2×2 and 3×3 determinants, cofactor expansion, Cramer's Rule, matrix inverses, triangle area, invertibility tests, and an introduction to eigenvalues — all in one comprehensive guide.
The determinant is a single number computed from a square matrix. It encodes fundamental information: whether the matrix is invertible, how it scales area or volume, and the solution structure of associated linear systems. Every square matrix has exactly one determinant.
det(A) ≠ 0 ↔ A is invertible. det(A) = 0 ↔ A is singular (no inverse). This is the fastest test for invertibility.
The absolute value |det(A)| gives the factor by which A scales areas (2D) or volumes (3D). A negative determinant means the transformation reverses orientation.
For Ax = b, if det(A) ≠ 0 then the system has a unique solution. Cramer's Rule expresses each variable as a ratio of determinants.
For a 2×2 matrix, the determinant is the difference of the diagonal products — the simplest determinant to compute and one you should know instantly.
A = [[a, b], [c, d]]
det(A) = ad − bc
Main diagonal product minus anti-diagonal product
A = [[3, 5], [1, 4]]
det(A) = (3)(4) − (5)(1) = 12 − 5 = 7
det(A) = 7 → A is invertible
B = [[2, −3], [−4, 6]]
det(B) = (2)(6) − (−3)(−4) = 12 − 12 = 0
det(B) = 0 → B is singular (not invertible)
C = [[0, −2], [5, 3]]
det(C) = (0)(3) − (−2)(5) = 0 + 10 = 10
det(C) = 10 → C is invertible
For a 3×3 matrix, expand along any row or column. The first-row expansion is the standard starting point. Signs alternate in the checkerboard pattern below.
The sign of cofactor (i, j) is (−1)^(i+j)
Expanding along Row 1:
det(A) = a·(ei − fh) − b·(di − fg) + c·(dh − eg)
where A = [[a,b,c],[d,e,f],[g,h,i]]
A = [[2, −1, 3], [0, 4, −2], [1, 3, 5]]
Expand along Row 1:
= 2 · det([[4,−2],[3,5]]) − (−1) · det([[0,−2],[1,5]]) + 3 · det([[0,4],[1,3]])
= 2(4·5 − (−2)·3) + 1(0·5 − (−2)·1) + 3(0·3 − 4·1)
= 2(20 + 6) + 1(0 + 2) + 3(0 − 4)
= 2(26) + 1(2) + 3(−4)
= 52 + 2 − 12
det(A) = 42
These properties let you simplify matrices before computing, and they underlie proofs about invertibility, eigenvalues, and linear independence.
Transpose Invariance
det(Aᵀ) = det(A)
Transposing a matrix leaves the determinant unchanged. Rows and columns are interchangeable for all determinant purposes.
Row Swap
Swapping two rows → det changes sign
Swap rows i and j: the new determinant equals −det(A). Useful in row reduction.
Row Scaling
Multiply row by k → det multiplied by k
If every entry in one row is scaled by k, the determinant scales by k. Factoring out k can simplify computation.
Row Addition
Adding multiple of one row to another → det unchanged
This is the key row operation that leaves det unchanged. Gaussian elimination exploits this to create zeros.
Multiplicativity
det(AB) = det(A) · det(B)
The determinant of a product equals the product of the determinants. Applies to any two square matrices of the same size.
Zero Rows/Columns
A row or column of zeros → det = 0
Also: if two rows (or columns) are identical or proportional, det = 0.
Inverse
det(A⁻¹) = 1/det(A)
Follows from det(AA⁻¹) = det(I) = 1 and the multiplicativity property.
Cramer's Rule solves a square linear system Ax = b by expressing each unknown as a ratio of determinants. It requires det(A) ≠ 0.
System: ax + by = e | cx + dy = f
D = det(A) = ad − bc
Dₓ = det([[e,b],[f,d]]) = ed − bf (replace column 1 with constants)
D_y = det([[a,e],[c,f]]) = af − ec (replace column 2 with constants)
x = Dₓ / D y = D_y / D
Worked Example
3x + 2y = 8
x − y = 1
D = (3)(−1) − (2)(1) = −3 − 2 = −5
Dₓ = (8)(−1) − (2)(1) = −8 − 2 = −10
D_y = (3)(1) − (8)(1) = 3 − 8 = −5
x = −10/−5 = 2 y = −5/−5 = 1
Verify: 3(2)+2(1)=8 ✓ 2−1=1 ✓
For Ax = b with A being 3×3, compute D = det(A). Then for each variable xₙ, form matrix Aₙ by replacing the n-th column of A with the constant vector b. Then xₙ = det(Aₙ)/D.
x = det(A₁)/D y = det(A₂)/D z = det(A₃)/D
The adjugate (classical adjoint) method gives a closed-form formula for the inverse using cofactors and the determinant. It is most practical for 2×2 matrices and pedagogically important for 3×3.
A = [[a, b], [c, d]] det(A) = ad − bc
A⁻¹ = (1/det(A)) · [[d, −b], [−c, a]]
Swap main diagonal, negate off-diagonal, divide by det(A)
Worked Example
A = [[4, 3], [2, 1]]
det(A) = (4)(1) − (3)(2) = 4 − 6 = −2
Adjugate = [[1, −3], [−2, 4]]
A⁻¹ = (1/−2) · [[1, −3], [−2, 4]]
A⁻¹ = [[−1/2, 3/2], [1, −2]]
Verify: A · A⁻¹ = [[4·(−1/2)+3·1, 4·(3/2)+3·(−2)], [2·(−1/2)+1·1, 2·(3/2)+1·(−2)]] = [[1,0],[0,1]] ✓
Compute det(A)
If det(A) = 0, stop — the matrix is singular and has no inverse.
Find the matrix of cofactors
Compute Cᵢⱼ = (−1)^(i+j) · Mᵢⱼ for every entry, where Mᵢⱼ is the minor.
Transpose the cofactor matrix
The adjugate adj(A) is the transpose of the cofactor matrix.
Divide by the determinant
A⁻¹ = adj(A) / det(A). Each entry in the adjugate is divided by det(A).
Given any three vertices in the coordinate plane, a determinant formula computes the triangle's area without needing to draw or measure it. It also tests collinearity.
Vertices: (x₁,y₁), (x₂,y₂), (x₃,y₃)
Area = (1/2) · |det([[x₁,y₁,1],[x₂,y₂,1],[x₃,y₃,1]])|
Absolute value ensures a positive area
Vertices: (0,0), (4,0), (1,3)
det = 0(0·1−3·1) − 0(4·1−1·1) + 1(4·3−0·1)
= 0 − 0 + 1(12)
= 12
Area = (1/2)|12| = 6 sq units
If the determinant equals 0, the three points are collinear — they all lie on the same line and form a triangle with zero area.
Points: (1,1), (3,3), (5,5)
All on y = x → det = 0 → collinear ✓
The determinant is the canonical tool for checking whether a matrix is invertible. All of the following conditions are equivalent — if one is true, they are all true.
| Condition | What It Means |
|---|---|
| det(A) ≠ 0 | The determinant is nonzero |
| A⁻¹ exists | There is a unique inverse matrix |
| Ax = b has unique solution | For every vector b, there is exactly one x |
| Ax = 0 has only x = 0 | The null space is trivial |
| Rows of A are linearly independent | No row is a linear combination of others |
| Columns of A span ℝⁿ | The column space equals ℝⁿ |
| A row-reduces to the identity I | RREF of A is the identity matrix |
Eigenvalues measure how a matrix stretches or compresses special directions called eigenvectors. They arise in differential equations, data science, physics, and computer graphics.
Definition
Av = λv (v ≠ 0)
λ = eigenvalue v = eigenvector
The matrix A acts on eigenvector v by scaling it by λ — the direction of v is preserved (or reversed if λ < 0). To find eigenvalues, rearrange to (A − λI)v = 0. For a nontrivial solution v ≠ 0 to exist, the matrix A − λI must be singular, i.e., det(A − λI) = 0.
A = [[a, b], [c, d]]
A − λI = [[a−λ, b], [c, d−λ]]
det(A − λI) = (a−λ)(d−λ) − bc = 0
λ² − (a+d)λ + (ad−bc) = 0
tr(A) = trace = sum of diagonal entries = a + d
A = [[5, 2], [2, 5]]
tr(A) = 10 det(A) = 25 − 4 = 21
Characteristic equation: λ² − 10λ + 21 = 0
Factor: (λ − 7)(λ − 3) = 0
Eigenvalues: λ₁ = 7, λ₂ = 3
Finding eigenvector for λ = 7:
(A − 7I)v = 0 → [[-2,2],[2,-2]]v = 0 → v = [1, 1]
Finding eigenvector for λ = 3:
(A − 3I)v = 0 → [[2,2],[2,2]]v = 0 → v = [1, −1]
Symmetric matrices always have real eigenvalues; their eigenvectors are perpendicular.
x + y + z = 6
2x − y + z = 3
x + 2y − z = 2
A = [[1,1,1],[2,−1,1],[1,2,−1]] b = [6,3,2]
det(A) = 1(1−2) − 1(−2−1) + 1(4+1) = −1+3+5 = 7
For x: replace col 1 with b → det(A₁) = 6(1−2)−1(−3−2)+1(6+2) = −6+5+8 = 7
For y: replace col 2 with b → det(A₂) = 1(−3−2)−6(−2−1)+1(4−3) = −5+18+1 = 14
For z: replace col 3 with b → det(A₃) = 1(−2−6)−1(−4−3)+6(4+1) = −8+7+30 = 21
x = 7/7 = 1 y = 14/7 = 2 z = 21/7 = 3
Verify: 1+2+3=6 ✓ 2−2+3=3 ✓ 1+4−3=2 ✓
Compute det([[2,4,6],[1,3,5],[3,5,7]])
Factor 2 from Row 1 → det = 2 · det([[1,2,3],[1,3,5],[3,5,7]])
R2 = R2 − R1: [[1,2,3],[0,1,2],[3,5,7]]
R3 = R3 − 3R1: [[1,2,3],[0,1,2],[0,−1,−2]]
R3 = R3 + R2: [[1,2,3],[0,1,2],[0,0,0]]
Upper triangular with a zero row → inner det = 0
det = 2 · 0 = 0 (Matrix is singular)
Set up: det([[2,1,1],[5,3,1],[3,7,1]])
= 2(3·1−1·7) − 1(5·1−1·3) + 1(5·7−3·3)
= 2(3−7) − 1(5−3) + 1(35−9)
= 2(−4) − 1(2) + 1(26)
= −8 − 2 + 26 = 16
Area = (1/2)|16| = 8 square units
| Formula | Expression |
|---|---|
| 2×2 determinant | det([[a,b],[c,d]]) = ad − bc |
| 3×3 (row 1 expansion) | a(ei−fh) − b(di−fg) + c(dh−eg) |
| 2×2 inverse | A⁻¹ = (1/det(A))·[[d,−b],[−c,a]] |
| Cramer x (2×2) | x = det([[e,b],[f,d]]) / det(A) |
| Triangle area | (1/2)|det([[x₁,y₁,1],[x₂,y₂,1],[x₃,y₃,1]])| |
| Characteristic eq (2×2) | λ² − tr(A)λ + det(A) = 0 |
| Cofactor sign | (−1)^(i+j) |
| det(AB) | det(A) · det(B) |
| det(A⁻¹) | 1 / det(A) |
For a 2×2 matrix A = [[a, b], [c, d]], the determinant is det(A) = ad − bc. Multiply the entries along the main diagonal (top-left to bottom-right) and subtract the product of the entries along the anti-diagonal (top-right to bottom-left). Example: det([[3, 5], [1, 4]]) = (3)(4) − (5)(1) = 12 − 5 = 7.
Expand along the first row. For matrix A = [[a, b, c], [d, e, f], [g, h, i]], the determinant is: det(A) = a(ei − fh) − b(di − fg) + c(dh − eg). Each term is an entry in the first row times the determinant of the 2×2 submatrix formed by deleting that entry's row and column. Signs alternate: +, −, + along any row or column. Example: det([[2,−1,3],[0,4,−2],[1,3,5]]) = 2(4·5−(−2)·3) − (−1)(0·5−(−2)·1) + 3(0·3−4·1) = 2(26)+1(2)+3(−4) = 52+2−12 = 42.
Important determinant properties: (1) det(Aᵀ) = det(A) — the determinant is unchanged by transposing. (2) Swapping two rows changes the sign of the determinant. (3) Multiplying a row by scalar k multiplies the determinant by k. (4) Adding a multiple of one row to another row leaves the determinant unchanged. (5) det(AB) = det(A)·det(B). (6) det(A⁻¹) = 1/det(A). (7) If two rows are identical or proportional, det(A) = 0. (8) A matrix is invertible if and only if its determinant is nonzero.
Cramer's Rule uses determinants to solve a linear system Ax = b. For a 2×2 system ax + by = e and cx + dy = f, let D = det(A) = ad − bc. Then x = Dₓ/D where Dₓ replaces the first column of A with [e, f], and y = D_y/D where D_y replaces the second column with [e, f]. Cramer's Rule applies only when D ≠ 0 (the system has a unique solution). For large systems, Gaussian elimination is usually more efficient, but Cramer's Rule is valuable for small systems and for theoretical purposes.
For a 2×2 matrix A = [[a, b], [c, d]] with det(A) = ad − bc ≠ 0, the inverse is A⁻¹ = (1/det(A)) · [[d, −b], [−c, a]]. The adjugate (or classical adjoint) swaps the main diagonal entries and negates the off-diagonal entries. Example: A = [[4, 3], [2, 1]]. det(A) = 4−6 = −2. A⁻¹ = (1/−2)·[[1,−3],[−2,4]] = [[−1/2, 3/2],[1, −2]]. Verify: AA⁻¹ should equal the identity matrix I.
Given three vertices (x₁, y₁), (x₂, y₂), (x₃, y₃), the area of the triangle is: Area = (1/2)|det([[x₁, y₁, 1],[x₂, y₂, 1],[x₃, y₃, 1]])|. The absolute value ensures a positive area regardless of vertex order. If the determinant equals zero, the three points are collinear (they lie on a single line). Example: vertices (0,0), (4,0), (0,3) → det = 0(0−3) − 0(4−0) + 1(12−0) = 12. Area = (1/2)|12| = 6 square units.
A matrix with determinant zero is called a singular matrix. A singular matrix is NOT invertible — it has no inverse. Geometrically, it means the matrix transformation collapses space (squashes it into a lower dimension). For a system of equations, a zero determinant means the system either has no solution or infinitely many solutions — it cannot have a unique solution. You can check for singularity before attempting to solve a system or find an inverse.
The eigenvalues of a square matrix A are the scalars λ satisfying Av = λv for a nonzero vector v. To find eigenvalues, solve the characteristic equation det(A − λI) = 0, where I is the identity matrix. For a 2×2 matrix A = [[a, b], [c, d]], this gives the quadratic λ² − (a+d)λ + (ad−bc) = 0, i.e., λ² − tr(A)λ + det(A) = 0. Eigenvalues can be real or complex. Each eigenvalue λ has corresponding eigenvectors found by solving (A − λI)v = 0.
Cofactor expansion (also called Laplace expansion) expresses the determinant as a sum of entries times their cofactors. The cofactor Cᵢⱼ = (−1)^(i+j) · Mᵢⱼ, where Mᵢⱼ is the minor (the determinant of the submatrix formed by deleting row i and column j). You can expand along ANY row or column — the result is always the same determinant. Strategy: choose the row or column with the most zeros to minimize calculations. For a 3×3 matrix, expanding along a row with a zero eliminates one term entirely.
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