haya.Learning made clear
Unit IV · Determinants
9 min readConcept sheet1-mark MCQ and assertion-reason; 2 and 3-mark questions on whether a system is consistent; also a JEE favourite

Consistency of a system and the number of solutions

A system is consistent if it has at least one solution and inconsistent if it has none. Test |A| first, then (adj A)B.

Matrix forma₁x + b₁y + c₁z = d₁ (and two more) becomes AX = B: A holds the coefficients, X = [x; y; z] and B = [d₁; d₂; d₃]
|A| ≠ 0Consistent, with a unique solution
|A| = 0 and (adj A)B ≠ OInconsistent: no solution
|A| = 0 and (adj A)B = OEither infinitely many solutions (consistent) or no solution (inconsistent); check the equations

What consistent means

A system of linear equations is consistent if it has at least one solution, and inconsistent if it has none. Think of two equations in x and y as two lines. If they cross once, there is a unique solution. If they are the same line, there are infinitely many. If they are parallel, there is no solution. With three variables, the matrix test below decides.

Decision table for AX = B

|A|(adj A)BNumber of solutionsType
≠ 0Not neededExactly one, X = A⁻¹BConsistent
= 0≠ ONoneInconsistent
= 0= OInfinitely many, or noneConsistent if infinitely many; inconsistent if none

Example 1: a system with no solution

Problem

Check whether the system 2x − y = 5, 4x − 2y = 7 is consistent.

Working
  1. A = [2, −1; 4, −2], X = [x; y], B = [5; 7]
  2. |A| = 2 × (−2) − (−1) × 4 = −4 + 4 = 0, so there is no unique solution.
  3. adj A = [−2, 1; −4, 2]
  4. (adj A)B = [−2 × 5 + 1 × 7; −4 × 5 + 2 × 7] = [−3; −6]
  5. (adj A)B ≠ O, so the system has no solution.
  6. On a graph the lines are parallel: 2x − y = 5 and 2x − y = 3.5.
Answer

The system is inconsistent (no solution).

Example 2: a system with infinitely many solutions

Problem

Examine the system x + y + z = 3, x − y + z = 1, 2x + 2z = 4.

Working
  1. A = [1, 1, 1; 1, −1, 1; 2, 0, 2], B = [3; 1; 4]
  2. |A| = 1(−2 − 0) − 1(2 − 2) + 1(0 + 2) = −2 − 0 + 2 = 0
  3. adj A = [−2, −2, 2; 0, 0, 0; 2, 2, −2]
  4. (adj A)B = [−6 − 2 + 8; 0; 6 + 2 − 8] = [0; 0; 0] = O
  5. So compare the equations: the third is the sum of the first two, so it adds nothing new.
  6. Subtracting the second from the first: 2y = 2, so y = 1 and x + z = 2. Put x = k: z = 2 − k.
Answer

Consistent, with infinitely many solutions: x = k, y = 1, z = 2 − k for any real k.

Watch out

  • Find |A| first. If |A| ≠ 0, stop: the system is consistent with one solution.
  • If |A| = 0, you must work out (adj A)B before deciding.
  • (adj A)B = O does not prove infinitely many solutions. Solve or compare the equations to be sure.
  • Inconsistent means no solution; it never means 'many solutions'.
  • Write B as a column matrix, so that (adj A)B is defined.
In real life

If two bills for the same items give equations with no common solution, the system is inconsistent, so at least one bill is wrong. Accountants and shopkeepers catch billing errors with exactly this kind of check.

Exam angle

2 marks: “Examine the consistency of the system 2x + 3y = 5, 6x + 9y = 12.” |A| = 18 − 18 = 0. adj A = [9, −3; −6, 2]. (adj A)B = [45 − 36; −30 + 24] = [9; −6] ≠ O. So the system is inconsistent.

2 marks: “Examine the consistency of the system x + y = 5, 2x + 2y = 10.” |A| = 2 − 2 = 0. adj A = [2, −1; −2, 1]. (adj A)B = [10 − 10; −10 + 10] = O. The second equation is twice the first, so there are infinitely many solutions: the system is consistent.

1 marks: “For what values of k does the system kx + 2y = 5, 3x + y = 1 have a unique solution?” A unique solution needs |A| ≠ 0: k − 6 ≠ 0. So every real k except k = 6.

Remember it

|A| ≠ 0: one answer. |A| = 0: work out (adj A)B; if it is not zero, there is no answer.