LINEAR-ALGEBRA · Unit 19 · Video 2 · Interactive Practice

Swap a Column, Solve the System: Cramer's Rule

IKey Formulas

Formula	Name	Description
$x_j = \dfrac{\det(B_j)}{\det(A)}$	Cramer's Rule	$B_j$ is $A$ with column $j$ replaced by $\mathbf{b}$
$C_{ij} = (-1)^{i+j}\det(M_{ij})$	Cofactor	$M_{ij}$ : delete row $i$ , column $j$
$A^{-1} = \dfrac{1}{\det(A)}C^T$	Inverse formula	Connects inverse to cofactors
Requires $\det(A) \neq 0$	Unique solution	System solvable iff $A$ invertible

IIVisualization 1 — The Column Swap

For the video's system $x+2y-z=3,\ 2x+y+z=1,\ 3x-y+2z=2$ , solving for variable $j$ means replacing column $j$ of $A$ with $\mathbf{b}$ , then taking $\det(B_j)/\det(A)$ .

The cofactors along column $j$ are identical in $A$ and $B_j$ — they use only the other columns — so $\det(B_j)$ reproduces the numerator of $\mathbf{x}=\tfrac{1}{\det A}\,C^{T}\mathbf{b}$ .

IIIVisualization 2 — Two Lines, One Solution

Each equation is a line and the solution is their intersection — which flies to infinity as $\det(A)\to 0$ and the lines turn parallel.

a₁ 2 b₁ 1 a₂ 1 b₂ -1

When $\det(A)=0$ the lines are parallel (no solution) or coincident (infinitely many) — either way Cramer's ratio is undefined.

IVVisualization 3 — A Theorem, Not an Algorithm

Each $n\times n$ determinant costs about $n!$ operations, so Cramer needs $(n+1)$ of them versus Gaussian elimination's $\sim\tfrac{2n^3}{3}$ .

n 5

Reach for Cramer's Rule for symbolic answers, single-variable formulas, and small hand systems — never for large numerical ones.

VQuiz Questions

Question 1

To find $y$ (the second variable) in a 3×3 system $A\mathbf{x} = \mathbf{b}$ using Cramer's Rule, what do you compute?

✅ Correct! The $j$ -th variable uses column $j$ replaced by $\mathbf{b}$ .

❌ Not quite. The variable number tells you which column to replace, and $\det(B_j)$ goes in the numerator.

Show solution

Solution:

Cramer's Rule: $x_j = \dfrac{\det(B_j)}{\det(A)}$ , where $B_j$ is $A$ with column $j$ replaced by $\mathbf{b}$ .

Since $y$ is the second variable ( $j = 2$ ):

$y = x_2 = \frac{\det(B_2)}{\det(A)}$

where $B_2$ has column 2 replaced by $\mathbf{b}$ .

Note: the formula is $\det(B_j)/\det(A)$ , not the reciprocal.

Question 2

For the system from the video, we found: $\det(A) = 6$ , $\;\det(B_1) = 12$ , $\;\det(B_2) = -4$ , $\;\det(B_3) = -14$ .

What is the value of $z$ ?

✅ Correct! $z = \det(B_3)/\det(A) = -14/6 = -7/3$ .

❌ Not quite.

Show solution

Solution:

$z$ is the third variable, so use $B_3$ :

$z = \frac{\det(B_3)}{\det(A)} = \frac{-14}{6} = -\frac{7}{3}$

Why the other options are wrong:

$z = 2$ : That's $x = \det(B_1)/\det(A) = 12/6$ — wrong determinant!
$z = -2/3$ : That's $y = \det(B_2)/\det(A) = -4/6$ — also wrong determinant!
$z = -14$ : That's $\det(B_3)$ alone — forgot to divide by $\det(A)$ !

Question 3

True or False: Cramer's Rule can be applied to find a unique solution even when $\det(A) = 0$ .

✅ Correct! When $\det(A) = 0$ , the formula is undefined — no unique solution exists.

❌ Not quite. What happens when you divide by $\det(A) = 0$ ?

Show solution

Answer: False

Cramer's Rule requires $\det(A) \neq 0$ :

$x_j = \frac{\det(B_j)}{\det(A)}$

When $\det(A) = 0$ :

Division by zero makes the formula undefined
The matrix $A$ is singular (not invertible)
The system has no unique solution — either no solution or infinitely many

As the video states: "When det(A) equals zero, the formula correctly signals no unique solution exists."

Question 4

Which of the following is NOT a recommended use case for Cramer's Rule?

✅ Correct! Cramer's Rule shines for symbolic, theoretical, and small-scale problems — not large numerical computations.

❌ Not quite. Think about which scenario would make the factorial cost of Cramer's Rule impractical.

Show solution

Answer: Solving a 100×100 numerical system on a computer is NOT a good use of Cramer's Rule.

Why? For $n = 100$ , Cramer's Rule with cofactor expansion requires an astronomical number of operations, while Gaussian elimination needs only about 670,000. The factorial growth makes it completely impractical for large systems.

Good uses for Cramer's Rule:

✅ Symbolic systems — gives closed-form answers with variable coefficients
✅ Formula for one variable — find just $x_j$ without solving everything
✅ Theoretical proofs — reveals how solutions depend on every entry of $A$ and $\mathbf{b}$
✅ Small systems by hand — 2×2 and 3×3 are manageable

"The best formulas aren't always the fastest — sometimes they're the ones that let you see."

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