$$

$$> \text{adj}(A) = C^T = \begin{bmatrix} -11 & 2 & 2 \\ -4 & 0 & 1 \\ 6 & -1 & -1 \end{bmatrix} >$$

Since $\det(A) = 1$, $A^{-1} = \text{adj}(A)$. Q5: Using Cramer's rule with parameters

For what $k$ does Cramer's rule apply? Solve for $x$ and $y$:

$$> \begin{cases} kx + y = 1 \\ x + ky = 1 \end{cases} >$$

Solution: $\det(A) = k(k) - 1(1) = k^2 - 1 = (k-1)(k+1)$

Cramer's rule applies when $\det(A) \neq 0$, i.e., $k \neq \pm 1$.

$$> \det(A_1) = \begin{vmatrix} 1 & 1 \\ 1 & k \end{vmatrix} = k - 1 >$$

$$> \det(A_2) = \begin{vmatrix} k & 1 \\ 1 & 1 \end{vmatrix} = k - 1 >$$

$x = \frac{k-1}{(k-1)(k+1)} = \frac{1}{k+1}$ (for $k \neq \pm 1$) $y = \frac{k-1}{(k-1)(k+1)} = \frac{1}{k+1}$ (for $k \neq \pm 1$)

Check: When $k = 1$, the system is $x+y=1$, $x+y=1$ (infinitely many solutions). When $k = -1$, the system is $-x+y=1$, $x-y=1$ (inconsistent). Q6: Adjugate properties

Verify $A \cdot \text{adj}(A) = \det(A) \cdot I$ for

$$> A = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} >$$

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Solution: $\det(A) = 8 - 3 = 5$

$$> \text{adj}(A) = \begin{bmatrix} 4 & -1 \\ -3 & 2 \end{bmatrix} >$$

$$> A \cdot \text{adj}(A) = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 4 & -1 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} 8-3 & -2+2 \\ 12-12 & -3+8 \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} = 5I >$$

✓ Q7: Solving with inverse

Solve $A\mathbf{x} = \mathbf{b}$ using $A^{-1}$ where

$$> A = \begin{bmatrix} 1 & 2 \\ 2 & 5 \end{bmatrix} >$$

,

$$> \mathbf{b} = \begin{bmatrix} 3 \\ 7 \end{bmatrix} >$$

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Solution: $\det(A) = 5 - 4 = 1$

$$> A^{-1} = \frac{1}{1} \begin{bmatrix} 5 & -2 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} 5 & -2 \\ -2 & 1 \end{bmatrix} >$$

$$> \mathbf{x} = A^{-1}\mathbf{b} = \begin{bmatrix} 5 & -2 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} 3 \\ 7 \end{bmatrix} = \begin{bmatrix} 15 - 14 \\ -6 + 7 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} >$$

Check: $1(1) + 2(1) = 3$ ✓, $2(1) + 5(1) = 7$ ✓. Q8: Cramer's rule — word problem

A company makes two products. The profit equations are:

$$> \begin{cases} 5x + 3y = 190 \\ 2x + 4y = 120 \end{cases} >$$

where $x$ = units of product A, $y$ = units of product B. Find $x$ and $y$.

Solution: $\det(A) = 5(4) - 3(2) = 20 - 6 = 14$

$$> \det(A_1) = \begin{vmatrix} 190 & 3 \\ 120 & 4 \end{vmatrix} = 190(4) - 3(120) = 760 - 360 = 400 >$$

$$> \det(A_2) = \begin{vmatrix} 5 & 190 \\ 2 & 120 \end{vmatrix} = 5(120) - 190(2) = 600 - 380 = 220 >$$

$x = \frac{400}{14} \approx 28.57$, $y = \frac{220}{14} \approx 15.71$. Q9: Determinant of adjugate

If $A$ is $3 \times 3$ with $\det(A) = -4$, find $\det(\text{adj}(A))$.

Solution: $\det(\text{adj}(A)) = (\det(A))^{n-1} = (-4)^2 = 16$. Q10: When Cramer's rule fails

Explain why Cramer's rule cannot solve the system:

$$> \begin{cases} x + 2y = 5 \\ 2x + 4y = 10 \end{cases} >$$

and find all solutions.

Solution: $\det(A) = 1(4) - 2(2) = 4 - 4 = 0$, so Cramer's rule is inapplicable.

The second equation is twice the first, so we have $x + 2y = 5$. $x = 5 - 2y$, $y$ free. Infinitely many solutions.

Parametric:

$$> \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 5 \\ 0 \end{bmatrix} + y\begin{bmatrix} -2 \\ 1 \end{bmatrix} >$$

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🔗 Cross-References

• Next topic: Vector Spaces Definition
• Week 1 (Determinants): Foundation for Cramer's rule
• Week 5 (Rank-Nullity): $\det(A) \neq 0$ means full rank, zero nullity
• Week 12 (Eigenvalues): The characteristic polynomial uses $\det(A - \lambda I)$
• BSCS3004 (Deep Learning): Matrix inverses appear in normal equations for linear regression Join Discord Previous2.1 Gaussian EliminationNext3.1 Vector Spaces Definition