📐 Concept diagram

08-07 — Determinants (Deep)

Phase: 8 — Linear Algebra (Rigorous) Subject: 08-07 Prerequisites: 08-06 — Orthogonal Projections Next subject: 08-08 — Eigenvalues and Eigenvectors

Learning Objectives

By the end of this subject, you will be able to:

1. Define the determinant of an n × n matrix via permutations and parity (Leibniz formula), and compute small determinants by hand
2. Compute determinants using cofactor expansion (Laplace expansion) along any row or column
3. State and apply the key properties of determinants: multilinearity, alternating property, multiplicativity, and effect of row operations
4. Use Cramer's rule to solve small linear systems and understand when it applies
5. Interpret the determinant geometrically as signed volume/area in R^n and use it to test invertibility and orientation

Core Content

1. Definition via Permutations (Leibniz Formula)

The determinant of an n × n matrix A = [a_ij] is:

$det(A) = Σ_{σ ∈ S_n} sgn(σ) · a_{1,σ(1)} a_{2,σ(2)} ... a_{n,σ(n)}

> **CRITICAL — Foundational:** The determinant is uniquely defined by: (1) det(I) = 1, (2) swapping rows flips sign, (3) linear in each row. These three imply ALL other properties.
$

where: - S_n is the set of all n! permutations of {1, 2, ..., n} - sgn(σ) = (+1 if σ is even, −1 if σ is odd) - A permutation is even/odd based on the number of transpositions (swaps) needed to return it to the identity

Examples: - n = 1: det([a]) = a - n = 2: det([a b; c d]) = a·d − b·c (S₂ = {identity (even), swap (odd)}) - n = 3: det([a₁₁ a₁₂ a₁₃; a₂₁ a₂₂ a₂₃; a₃₁ a₃₂ a₃₃]) = a₁₁a₂₂a₃₃ + a₁₂a₂₃a₃₁ + a₁₃a₂₁a₃₂ − a₁₁a₂₃a₃₂ − a₁₂a₂₁a₃₃ − a₁₃a₂₂a₃₁ (6 terms: 3 even permutations, 3 odd)

Parity (sign) of permutations: An inversion is a pair (i, j) with i < j but σ(i) > σ(j). σ is even if it has an even number of inversions; odd otherwise. Equivalently, sgn(σ) = (−1)^(#inversions).

Example: σ = (2, 3, 1). Inversions: (2,1), (3,1) → 2 inversions → sgn = +1 (even).

2. Cofactor Expansion (Laplace Expansion)

The (i, j) minor M_ij is the determinant of the (n−1) × (n−1) submatrix obtained by deleting row i and column j.

The (i, j) cofactor is C_ij = (−1)^{i+j} M_ij.

Cofactor expansion along any row i:

$det(A) = Σ_{j=1}^n a_ij C_ij
$

Or along any column j:

$det(A) = Σ_{i=1}^n a_ij C_ij
$

Example (3 × 3):

$A = [1  2  3]
    [4  5  6]
    [7  8  9]
$

Expand along row 1: det(A) = 1·C₁₁ + 2·C₁₂ + 3·C₁₃ = 1·(+1)·det[5 6; 8 9] + 2·(−1)·det[4 6; 7 9] + 3·(+1)·det[4 5; 7 8] = 1·(45−48) − 2·(36−42) + 3·(32−35) = 1·(−3) − 2·(−6) + 3·(−3) = −3 + 12 − 9 = 0.

3. Properties of Determinants

Fundamental properties:

1. det(I) = 1

2. Row interchange: Swapping two rows multiplies det by −1.

3. Multilinearity (rows): det is linear in each row when others are fixed. det(..., c·row_i, ...) = c·det(..., row_i, ...) det(..., u+v, ...) = det(..., u, ...) + det(..., v, ...)

4. Alternating: If two rows are equal, det = 0.

5. Row operation (add multiple of row): Adding a multiple of one row to another leaves det unchanged.

6. Triangular matrix: det = product of diagonal entries.

7. Singularity: det(A) = 0 iff A is singular (not invertible). Equivalently, det(A) ≠ 0 iff columns/rows are independent.

8. Multiplicativity: det(AB) = det(A)·det(B)

9. Transpose: det(A^T) = det(A)

10. Inverse: det(A⁻¹) = 1/det(A) (when A is invertible)

11. Block triangular: det([A B; 0 C]) = det(A)·det(C)

Effect of row operations on determinant: - Row swap: det → −det - Row scaling by c: det → c·det - Row addition: det unchanged

Example using row operations:

$A = [1  2  3]
    [2  1  3]
    [3  3  6]
$

R₂−2R₁: [1 2 3; 0 −3 −3; 3 3 6] R₃−3R₁: [1 2 3; 0 −3 −3; 0 −3 −3] R₃−R₂: [1 2 3; 0 −3 −3; 0 0 0] Triangular: det = 1·(−3)·0 = 0. A is singular. ✓

4. Cramer's Rule

For an n × n invertible matrix A, the unique solution to Ax = b has components:

$x_i = det(A_i) / det(A)
$

where A_i is A with column i replaced by b.

Why it works: Write x = A⁻¹b. By the adjugate formula A⁻¹ = adj(A)/det(A). Then x_i = (1/det(A)) Σ_j adj(A)_ij b_j = det(A_i)/det(A). (The sum is precisely the cofactor expansion of A with column i replaced by b.)

Example:

$2x + y = 5
x − 3y = −1
$

A = [2 1; 1 −3], det(A) = −6 − 1 = −7.

x = det([5 1; −1 −3]) / (−7) = (−15 − (−1))/(−7) = (−14)/(−7) = 2. y = det([2 5; 1 −1]) / (−7) = (−2 − 5)/(−7) = (−7)/(−7) = 1.

Limitations: Cramer's rule requires O(n!) operations. It's computationally impractical for n > 4 but theoretically useful.

Common Pitfall: Cramer's rule is elegant but EXPENSIVE — never use it for n > 4. Use Gaussian elimination (O(n³)) for practical computation.

5. Geometric Interpretation

In R^n, the absolute value |det(A)| is the n-dimensional volume of the parallelepiped spanned by the column vectors of A.

• n = 2: |det([a b; c d])| = |ad − bc| = area of parallelogram spanned by (a,c) and (b,d)
• n = 3: |det(A)| = volume of parallelepiped

Sign: det(A) > 0 means the columns form a right-handed (positively oriented) basis; det(A) < 0 means left-handed.

Area/volume after linear transformation: If T(x) = Ax, then volume(T(S)) = |det(A)|·volume(S) for any region S.

Connection to independence: det(A) ≠ 0 iff columns are independent iff the parallelepiped has nonzero n-dimensional volume.

Key Terms

• Cofactor expansion

Worked Examples

Example 1: Determinant by Cofactor Expansion

Problem: Compute det(A) where A = [2 −1 0; 3 1 2; 1 0 −1].

Solution: Expand along row 3 (has a zero for convenience).

$det(A) = 1·C₃₁ + 0·C₃₂ + (−1)·C₃₃
$

C₃₁ = (+)det[−1 0; 1 2] = 1·(−1·2 − 0·1) = −2. (sign: (−1)^{3+1} = +1) C₃₃ = (+)det[2 −1; 3 1] = 1·(2·1 − (−1)·3) = 5. (sign: (−1)^{3+3} = +1)

det(A) = 1·(−2) + (−1)·5 = −2 − 5 = −7.

Check via row operations: R₂ − (3/2)R₁: [2 −1 0; 0 5/2 2; 1 0 −1] R₃ − (1/2)R₁: [2 −1 0; 0 5/2 2; 0 1/2 −1] R₃ − (1/5)R₂: [2 −1 0; 0 5/2 2; 0 0 −7/5] det = 2 · (5/2) · (−7/5) = −7. ✓

Example 2: Determinant Properties in Action

Problem: Without computing, find the relationship between det(A) and det(B): A = [a b c; d e f; g h i] B = [a b c; 2d 2e 2f; g h i]

Solution: Row 2 of B is 2 × row 2 of A. By multilinearity, det(B) = 2·det(A).

Another: If C = [a b c; d+a e+b f+c; g h i] (row 2 = original row 2 + row 1 of A), then det(C) = det(A) since adding a multiple of one row to another doesn't change the determinant.

Another: If D = [a b c; g h i; d e f] (rows 2 and 3 swapped), then det(D) = −det(A).

Example 3: Cramer's Rule

Problem: Solve using Cramer's rule:

$x + 2y + z = 5
2x − y + 3z = 7
x + y + z = 4
$

Solution: A = [1 2 1; 2 −1 3; 1 1 1].

det(A) = 1·det[−1 3; 1 1] − 2·det[2 3; 1 1] + 1·det[2 −1; 1 1] = 1·(−1−3) − 2·(2−3) + 1·(2−(−1)) = −4 − 2·(−1) + 3 = −4 + 2 + 3 = 1.

x = det([5 2 1; 7 −1 3; 4 1 1])/1 = 5·det[−1 3; 1 1] − 2·det[7 3; 4 1] + 1·det[7 −1; 4 1] = 5·(−4) − 2·(7−12) + 1·(7−(−4)) = −20 − 2·(−5) + 11 = −20 + 10 + 11 = 1.

y = det([1 5 1; 2 7 3; 1 4 1])/1 = 1·det[7 3; 4 1] − 5·det[2 3; 1 1] + 1·det[2 7; 1 4] = 1·(7−12) − 5·(2−3) + 1·(8−7) = −5 − 5·(−1) + 1 = −5 + 5 + 1 = 1.

z = det([1 2 5; 2 −1 7; 1 1 4])/1 = 1·det[−1 7; 1 4] − 2·det[2 7; 1 4] + 5·det[2 −1; 1 1] = 1·(−4−7) − 2·(8−7) + 5·(2+1) = −11 − 2 + 15 = 2.

Solution: (x, y, z) = (1, 1, 2). Check: 1+2+2=5, 2−1+6=7, 1+1+2=4. ✓

Example 4: Geometric Interpretation

Problem: Find the area of the triangle with vertices (0,0), (3,1), (1,4).

Solution: The triangle has half the area of the parallelogram spanned by v = (3,1) and w = (1,4).

Area = (1/2) |det([3 1; 1 4])| = (1/2) |3·4 − 1·1| = (1/2)·11 = 5.5.

The parallelogram has area 11.

Quiz

Q1: How many terms are in the Leibniz expansion of a 4 × 4 determinant?

A) 4 B) 16 C) 24 D) 256

Correct: C)

• If you chose C: The determinant sums over all n! permutations: 4! = 24 terms. Correct!
• If you chose A: 4 is n, not n!.
• If you chose B: 16 = 4² = n², not n!.
• If you chose D: 256 = 4⁴, not relevant to determinants.

Q2: The cofactor C_ij of matrix entry a_ij includes the sign factor:

A) (−1)^i B) (−1)^j C) (−1)^{i+j} D) (−1)^{i·j}

Correct: C)

• If you chose C: C_ij = (−1)^{i+j} M_ij, creating the checkerboard pattern: [+ − +; − + −; + − +]. Correct!
• If you chose A or B: The sign depends on both row and column indices.
• If you chose D: i·j does not produce the correct checkerboard pattern.

Q3: If you swap two rows of a matrix, what happens to its determinant?

A) It stays the same B) It is multiplied by −1 C) It becomes 0 D) It doubles

Correct: B)

• If you chose B: Row swaps reverse orientation, multiplying the determinant by −1. This is the alternating property. Correct!
• If you chose A: Swapping rows changes the sign, unlike row addition.
• If you chose C: Swapping doesn't make the determinant zero unless rows were identical.
• If you chose D: Swapping flips the sign, it doesn't scale by 2.

Q4: For an n × n matrix A, det(cA) equals:

A) c·det(A) B) c^n·det(A) C) n·det(A) D) det(A)/c

Correct: B)

• If you chose B: Each of the n rows is multiplied by c, and by multilinearity each contributes a factor c, so det(cA) = c^n det(A). Correct!
• If you chose A: Forgets that all n rows are scaled; only correct for n = 1.
• If you chose C: Confuses the dimension n with the scaling factor.
• If you chose D: Division would make the determinant smaller, not larger.

Q5: Cramer's rule for solving Ax = b gives the i-th component as:

A) det(A_i)/det(A) where A_i is A with column i replaced by b B) det(A)/det(A_i) C) det(A)·det(A_i) D) det(A_i) − det(A)

Correct: A)

• If you chose A: x_i = det(A_i)/det(A). The numerator replaces column i of A with b and computes the determinant. Correct!
• If you chose B: The ratio is inverted; det(A_i) goes in the numerator.
• If you chose C: The product is not Cramer's rule.
• If you chose D: The difference does not give the solution.

Q6: Geometrically, |det(A)| for a 2 × 2 matrix A represents:

A) The perimeter of the parallelogram spanned by its columns B) The area of the parallelogram spanned by its columns C) The angle between its columns D) The sum of the lengths of its columns

Correct: B)

• If you chose B: |ad − bc| is the area of the parallelogram with sides (a,c) and (b,d). In n dimensions, |det| gives the n-dimensional volume spanned by the columns. Correct!
• If you chose A: The determinant gives area, not perimeter.
• If you chose C: The determinant encodes angle information (det = ‖a‖‖b‖ sin θ) but isn't the angle itself.
• If you chose D: The determinant is not a sum of lengths.

Practice Problems

(Answers are below. Try each problem before checking.)

Problem 1: State the sign of each permutation of {1, 2, 3}: (a) (1, 2, 3), (b) (2, 3, 1), (c) (3, 1, 2), (d) (3, 2, 1).

Problem 2: Compute det([2 1 3; 4 −1 2; 1 0 5]) using cofactor expansion.

Problem 3: Use determinant properties to compute det([a b c; d e f; a+g b+h c+i]) in terms of other determinants.

Problem 4: Solve using Cramer's rule: 3x + y = 5, 2x − y = 0.

Problem 5: Find the volume of the parallelepiped spanned by (1, 0, 1), (2, 1, 0), and (0, 1, 3).

Problem 6: Prove that det(A⁻¹) = 1/det(A) for invertible A.

Problem 7: Show that det(cA) = c^n det(A) for an n × n matrix A.

Summary

1. The determinant is defined by the Leibniz formula summing over all n! permutations with signs determined by parity; it's the unique multilinear alternating form with det(I) = 1
2. Cofactor expansion provides a recursive computation method: pick a row or column, compute minors and cofactors with signs (−1)^{i+j}
3. Determinants have powerful algebraic properties: det(AB) = det(A)det(B), row swap negates det, scaling a row scales det, and two equal rows give det = 0
4. Cramer's rule gives explicit formulas for solving Ax = b when A is invertible: x_i = det(A_i)/det(A); elegant in theory but computationally expensive
5. Geometrically, |det(A)| is the n-volume of the parallelepiped spanned by its columns; det(A) = 0 ⇔ columns are linearly dependent ⇔ zero n-volume

Pitfalls

1. Assuming det(A + B) = det(A) + det(B). The determinant is multiplicative, NOT additive. In general det(A + B) ≠ det(A) + det(B). A simple counterexample: A = I, B = −I, where det(I) + det(−I) = 1 + 1 = 2 but det(I + (−I)) = det(0) = 0.

2. Getting cofactor signs wrong. The cofactor sign is (−1)^{i+j}, creating a checkerboard pattern. A common error is using (−1)^i or (−1)^j alone. For a 3×3 expansion, misplacing a single sign flips the sign of the entire determinant.

3. Forgetting that det(cA) = c^n det(A), not c·det(A). Each of the n rows is scaled by c, so the determinant picks up a factor of c^n. For a 3×3 matrix, det(2A) = 8·det(A), not 2·det(A).

4. Using Cramer's rule numerically for n > 3. Cramer's rule requires computing n+1 determinants of size n, each O(n!). For n = 10, this is ~4×10^8 operations vs. ~10^3 for Gaussian elimination. Cramer's rule is a theoretical tool; use elimination for computation.

5. Confusing the effect of row operations on the determinant. Row swaps multiply det by −1. Row scaling multiplies det by the scale factor. Row addition (adding a multiple of one row to another) leaves det UNCHANGED — but only if you add to a DIFFERENT row. Adding a multiple of a row to itself doubles it, which scales det.

Next Steps

Move on to 08-08 — Eigenvalues and Eigenvectors to discover the characteristic equation, how to find eigenvalues and eigenvectors, the distinction between algebraic and geometric multiplicity, and the conditions for diagonalization.