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These are complete exercise solutions for Samacheer Kalvi 12th Business Maths — Applications of Matrices and Determinants (Chapter 1). Solve each question yourself first, then verify your method and answer below. Focus on understanding the step-by-step method — the same patterns repeat in exams.
Key Concepts (Quick Recap)
- Matrix: A rectangular array of numbers arranged in rows and columns. Order is m × n.
- Determinant: A scalar value computed from a square matrix. For 2×2: |A| = ad − bc. For 3×3: expand along any row/column using cofactors.
- Rank of a matrix: The number of linearly independent rows (or columns). Equals the number of non-zero rows in row-echelon form. Equals the highest order of a non-zero minor. Denoted ρ(A).
- Consistency of a system AX = B:
- If ρ(A) = ρ(A|B) = n (number of variables) → unique solution
- If ρ(A) = ρ(A|B) < n → infinitely many solutions
- If ρ(A) ≠ ρ(A|B) → no solution (inconsistent)
- Inverse of a matrix: A⁻¹ = (1/|A|) · adj(A). Exists only if |A| ≠ 0.
- Adjoint of a matrix: adj(A) = (Cofactor matrix)ᵀ. Key property: A · adj(A) = adj(A) · A = |A| · I.
- Cramer's Rule: For AX = B, x = Δₓ/Δ, y = Δᵧ/Δ, z = Δ_z/Δ where Δ = |A|.
- Minor and Cofactor: Minor Mᵢⱼ is the determinant of the submatrix after deleting row i and column j. Cofactor Cᵢⱼ = (−1)^(i+j) · Mᵢⱼ.
- Elementary row operations: Rᵢ ↔ Rⱼ, Rᵢ → kRᵢ, Rᵢ → Rᵢ + kRⱼ — these preserve rank.
- Row-echelon form: All zero rows at the bottom, leading entry of each non-zero row is to the right of the one above.
I. Choose the Best Answer (MCQs)
(a) 5 (b) 25 ✓ (c) 1/5 (d) 125
| **Q1. If A is a 3×3 matrix and | A | = 5, then | adj(A) | equals:** |
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| Answer: (b) ✓ — For an n×n matrix, | adj(A) | = | A | ^(n−1). Here n = 3, so | adj(A) | = | A | ² = 5² = 25. |
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Q2. The rank of the matrix [[1,2],[2,4]] is: (a) 1 ✓ (b) 2 (c) 0 (d) 3
Answer(a) ✓ — R₂ = 2R₁, so in row-echelon form we get [[1,2],[0,0]]. Only one non-zero row, so rank = 1.
Q3. The system x + y = 3, 2x + 2y = 6 has: (a) No solution (b) Unique solution (c) Infinitely many solutions ✓ (d) Cannot determine
| Answer: (c) ✓ — The second equation is 2 times the first. ρ(A) = ρ(A | B) = 1 < 2 (number of variables). So infinitely many solutions. |
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Q4. If A⁻¹ = [[2,0],[0,2]], then A equals: (a) [[2,0],[0,2]] (b) [[0.5,0],[0,0.5]] ✓ (c) [[4,0],[0,4]] (d) [[1,0],[0,1]]
| Answer: (b) ✓ — A = (A⁻¹)⁻¹. Invert [[2,0],[0,2]]: | A⁻¹ | = 4, adj = [[2,0],[0,2]], so A = (1/4)[[2,0],[0,2]] = [[0.5,0],[0,0.5]]. |
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Q5. Cramer's rule is applicable only when:
Answer
| (a) | A | = 0 (b) | A | ≠ 0 ✓ (c) A is singular (d) A is rectangular |
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| Answer: (b) ✓ — Cramer's rule requires Δ = | A | ≠ 0, i.e., the coefficient matrix must be non-singular. |
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Q6. The cofactor C₂₃ of the matrix [[1,2,3],[4,5,6],[7,8,9]] is: (a) −6 ✓ (b) 6 (c) −3 (d) 3
| Answer: (a) ✓ — Delete row 2, column 3: submatrix [[1,2],[7,8]]. Minor M₂₃ = | 1,2;7,8 | = 8−14 = −6. Cofactor C₂₃ = (−1)^(2+3) · (−6) = (−1)⁵ · (−6) = (−1)(−6) = 6. Wait — let me recalculate: C₂₃ = (−1)^(2+3) · M₂₃ = (−1) · (−6) = 6. Actually, M₂₃ = 1×8 − 2×7 = 8−14 = −6. C₂₃ = (−1)⁵(−6) = (−1)(−6) = 6. The answer is (b) = 6. |
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Correction: Answer is (b) ✓ — C₂₃ = (−1)^(2+3) · M₂₃ = −1 · (−6) = 6.
Q7. If A is an orthogonal matrix, then A⁻¹ equals: (a) A (b) −A (c) Aᵀ ✓ (d) adj(A)
Answer(c) ✓ — An orthogonal matrix satisfies AᵀA = I, so A⁻¹ = Aᵀ.
Q8. The system of 3 equations in 2 unknowns is: (a) Always consistent (b) Always inconsistent (c) May or may not be consistent ✓ (d) Has a unique solution
Answer(c) ✓ — A system with more equations than unknowns may be consistent (if equations are dependent) or inconsistent (if contradictory). Consistency depends on the specific equations.
II. Fill in the Blanks
| 1. For a 2×2 matrix A, | adj(A) | = ______. ** | A | ^(2−1) = | A | ** |
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2. A matrix is said to be singular if its ______ is zero. determinant
3. The rank of an identity matrix of order 3 is ______. 3
| 4. If | A | ≠ 0, then the system AX = B has a ______ solution. unique |
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5. In Cramer's rule, if Δ = 0 and at least one of Δₓ, Δᵧ, Δ_z is non-zero, the system is ______. inconsistent
| 6. The product A · adj(A) equals ______. ** | A | · I** |
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7. The rank of a null matrix (zero matrix) of any order is ______. 0
| 8. If A is a non-singular matrix of order 3, then A⁻¹ = ______. **(1/ | A | ) · adj(A)** |
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III. True or False
1. The rank of a matrix cannot exceed the smaller of its number of rows and columns. → True. Rank ≤ min(m, n) for an m×n matrix.
| 2. Cramer's rule can solve a system of equations when the determinant of the coefficient matrix is zero. → False. Cramer's rule requires | A | ≠ 0. When | A | = 0, the system is either inconsistent or has infinitely many solutions — Cramer's rule does not apply. |
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3. The adjoint of a symmetric matrix is symmetric. → True. If A = Aᵀ, then the cofactor matrix is symmetric, and its transpose (the adjoint) equals itself.
| 4. A square matrix with zero determinant has no inverse. → True. A⁻¹ exists only when | A | ≠ 0. If | A | = 0, the matrix is singular and non-invertible. |
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5. Elementary row operations change the rank of a matrix. → False. Elementary row operations do not change the rank of a matrix — they preserve row space and linear independence.
| 6. **If ρ(A) = ρ(A | B) = r < n, the system has a unique solution. → False.** When ρ(A) = ρ(A | B) = r < n, there are infinitely many solutions (n − r free variables). Unique solution requires ρ(A) = ρ(A | B) = n. |
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| 7. **The determinant of adj(A) equals | A | ^(n−1) for an n×n matrix. → True.** This is a standard property: | adj(A) | = | A | ^(n−1). |
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IV. Match the Following
| Left | Right |
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| 1. Non-singular matrix | d. | A | ≠ 0 |
| 2. Cramer's rule | e. x = Δₓ/Δ |
| 3. Rank of identity matrix Iₙ | a. n |
| 4. A · adj(A) | c. | A | · I |
| 5. Singular matrix | b. | A | = 0 |
Answer1→d, 2→e, 3→a, 4→c, 5→b
V. Answer Briefly
Q1. Find the rank of A = [[1,2,3],[2,4,6],[1,1,1]].
Answer
R₂ → R₂ − 2R₁: [[1,2,3],[0,0,0],[1,1,1]] R₃ → R₃ − R₁: [[1,2,3],[0,0,0],[0,−1,−2]] Swap R₂ and R₃: [[1,2,3],[0,−1,−2],[0,0,0]] Two non-zero rows → ρ(A) = 2
Q2. Find the cofactor matrix of A = [[1,2],[3,4]].
Answer
C₁₁ = (−1)^(1+1) · 4 = 4 C₁₂ = (−1)^(1+2) · 3 = −3 C₂₁ = (−1)^(2+1) · 2 = −2 C₂₂ = (−1)^(2+2) · 1 = 1 Cofactor matrix = [[4,−3],[−2,1]]
Q3. Find adj(A) for A = [[1,2],[3,4]].
Answer
Cofactor matrix = [[4,−3],[−2,1]] adj(A) = (Cofactor matrix)ᵀ = [[4,−2],[−3,1]]
| **Q4. Verify A · adj(A) = | A | · I for A = [[1,2],[3,4]].** |
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|A| = 1×4 − 2×3 = 4 − 6 = −2 adj(A) = [[4,−2],[−3,1]] A · adj(A) = [[1,2],[3,4]] · [[4,−2],[−3,1]] = [[4−6,−2+2],[12−12,−6+4]] = [[−2,0],[0,−2]] = −2 · [[1,0],[0,1]] = |A| · I ✓
Q5. Using Cramer's rule, solve x + 2y = 5, 3x − y = 1.
Answer
Δ = |1,2;3,−1| = −1 − 6 = −7 Δₓ = |5,2;1,−1| = −5 − 2 = −7 → x = Δₓ/Δ = −7/−7 = 1 Δᵧ = |1,5;3,1| = 1 − 15 = −14 → y = Δᵧ/Δ = −14/−7 = 2
Q6. Check the consistency of the system: x + y + z = 6, x + 2y + 3z = 14, x + 4y + 7z = 30.
Answer
Augmented matrix: [[1,1,1,|6],[1,2,3,|14],[1,4,7,|30]] R₂→R₂−R₁: [[1,1,1,|6],[0,1,2,|8],[1,4,7,|30]] R₃→R₃−R₁: [[1,1,1,|6],[0,1,2,|8],[0,3,6,|24]] R₃→R₃−3R₂: [[1,1,1,|6],[0,1,2,|8],[0,0,0,|0]] ρ(A) = 2, ρ(A|B) = 2 → Consistent with infinitely many solutions (ρ < n = 3).
Q7. Find the determinant of A = [[2,−1,3],[1,2,−1],[3,1,2]].
Answer
Expanding along R₁: |A| = 2[2×2 − (−1)×1] − (−1)[1×2 − (−1)×3] + 3[1×1 − 2×3] = 2[4+1] + 1[2+3] + 3[1−6] = 2(5) + 1(5) + 3(−5) = 10 + 5 − 15 = 0
VI. Answer in Detail
Q1. Find the inverse of A = [[2,1,1],[1,2,1],[1,1,2]] using the adjoint method. Show all steps.
Answer
Step 1: Find |A| |A| = 2[2×2 − 1×1] − 1[1×2 − 1×1] + 1[1×1 − 2×1] = 2[4−1] − 1[2−1] + 1[1−2] = 2(3) − 1(1) + 1(−1) = 6 − 1 − 1 = 4 Since |A| = 4 ≠ 0, A⁻¹ exists. Step 2: Find cofactors C₁₁ = +(4−1) = 3, C₁₂ = −(2−1) = −1, C₁₃ = +(1−2) = −1 C₂₁ = −(2−1) = −1, C₂₂ = +(4−1) = 3, C₂₃ = −(2−1) = −1 C₃₁ = +(1−2) = −1, C₃₂ = −(2−1) = −1, C₃₃ = +(4−1) = 3 Cofactor matrix = [[3,−1,−1],[−1,3,−1],[−1,−1,3]] Step 3: Find adj(A) = transpose of cofactor matrix adj(A) = [[3,−1,−1],[−1,3,−1],[−1,−1,3]] (matrix is symmetric, so adj = cofactor matrix) Step 4: A⁻¹ = (1/|A|) · adj(A) A⁻¹ = (1/4) · [[3,−1,−1],[−1,3,−1],[−1,−1,3]] = [[3/4, −1/4, −1/4],[−1/4, 3/4, −1/4],[−1/4, −1/4, 3/4]] Verification: A · A⁻¹ should equal I₃.
Q2. Solve the system using the matrix inverse method: 2x + y + z = 5, 4x − 6y = −2, −2x + 7y + 2z = 9.
Answer
Writing as AX = B: A = [[2,1,1],[4,−6,0],[−2,7,2]], X = [[x],[y],[z]], B = [[5],[−2],[9]] Step 1: Find |A| |A| = 2(−6×2 − 0×7) − 1(4×2 − 0×(−2)) + 1(4×7 − (−6)(−2)) = 2(−12) − 1(8) + 1(28 − 12) = −24 − 8 + 16 = −16 Step 2: Find adj(A) — cofactors: C₁₁ = −12 − 0 = −12, C₁₂ = −(8−0) = −8, C₁₃ = 28 − 12 = 16 C₂₁ = −(2−7) = 5, C₂₂ = 4−(−2) = 6, C₂₃ = −(14−(−2)) = −16 Wait, let me recalculate carefully. C₁₁ = +(−6×2 − 0×7) = −12 C₁₂ = −(4×2 − 0×(−2)) = −(8) = −8 C₁₃ = +(4×7 − (−6)(−2)) = 28 − 12 = 16 C₂₁ = −(1×2 − 1×7) = −(2−7) = 5 C₂₂ = +(2×2 − 1×(−2)) = 4+2 = 6 C₂₃ = −(2×7 − 1×(−2)) = −(14+2) = −16 C₃₁ = +(1×0 − 1×(−6)) = 0+6 = 6 C₃₂ = −(2×0 − 1×4) = −(0−4) = 4 C₃₃ = +(2×(−6) − 1×4) = −12−4 = −16 Cofactor matrix = [[−12,−8,16],[5,6,−16],[6,4,−16]] adj(A) = transpose = [[−12,5,6],[−8,6,4],[16,−16,−16]] Step 3: X = A⁻¹B = (1/−16) · adj(A) · B adj(A) · B = [[−12×5+5×(−2)+6×9],[−8×5+6×(−2)+4×9],[16×5+(−16)×(−2)+(−16)×9]] = [[−60−10+54],[−40−12+36],[80+32−144]] = [[−16],[−16],[−32]] X = (1/−16) · [[−16],[−16],[−32]] = [[1],[1],[2]] Therefore x = 1, y = 1, z = 2
Q3. Using Cramer's rule, solve: x + y + z = 6, x − y + z = 2, 2x + y − z = 1.
Answer
A = [[1,1,1],[1,−1,1],[2,1,−1]], B = [[6],[2],[1]] Δ = |A| = 1(1−1) − 1(−1−2) + 1(1+2) = 0 + 3 + 3 = 6 Δₓ = |6,1,1;2,−1,1;1,1,−1| = 6(1−1) − 1(−2−1) + 1(2+1) = 0 + 3 + 3 = 6 Δᵧ = |1,6,1;1,2,1;2,1,−1| = 1(−2−1) − 6(−1−2) + 1(1−4) = −3 + 18 − 3 = 12 Δ_z = |1,1,6;1,−1,2;2,1,1| = 1(−1−2) − 1(1−4) + 6(1+2) = −3 + 3 + 18 = 18 x = Δₓ/Δ = 6/6 = 1 y = Δᵧ/Δ = 12/6 = 2 z = Δ_z/Δ = 18/6 = 3 Solution: x = 1, y = 2, z = 3
Q4. Find the rank of A = [[1,2,3,4],[2,4,6,8],[3,6,9,12]] using the row-echelon method.
Answer
Step 1: R₂ → R₂ − 2R₁ [[1,2,3,4],[0,0,0,0],[3,6,9,12]] Step 2: R₃ → R₃ − 3R₁ [[1,2,3,4],[0,0,0,0],[0,0,0,0]] Only 1 non-zero row → ρ(A) = 1 Note: Every row is a multiple of the first row, so there is only 1 linearly independent row.
Q5. Test the consistency and solve if possible: x + y + z = 3, x + 2y + 3z = 4, 2x + 3y + 4z = 7.
Answer
Augmented matrix: [[1,1,1,|3],[1,2,3,|4],[2,3,4,|7]] R₂ → R₂ − R₁: [[1,1,1,|3],[0,1,2,|1],[2,3,4,|7]] R₃ → R₃ − 2R₁: [[1,1,1,|3],[0,1,2,|1],[0,1,2,|1]] R₃ → R₃ − R₂: [[1,1,1,|3],[0,1,2,|1],[0,0,0,|0]] ρ(A) = 2, ρ(A|B) = 2, n = 3 ρ(A) = ρ(A|B) = 2 < 3 → Consistent with infinitely many solutions Let z = t (parameter). From R₂: y + 2t = 1 → y = 1 − 2t From R₁: x + (1−2t) + t = 3 → x = 2 + t Solution: x = 2+t, y = 1−2t, z = t for any real t.
Q6. A company produces three products P₁, P₂, P₃ using resources R₁, R₂. The requirements per unit are given by matrix A = [[2,3,1],[1,2,4]]. The available resources are B = [[50],[60]]. If the production vector is X = [[x],[y],[z]], set up the system and explain how Cramer's rule would apply.
Answer
System: 2x + 3y + z = 50, x + 2y + 4z = 60 |A| (coefficient determinant) for the augmented matrix [[2,3,1],[1,2,4]] is 2×4 − 3×1 = 5 for a 2×3 system. But Cramer's rule directly applies only for square coefficient matrices (n equations, n unknowns). Since we have 2 equations in 3 unknowns, the system is underdetermined. We would need an additional equation or must express the solution in terms of a parameter. If we had a 3rd equation (e.g., a constraint like x + y + z = 25), we could form a 3×3 system and apply Cramer's rule with Δ, Δₓ, Δᵧ, Δ_z as usual.
Common Mistakes to Avoid
1. Sign errors in cofactor expansion: Always use (−1)^(i+j) correctly. A common error is forgetting the sign for off-diagonal cofactors. Example: C₂₁ = −M₂₁, not +M₂₁.
2. Confusing adjoint with cofactor matrix: adj(A) is the transpose of the cofactor matrix. Don't forget to transpose after computing cofactors.
| 3. Applying Cramer's rule when Δ = 0: Cramer's rule requires | A | ≠ 0. If Δ = 0, check consistency using rank instead — don't just write "no solution." |
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4. Row operation errors during rank calculation: Always apply row operations correctly. A common mistake is adding rows in the wrong direction (Rᵢ → Rᵢ − Rⱼ vs Rⱼ → Rⱼ − Rᵢ).
5. Wrong determinant expansion for 3×3 matrices: When expanding along a row/column, each element multiplies its cofactor (not just its minor). Use Cᵢⱼ = (−1)^(i+j) · Mᵢⱼ.
| 6. Incorrect consistency check: Compare ρ(A) with ρ(A | B), not just | A | with 0. A zero determinant does not automatically mean "no solution" — the system could have infinitely many solutions. |
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Exam Checklist
- [ ] I can find the determinant of 2×2 and 3×3 matrices correctly
- [ ] I can compute minors, cofactors, and the adjoint of a matrix
- [ ] I can verify the property A · adj(A) = |A| · I
- [ ] I can find the inverse of a 3×3 matrix using A⁻¹ = (1/|A|) · adj(A)
- [ ] I can apply Cramer's rule to solve 2×2 and 3×3 systems
- [ ] I can determine the rank of a matrix using row-echelon form
- [ ] I can test the consistency of a system AX = B using rank
- [ ] I can distinguish between unique solution, no solution, and infinitely many solutions based on rank
