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CBSE · Class 12 · Mathematics

Systems of Linear Equations by the Matrix Method (Ch 4: 4.6)

2 concepts · 16 practice questions with answers · Free to read

Solving two equations by the inverse

From daily life

A shop sells pens and books. Two bills give two equations. Writing them as one matrix equation lets the inverse undo the mixing and reveal the prices.

Simple definition

A system a₁x + b₁y = c₁, a₂x + b₂y = c₂ is written as AX = B, where A holds the coefficients, X = [x ; y] and B = [c₁ ; c₂]. If |A| ≠ 0, the unique solution is X = A⁻¹B. If |A| = 0 look at (adj A)B: if it is not zero the system has no solution.

Formula: AX = B ⇒ X = A⁻¹ B

Rule 1: Solve by X = A⁻¹B — Find A⁻¹, then multiply by B.

  • Easy: 2x + y = 7, x + y = 4: A⁻¹ = [1 −1 ; −1 2], X = [3 ; 1].
  • Medium: 3x + 2y = 12, 4x + 3y = 17: A⁻¹ = [3 −2 ; −4 3], X = [2 ; 3].
  • Exam-type: 2x + 4y = 10, x + 3y = 7: |A| = 2, A⁻¹ = ½ [3 −4 ; −1 2], X = [1 ; 2].

Rule 2: When |A| = 0 — Test the system with (adj A)B.

  • Easy: x + 2y = 3, 2x + 4y = 7: |A| = 0 and (adj A)B = [−2 ; 1] is not zero, so there is no solution.
  • Medium: x + 2y = 3, 2x + 4y = 6: |A| = 0 and (adj A)B = 0; the two equations are the same line, so there are infinitely many solutions.
  • Exam-type: Substitute back: for X = [2 ; 3] we have 3(2) + 2(3) = 12 and 4(2) + 3(3) = 17.
Remember

The matrix method needs |A| ≠ 0; otherwise use another approach.

Common mistake

Writing X = BA⁻¹; the order must be A⁻¹B.

Memory clue

AX = B, so multiply by A⁻¹ on the left of both sides.

Practice questions

Easy
  1. 1Write x + 2y = 5, 3x − y = 1 in the form AX = B.
  2. 2Solve x + y = 5, x − y = 1 by the matrix method.
  3. 3Solve 3x + y = 10, x + y = 6.
Medium
  1. 4Solve 4x + 3y = 11, x + y = 3.
  2. 5Solve x + 3y = 11, 2x + 5y = 19.
  3. 6Does 2x + 3y = 5, 4x + 6y = 7 have a solution?
Challenging
  1. 7For which k does kx + y = 3, 2x + y = 5 fail to have a unique solution?
  2. 8Two pens and three books cost ₹130; one pen and two books cost ₹80. Find the price of each.
Show answers (try them yourself first)
  1. 1A = [1 2 ; 3 −1], X = [x ; y], B = [5 ; 1]
  2. 2x = 3, y = 2
  3. 3|A| = 2, A⁻¹ = ½ [1 −1 ; −1 3], X = [2 ; 4], so x = 2, y = 4
  4. 4A⁻¹ = [1 −3 ; −1 4], X = [2 ; 1], so x = 2, y = 1
  5. 5|A| = −1, A⁻¹ = [−5 3 ; 2 −1], so x = 2, y = 3
  6. 6No: |A| = 0 and (adj A)B = [9 ; −6] is not zero
  7. 7|A| = k − 2 = 0 gives k = 2
  8. 8Pen ₹20, book ₹30 (check: 40 + 90 = 130, 20 + 60 = 80)

Solving three equations by the inverse

From daily life

Three friends tell you the sums of pairs of their ages. From the three sums you can recover each age. Three equations are written as one matrix equation.

Simple definition

For three equations in x, y, z, A is a 3 × 3 matrix of coefficients, X = [x ; y ; z] and B holds the constants. When |A| ≠ 0, X = A⁻¹B with A⁻¹ = adj A / |A|. If |A| = 0 this method does not give a unique solution.

Formula: X = A⁻¹B = (adj A) B / |A|

Rule 1: Solve a 3 × 3 system — Find |A|, adj A, then multiply.

  • Easy: x + y = 3, y + z = 5, x + z = 4 with A⁻¹ = ½ [1 −1 1 ; 1 1 −1 ; −1 1 1] gives X = [1 ; 2 ; 3].
  • Medium: With A = [2 1 0 ; 1 3 2 ; 0 1 4] and B = [4 ; 9 ; 6], X = (1/16) [16 ; 32 ; 16] = [1 ; 2 ; 1].
  • Exam-type: Check x = 1, y = 2, z = 1 in 2x + y = 4; x + 3y + 2z = 9; y + 4z = 6.

Rule 2: Check |A| first — A zero determinant means the method fails.

  • Easy: [1 1 1 ; 2 2 2 ; 1 −1 0] has |A| = 0 because rows 1 and 2 are proportional.
  • Medium: A system with |A| = 0 and (adj A)B not zero has no solution.
  • Exam-type: A system whose first two equations have the same left side but different right sides is inconsistent.
Remember

Always substitute your answer back into all three equations.

Common mistake

Starting to multiply before checking |A| ≠ 0.

Memory clue

|A| first, adj A second, multiply third.

Practice questions

Easy
  1. 1For x − y + z = 2, y + z = 4, x + z = 5, find |A| and solve.
  2. 2Can x + y + z = 6, 2x + 2y + 2z = 12, x − y = 0 be solved by the inverse method? Why?
  3. 3Solve AX = [5 ; 11 ; 13] for A = [2 1 0 ; 1 3 2 ; 0 1 4].
Medium
  1. 4Solve x + y = 7, y + z = 9, x + z = 10.
  2. 5A pen and a pencil cost ₹9, a pencil and an eraser ₹7, a pen and an eraser ₹8. Find each price.
  3. 6Show that A = [1 1 0 ; 0 1 1 ; 1 0 1] gives a unique solution for every B.
Challenging
  1. 7If |A| = 0 and (adj A)B is not the zero matrix, what can you say about AX = B?
  2. 8Decide whether x + y + z = 1, x + y + z = 2, x − y = 0 has a solution.
Show answers (try them yourself first)
  1. 1|A| = −1; the solution is x = 4, y = 3, z = 1
  2. 2No: |A| = 0 because rows 1 and 2 are proportional
  3. 3X = [2 ; 1 ; 3]
  4. 4x = 4, y = 3, z = 6
  5. 5Pen ₹5, pencil ₹4, eraser ₹3
  6. 6|A| = 2, which is not 0, so A⁻¹ exists and X = A⁻¹B is unique for every B
  7. 7The system is inconsistent (no solution)
  8. 8No solution: the first two equations contradict each other