Matrices and Determinants
This domain tests matrix algebra, evaluation of 2×2 and 3×3 determinants, the adjoint and inverse, and solving two simultaneous equations by the matrix or Cramer method. Marks are lost mainly on sign handling, so write the cofactor signs explicitly.
Types and algebra
Row, column, square, diagonal, scalar, identity, zero, symmetric (Aᵀ = A) and skew-symmetric (Aᵀ = −A, diagonal entries zero) matrices. Addition and scalar multiplication are entry-wise. AB needs columns of A = rows of B, and generally AB ≠ BA. Also (AB)ᵀ = BᵀAᵀ.
Determinants
For a 2×2 matrix, determinant = ad − bc. Expand a 3×3 determinant along any row or column using cofactors Cᵢⱼ = (−1)^(i+j) Mᵢⱼ. Properties: |Aᵀ| = |A|; swapping two rows changes the sign; multiplying one row by k multiplies the determinant by k, so |kA| = kⁿ|A| for order n; |AB| = |A||B|; two identical rows give 0.
Inverse and solving systems
adj A is the transpose of the cofactor matrix, and A·adj A = |A|I. Hence A⁻¹ = adj A/|A|, which exists only when |A| ≠ 0. For AX = B, X = A⁻¹B; Cramer's rule gives x = Dₓ/D, y = D_y/D with D ≠ 0.
Worked example
Solve 2x + 3y = 13, x + 2y = 8. Here A = [2 3; 1 2] and |A| = 4 − 3 = 1. The cofactor matrix is [2 −1; −3 2], so adj A = [2 −3; −1 2] and A⁻¹ = adj A. Then X = A⁻¹B = [2 −3; −1 2] × [13; 8] = [26 − 24; −13 + 16] = [2; 3]. So x = 2, y = 3; check 2(2) + 3(3) = 13.
Traps
- Dropping the (−1)^(i+j) sign in a cofactor.
- Writing adj A as the cofactor matrix instead of its transpose.
- Using |kA| = k|A| rather than kⁿ|A|.
- Dividing by |A| when it is 0, where no unique solution exists.