Matrices and systems of linear equations
Matrices and systems of linear equations open the PAU Applied Mathematics for Social Sciences II paper in the Canary Islands. The exam asks you to multiply and invert matrices, compute a determinant and a rank, solve a matrix equation AX = B, and above all to discuss a linear system containing a parameter and then solve the consistent cases.
Core rules
- The product AB exists when columns of A equal rows of B; generally AB ≠ BA and (AB)^T = B^T A^T.
- det(AB) = det A · det B, det(A^T) = det A, and det(kA) = k^n det A for an n×n matrix. A is invertible exactly when det A ≠ 0.
- rank(A) is the order of the largest non-zero minor; it also equals the number of non-zero rows left after Gaussian elimination.
- Rouché–Capelli: the system is consistent if and only if rank(A) = rank(A|b). If that value equals the number of unknowns n, the solution is unique; if it is r < n, there are infinitely many solutions with n − r free parameters.
- Cramer: x_i = det(A_i)/det A, where A_i is A with column i replaced by b; it requires det A ≠ 0.
Worked example
Discuss and solve for the parameter k: x + y + z = 1, x + 2y + 3z = 2, x + 4y + kz = 4.
Expanding along the first row: det A = (2k − 12) − (k − 3) + (4 − 2) = k − 7. If k ≠ 7 the determinant is non-zero, so the system is Cramer-type and has a unique solution. If k = 7, det A = 0, but the minor from rows 1–2 and columns 1–2 equals 1 ≠ 0, so rank A = 2. All 3×3 minors of the augmented matrix also vanish, so rank(A|b) = 2 and the system is consistent with 3 − 2 = 1 parameter. The first two equations give y = 1 − 2z and x = z, hence the solutions (t, 1 − 2t, t); substituting into the third equation confirms consistency when k = 7.
Traps that cost marks
- Stopping at det A = 0 without comparing ranks.
- Using det(kA) = k det A instead of k^n det A.
- Applying Cramer after finding det A = 0.
- A sign slip in the cofactor expansion: re-expand along another row to check.
- From AX = B concluding X = BA⁻¹ instead of X = A⁻¹B.