What is a determinant?
Every square matrix A has a number associated with it, called its determinant, written |A| or det A.
For a 2 × 2 matrix:
|a b; c d| = ad − bc
Example. |3 2; 1 4| = 3 × 4 − 2 × 1 = 10.
Minors and cofactors
The minor Mᵢⱼ of element aᵢⱼ is the determinant left after deleting row i and column j.
The cofactor is Aᵢⱼ = (−1)ⁱ⁺ʲ Mᵢⱼ. The signs follow a chessboard pattern: + − + / − + − / + − +.
Evaluating a 3 × 3 determinant
Expand along any row or column: multiply each element by its cofactor and add. Expanding along the first row:
|A| = a₁₁A₁₁ + a₁₂A₁₂ + a₁₃A₁₃
Example. Evaluate |1 2 3; 0 1 4; 5 6 0|.
= 1 × (1·0 − 4·6) − 2 × (0·0 − 4·5) + 3 × (0·6 − 1·5) = 1 × (−24) − 2 × (−20) + 3 × (−5) = −24 + 40 − 15 = 1.
Tip: expand along the row or column with the most zeros to save work.
An important fact: if you multiply the elements of a row by the cofactors of a different row and add, the result is always 0.
Area of a triangle
The area of a triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) is
Δ = ½ |x₁ y₁ 1; x₂ y₂ 1; x₃ y₃ 1|
Take the absolute value, since area cannot be negative. If Δ = 0, the three points are collinear (on one straight line).
Example. Find the area of the triangle with vertices (1, 0), (6, 0) and (4, 3).
Δ = ½ |1 0 1; 6 0 1; 4 3 1|. Expanding along the second column (two zeros): only the 3 contributes, with cofactor sign (−1)³⁺² = −1, so the determinant is −3 × (1·1 − 1·6) = −3 × (−5) = 15. The area is ½ × 15 = 7.5 square units.
Adjoint and inverse
The adjoint of A, adj A, is the transpose of the matrix of cofactors.
A key identity: A (adj A) = (adj A) A = |A| I. From this:
A⁻¹ = (1/|A|) adj A, which exists only when |A| ≠ 0
If |A| ≠ 0, A is non-singular (invertible).
If |A| = 0, A is singular and has no inverse.
For a 2 × 2 matrix, the adjoint is easy: swap the diagonal elements and change the signs of the off-diagonal ones.
If A = [a b; c d], then A⁻¹ = (1/(ad − bc)) [d −b; −c a]
Example. A = [3 2; 1 4], |A| = 10, so A⁻¹ = (1/10)[4 −2; −1 3].
Useful facts for an n × n matrix: |AB| = |A||B|, |Aᵀ| = |A|, |kA| = kⁿ|A|, and |adj A| = |A|ⁿ⁻¹.
Solving linear equations with matrices
A system of linear equations can be written as AX = B, where A is the coefficient matrix, X is the column of unknowns, and B is the column of constants.
If |A| ≠ 0, the system has a unique solution: X = A⁻¹B. The system is consistent.
If |A| = 0 and (adj A)B ≠ O, the system has no solution (it is inconsistent).
If |A| = 0 and (adj A)B = O, it has either infinitely many solutions or none.
Example. Solve 3x + 2y = 7 and x + 4y = 9.
A = [3 2; 1 4], X = [x; y], B = [7; 9]. Using A⁻¹ from above:
X = (1/10)[4 −2; −1 3][7; 9] = (1/10)[28 − 18; −7 + 27] = (1/10)[10; 20] = [1; 2].
So x = 1, y = 2. Check: 3 + 4 = 7 ✓ and 1 + 8 = 9 ✓.