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Class 12 Maths NCERT Solutions

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Matrices Class 12 Maths NCERT Solutions

The complete NCERT exercise solutions for Chapter 3, Matrices — 61 questions from Ex 3.1 to Ex 3.4, each worked through step by step in the CBSE marking pattern. Types of matrices, operations, transpose, symmetric and skew-symmetric forms, and elementary operations.

Class:12Subject:MathsChapter:3
4 Key Formulas
DWritten byDeep Narayan
Updated
Key Concept Summary

How many questions are in NCERT Class 12 Maths Chapter 3?

Chapter 3 carries 4 exercise questions, numbered Ex 3.1 to Ex 3.4. All of them are solved step by step on this page, along with the chapter's key formulas and exam pointers.

01

Chapter Overview

Matrices are rectangular arrays of numbers arranged in rows and columns, enclosed in brackets. They provide a compact way to represent and solve systems of linear equations, and are indispensable in physics, engineering, economics and computer science. This chapter builds the vocabulary — order, equality, transpose, symmetry — and the arithmetic (addition, scalar multiplication, product) you will use throughout determinants, linear transformations and beyond.

Board pattern

Marks are awarded for showing every intermediate step: state the order before multiplying, write each element computation, and verify both sides of a matrix identity separately. For transpose questions, always write A′ explicitly before comparing. A bare numeric answer without the working earns at most 1 mark.
02

Exercise 3.1 — Basics of Matrices

10Exercise questions

Step-by-step solution

  1. 1A has 3 rows and 4 columns.
  2. 2Number of elements = 3 × 4 = 12.
  3. 3a₁₃ = 19, a₂₁ = 35, a₃₃ = −5, a₂₄ = 12, a₂₃ = 5/3.

Final answer

Order: 3 × 4; elements: 12; a₁₃ = 19, a₂₁ = 35, a₃₃ = −5, a₂₄ = 12, a₂₃ = 5/3.

Step-by-step solution

  1. 124 = 1×24, 2×12, 3×8, 4×6, 6×4, 8×3, 12×2, 24×1 — eight possible orders.
  2. 213 is prime, so 13 = 1×13 or 13×1 — two possible orders.

Final answer

24 elements: orders 1×24, 2×12, 3×8, 4×6, 6×4, 8×3, 12×2, 24×1 (8 orders). 13 elements: orders 1×13, 13×1.

Step-by-step solution

  1. 118 = 1×18, 2×9, 3×6, 6×3, 9×2, 18×1 — six possible orders.
  2. 25 is prime, so 5 = 1×5 or 5×1 — two possible orders.

Final answer

18 elements: orders 1×18, 2×9, 3×6, 6×3, 9×2, 18×1. 5 elements: orders 1×5, 5×1.

Step-by-step solution

  1. 1a₁₁ = (1+1)²/2 = 2, a₁₂ = (1+2)²/2 = 9/2, a₂₁ = (2+1)²/2 = 9/2, a₂₂ = (2+2)²/2 = 8.

Final answer

Step-by-step solution

  1. 1Compute each element:
  2. 2a₁₁ = 4/2 = 2, a₁₂ = 5/2, a₁₃ = 3, a₁₄ = 7/2.
  3. 3a₂₁ = 7/2, a₂₂ = 4, a₂₃ = 9/2, a₂₄ = 5.
  4. 4a₃₁ = 5, a₃₂ = 11/2, a₃₃ = 6, a₃₄ = 13/2.

Final answer

Step-by-step solution

  1. 1Compare corresponding elements: 4 = y and x = 2.

Final answer

x = 2, y = 4.

Step-by-step solution

  1. 1Compare: a + b = 6, c + d = 8. No further constraint from the given equations, so infinitely many solutions exist.
  2. 2For example a = 2, b = 4, c = 3, d = 5 is one solution.

Final answer

a + b = 6 and c + d = 8 (infinitely many solutions, e.g. a=2, b=4, c=3, d=5).

Step-by-step solution

  1. 1Compare (1,1): 2a + b = 3.
  2. 2Compare (1,2): 3a − b = 1.
  3. 3Add: 5a = 4, so a = 4/5.
  4. 4Then b = 3 − 2(4/5) = 7/5.

Final answer

a = 4/5, b = 7/5.

Step-by-step solution

  1. 1Compare entries: x + y = 3, z = 2, 5 + w = 7, y + z = 6.
  2. 2From z = 2 and y + z = 6: y = 4.
  3. 3From x + y = 3: x = −1.
  4. 4From 5 + w = 7: w = 2.

Final answer

x = −1, y = 4, z = 2, w = 2.

Step-by-step solution

  1. 1Compare the four entries: a − b = −1, 2a − c = 5, 2a − b = 0, 3c + d = 13.
  2. 2Subtract (1,1) from (2,1): (2a − b) − (a − b) = 0 − (−1), so a = 1.
  3. 3From a − b = −1: b = a + 1 = 2.
  4. 4From 2a − c = 5: c = 2(1) − 5 = −3.
  5. 5From 3c + d = 13: d = 13 − 3(−3) = 22.

Final answer

a = 1, b = 2, c = −3, d = 22.

03

Exercise 3.2 — Matrix Algebra

28Exercise questions

Step-by-step solution

  1. 1(i) A + B: add corresponding entries.
  2. 2(ii) A − B: subtract corresponding entries.
  3. 3(iii) 3A − C: compute 3A first, then subtract C.
  4. 4(iv) 3A + 4B: compute 3A and 4B, then add.

Final answer

Step-by-step solution

  1. 1(i) Multiply row by column for each entry.
  2. 2(ii) A 3×1 times 1×3 gives a 3×3 matrix.
  3. 3(iii) Multiply each row of the first matrix by each column of the second.
  4. 4(iv) A 2×3 times 3×1 gives a 2×1 matrix.

Final answer

Step-by-step solution

  1. 1(i) Compute A + B and B + A separately; both give the same result.
  2. 2(ii) Compute A − B and B − A; they are negatives of each other, hence not equal unless zero.

Final answer

Step-by-step solution

  1. 1Compute A²: multiply A by itself.
  2. 2Compute B²: multiply B by itself.
  3. 3Subtract: A² − B².

Final answer

Step-by-step solution

  1. 1From Q3: A + B and A − B are known.
  2. 2Multiply (A+B)(A−B) using the matrix product rule.
  3. 3Compare with the result from Q4.

Final answer

(A+B)(A−B) = [−10,10; −10,10]. This is NOT equal to A² − B², so the identity (A+B)(A−B) = A² − B² fails for matrix multiplication.

Step-by-step solution

  1. 1Multiply the first equation by 3 and the second by 2: 6X + 9Y = 3A and 6X + 4Y = 2B.
  2. 2Subtract: 5Y = 3A − 2B = [6,9;12,0] − [4,−4;−2,10] = [2,13;14,−10]. So Y = (1/5)[2,13;14,−10].
  3. 3Substitute: 2X = A − 3Y = [2,3;4,0] − [6/5,39/5;42/5,−6] = [4/5,−24/5;−22/5,6]. So X = [2/5,−12/5;−11/5,3].
  4. 4Check: 2X + 3Y = [2,3;4,0] ✓ and 3X + 2Y = [2,−2;−1,5] ✓.

Final answer

Step-by-step solution

  1. 1The right side is the identity matrix I₂, so X = A⁻¹ where A = [3,7; 2,5].
  2. 2det(A) = 3(5) − 7(2) = 15 − 14 = 1.
  3. 3A⁻¹ = (1/det(A)) adj(A) = [5, −7; −2, 3].

Final answer

Step-by-step solution

  1. 1Compute A²: [3,1; −1,2] × [3,1; −1,2] = [8,5; −5,3].
  2. 2Compute 5A = [15,5; −5,10].
  3. 3Compute 7I = [7,0; 0,7].
  4. 4A² − 5A + 7I = [8−15+7, 5−5+0; −5+5+0, 3−10+7] = [0,0; 0,0].

Final answer

A² − 5A + 7I = O, verified.

Step-by-step solution

  1. 1Compute A², then A³ by successive matrix multiplication.
  2. 2Substitute into A³ − 6A² + 7A + 2I and show each entry equals 0.

Final answer

A³ − 6A² + 7A + 2I = O, verified.

Step-by-step solution

  1. 1(i) Compute AB, then (AB)'. Compute B'A' and compare.
  2. 2(ii) Compute 2A − B, then its transpose. Compare with 2A' − B'.

Final answer

Both identities verified: (AB)′ = B′A′ and (2A − B)′ = 2A′ − B′ hold.

Step-by-step solution

  1. 1B + C = [4,7; 2,4].
  2. 2A(B+C) = [1,2; 3,4][4,7; 2,4] = [4+4, 7+8; 12+8, 21+16] = [8,15; 20,37].
  3. 3AB = [1,2; 3,4][−1,0; 2,1] = [3,2; 5,4].
  4. 4AC = [1,2; 3,4][5,7; 0,3] = [5,13; 15,33].
  5. 5AB + AC = [8,15; 20,37] = A(B+C). ✓ Distributivity holds for matrices.

Final answer

A(B+C) = AB + AC = [8,15; 20,37]. Verified.

Step-by-step solution

  1. 1Compute AB, then (AB)C.
  2. 2Compute BC, then A(BC).
  3. 3Both give the same result.

Final answer

Step-by-step solution

  1. 1AI = A for any matrix A: AB = A × O = O.
  2. 2BA = O × A = O.
  3. 3Both are zero matrices, hence equal.

Final answer

AB = BA = O. They are equal.

Step-by-step solution

  1. 1A² = [0,−1; 1,0] × [0,−1; 1,0] = [−1,0; 0,−1] = −I.
  2. 2A⁴ = (−I)² = I.
  3. 3A²⁰ = (A⁴)⁵ = I⁵ = I.
  4. 4(A²)²⁰ = (−I)²⁰ = I.
  5. 5A⁴⁰ = (A⁴)¹⁰ = I. All equal I.

Final answer

(A²)²⁰ = A⁴⁰ = I₂, verified.

Step-by-step solution

  1. 1Base case n = 1: A¹ = [1,2; 0,1] = [1,2·1; 0,1]. True.
  2. 2Assume Aᵏ = [1,2k; 0,1]. Then Aᵏ⁺¹ = Aᵏ · A = [1,2k; 0,1][1,2; 0,1] = [1, 2k+2; 0, 1] = [1, 2(k+1); 0, 1].
  3. 3By induction, Aⁿ = [1, 2n; 0, 1] for all n.

Final answer

Step-by-step solution

  1. 1A² = [3,−2; 4,−3][3,−2; 4,−3] = [9−8, −6+6; 12−12, −8+9] = [1,0; 0,1] = I.
  2. 2Since A² = I, we have AA = I, so A⁻¹ = A.

Final answer

Step-by-step solution

  1. 1From Exercise 3.2 Q8: A² − 5A + 7I = O.
  2. 2Multiply by A⁻¹: A − 5I + 7A⁻¹ = O.
  3. 37A⁻¹ = 5I − A = [5,0; 0,5] − [3,1; −1,2] = [2,−1; 1,3].
  4. 4A⁻¹ = (1/7)[2,−1; 1,3].

Final answer

Step-by-step solution

  1. 1AB = [2,−3; 3,4][1,0; 2,1] = [2−6, 0−3; 3+8, 0+4] = [−4, −3; 11, 4].
  2. 2det(AB) = (−4)(4) − (−3)(11) = −16 + 33 = 17, so (AB)⁻¹ = (1/17)[4, 3; −11, −4].
  3. 3det A = 2(4) − (−3)(3) = 17 ⇒ A⁻¹ = (1/17)[4, 3; −3, 2]; det B = 1(1) − 0 = 1 ⇒ B⁻¹ = [1, 0; −2, 1].
  4. 4B⁻¹A⁻¹ = (1/17)[1,0; −2,1][4,3; −3,2] = (1/17)[4, 3; −8−3, −6+2] = (1/17)[4, 3; −11, −4].
  5. 5(AB)⁻¹ = B⁻¹A⁻¹, verified.

Final answer

Step-by-step solution

  1. 1A² = [1,2; 2,1][1,2; 2,1] = [1+4, 2+2; 2+2, 4+1] = [5, 4; 4, 5].
  2. 22A = [2,4; 4,2], 3I = [3,0; 0,3].
  3. 3A² − 2A − 3I = [5−2−3, 4−4−0; 4−4−0, 5−2−3] = [0,0; 0,0] = O. ✓
  4. 4Multiply A² − 2A − 3I = O by A⁻¹: A − 2I − 3A⁻¹ = O, so 3A⁻¹ = A − 2I.
  5. 5A − 2I = [−1,2; 2,−1], so A⁻¹ = (1/3)[−1,2; 2,−1].

Final answer

Step-by-step solution

  1. 1A′ = [1,0,a; 0,1,b; 0,0,1], B′ = [0,0,c; 1,0,d; 0,1,1].
  2. 2(i) A+B = [1,1,0; 0,1,1; a+c,b+d,2], so (A+B)′ = [1,0,a+c; 1,1,b+d; 0,1,2].
  3. 3A′ + B′ = [1,0,a; 0,1,b; 0,0,1] + [0,0,c; 1,0,d; 0,1,1] = [1,0,a+c; 1,1,b+d; 0,1,2]. Hence (A+B)′ = A′ + B′ ✓.
  4. 4(ii) AB = [0,1,0; 0,0,1; c,a+d,b+1], so (AB)′ = [0,0,c; 1,0,a+d; 0,1,b+1].
  5. 5B′A′ = [0,0,c; 1,0,d; 0,1,1]·[1,0,a; 0,1,b; 0,0,1] = [0,0,c; 1,0,a+d; 0,1,b+1] = (AB)′ ✓.

Final answer

(A+B)′ = A′ + B′ and (AB)′ = B′A′, both verified for these matrices.

Step-by-step solution

  1. 1A² = A × A.
  2. 2Row 3 of A²: [a·1 + b·0 + 1·a, a·0 + b·1 + 1·b, a·0 + b·0 + 1·1] = [2a, 2b, 1].
  3. 3So A² = [1,0,0; 0,1,0; 2a, 2b, 1].
  4. 4A² = I requires 2a = 0 and 2b = 0; for general a, b this fails, so A is not self-inverse.
  5. 5In fact Aⁿ = [1,0,0; 0,1,0; na, nb, 1] for n ∈ ℕ.

Final answer

Step-by-step solution

  1. 1det(A) = 4 − 2 = 2. A⁻¹ = (1/2)[1, −2; −1, 4].
  2. 2(A⁻¹)⁻¹: compute the inverse of A⁻¹.
  3. 3det(A⁻¹) = 1/2. (A⁻¹)⁻¹ = 2 · adj(A⁻¹) = 2 · (1/2)[4, 2; 1, 1] = [4,2; 1,1] = A.
  4. 4Alternatively, use the general property: (A⁻¹)⁻¹ = A always holds.

Final answer

(A⁻¹)⁻¹ = A, verified.

Step-by-step solution

  1. 1A² = [3,2; −2,−1][3,2; −2,−1] = [9−4, 6−2; −6+2, −4+1] = [5, 4; −4, −3].
  2. 22A = [6,4; −4,−2], I = [1,0; 0,1].
  3. 3A² − 2A + I = [5−6+1, 4−4+0; −4+4+0, −3+2+1] = [0,0; 0,0] = O. ✓
  4. 4Multiply by A⁻¹: A − 2I + A⁻¹ = O, so A⁻¹ = 2I − A.
  5. 5A⁻¹ = [2,0; 0,2] − [3,2; −2,−1] = [−1,−2; 2,3]. Verify: A·A⁻¹ = I. ✓

Final answer

Step-by-step solution

  1. 1If A and B are symmetric, A′ = A and B′ = B.
  2. 2(AB)′ = B′A′ = BA (reversal law for transposes).
  3. 3Given AB = BA: (AB)′ = AB.
  4. 4Therefore AB is symmetric.

Final answer

(AB)′ = B′A′ = BA = AB, so AB is symmetric.

Step-by-step solution

  1. 1(i) A+B = [1,1; 1,−1]. (A+B)² = [1,1; 1,−1][1,1; 1,−1] = [2,0; 0,2] = 2I.
  2. 2(ii) A² = [1,0; 0,1] = I. B² = [0,1; 1,0][0,1; 1,0] = [1,0; 0,1] = I.
  3. 3AB = [1,0; 0,−1][0,1; 1,0] = [0,1; −1,0]. BA = [0,1; 1,0][1,0; 0,−1] = [0,−1; 1,0]. AB ≠ BA.
  4. 42AB = [0,2; −2,0]. A²+B²+2AB = I+I+[0,2;−2,0] = [2,2; −2,2].
  5. 5(A+B)² = 2I = [2,0; 0,2] ≠ A²+B²+2AB = [2,2; −2,2].
  6. 6They are not equal because AB ≠ BA, so the binomial expansion (A+B)² = A²+AB+BA+B² ≠ A²+2AB+B².

Final answer

(A+B)² = 2I = [2,0; 0,2]. A²+B²+2AB = [2,2; −2,2]. They are NOT equal because AB ≠ BA, so (A+B)² = A²+AB+BA+B² ≠ A²+2AB+B².

Step-by-step solution

  1. 1Take A = [0,1; 0,0] and B = [1,0; 0,0].
  2. 2AB = [0,1; 0,0][1,0; 0,0] = [0,0; 0,0].
  3. 3BA = [1,0; 0,0][0,1; 0,0] = [0,1; 0,0].
  4. 4AB = O ≠ BA.

Final answer

A = [0,1; 0,0], B = [1,0; 0,0]: AB = O ≠ BA = [0,1; 0,0].

Step-by-step solution

  1. 1AB = [2·1+1·3+3·2, 2·2+1·4+3·1; −1·1+2·3+1·2, −1·2+2·4+1·1] = [11, 11; 7, 7].
  2. 2(AB)′ = [11,7; 11,7].
  3. 3A′ = [2,−1; 1,2; 3,1] (columns of A′ are the rows of A). B′ = [1,3,2; 2,4,1].
  4. 4B′A′: row 1 × columns of A′: [1·2+3·1+2·3, 1·(−1)+3·2+2·1] = [11, 7].
  5. 5Row 2: [2·2+4·1+1·3, 2·(−1)+4·2+1·1] = [11, 7]. So B′A′ = [11,7; 11,7].
  6. 6(AB)′ = B′A′ ✓.

Final answer

(AB)′ = B′A′ = [11,7; 11,7]. Verified.

Step-by-step solution

  1. 1A² = [1,−1; 1,−1][1,−1; 1,−1] = [1−1, −1+1; 1−1, −1+1] = [0, 0; 0, 0].

Final answer

A² = O. This is a nonzero matrix whose square is zero, illustrating that nonzero matrices can be nilpotent of index 2.

04

Exercise 3.3 — Transpose, Symmetric and Skew-Symmetric Matrices

19Exercise questions

Step-by-step solution

  1. 1(i) Transpose of a column matrix is a row matrix.
  2. 2(ii) Swap rows and columns.
  3. 3(iii) Swap rows and columns of the 2×3 matrix to get a 3×2 matrix.

Final answer

Step-by-step solution

  1. 1(i) Compute A+B, then (A+B)′. Compute A′ + B′ and compare.
  2. 2(ii) Compute A−B, then (A−B)′. Compute A′ − B′ and compare.

Final answer

Both identities verified by direct computation.

Step-by-step solution

  1. 1A = (A′)′ = [3,−1,0; 4,2,1].
  2. 2Compute BA, then (BA)′.
  3. 3Compute A′B′ and compare.

Final answer

(BA)′ = A′B′, verified by direct computation.

Step-by-step solution

  1. 1A′ = [−1,5,1; 2,3,−2; 3,0,1].
  2. 2A + A′ = [−2,7,4; 7,6,−2; 4,−2,2].
  3. 3Check symmetry: (A+A′)′ = A′ + (A′)′ = A′ + A = A + A′. ✓

Final answer

Step-by-step solution

  1. 1A is symmetric means A′ = A, which requires the (1,2) entry to equal the (2,1) entry.
  2. 2x = x is automatically true (the off-diagonal entries are both x).
  3. 3So A is symmetric for all real x, y. Any values of x and y work.
  4. 4More precisely, the symmetry condition gives x = x (always true) with no constraint on y beyond the given form.

Final answer

A is symmetric for all real x and y — the off-diagonal entries are both x, so A′ = A automatically.

Step-by-step solution

  1. 1A symmetric requires the (1,2) entry to equal the (2,1) entry.
  2. 2x + 1 = −2 gives x = −3.
  3. 3y is free (the diagonal entries impose no symmetry condition).

Final answer

x = −3; y can be any real number.

Step-by-step solution

  1. 1A skew-symmetric means A′ = −A, and every diagonal entry must be 0.
  2. 2Diagonal conditions: a + b = 0 and 4 = 0.
  3. 3But 4 ≠ 0, so no choice of a, b can make A skew-symmetric.

Final answer

No solution — the (2,2) entry is 4 ≠ 0, but every diagonal entry of a skew-symmetric matrix must be 0.

Step-by-step solution

  1. 1A skew-symmetric: A′ = −A, so all diagonal entries must be 0.
  2. 2The (2,2) entry is 1 ≠ 0, so A can never be skew-symmetric for any a, b.
  3. 3(The (1,3)/(3,1) entries 2 and −3 also violate skew-symmetry, since 2 = −(−3) = 3 is false.)

Final answer

No solution — the (2,2) entry is 1 ≠ 0 (diagonal of a skew-symmetric matrix must be 0).

Step-by-step solution

  1. 1Symmetric part: P = (B + B′)/2.
  2. 2B′ = [3, −4; −2, 3].
  3. 3B + B′ = [6, −6; −6, 6]. P = [3, −3; −3, 3].
  4. 4Skew-symmetric part: Q = (B − B′)/2.
  5. 5B − B′ = [0, 2; −2, 0]. Q = [0, 1; −1, 0].
  6. 6Verify: P + Q = [3, −2; −4, 3] = B. ✓

Final answer

Step-by-step solution

  1. 1A′ = [6, −2, 2; −2, 3, −1; 2, −1, 3] = A.
  2. 2Since A = A′, A is symmetric.
  3. 3P = (A + A′)/2 = A, Q = (A − A′)/2 = O.
  4. 4A = A + O.

Final answer

A is already symmetric (A′ = A), so the symmetric part is A itself and the skew-symmetric part is the zero matrix O. Hence A = A + O.

Step-by-step solution

  1. 1Assume AB is symmetric: (AB)′ = AB.
  2. 2(AB)′ = B′A′ = BA (since A, B symmetric).
  3. 3So BA = AB.
  4. 4Conversely, if AB = BA and A′ = A, B′ = B:
  5. 5(AB)′ = B′A′ = BA = AB. So AB is symmetric.
  6. 6Therefore AB is symmetric ⟺ AB = BA.

Final answer

(AB)′ = B′A′ = BA (using A′=A, B′=B). AB symmetric ⟺ (AB)′ = AB ⟺ BA = AB.

Step-by-step solution

  1. 1Let A be any square matrix. Define P = (A + A′)/2 and Q = (A − A′)/2.
  2. 2P′ = (A′ + A′′)/2 = (A′ + A)/2 = P, so P is symmetric.
  3. 3Q′ = (A′ − A′′)/2 = (A′ − A)/2 = −Q, so Q is skew-symmetric.
  4. 4P + Q = (A + A′ + A − A′)/2 = A.
  5. 5Uniqueness: if A = S₁ + K₁ = S₂ + K₂ with Sᵢ symmetric and Kᵢ skew-symmetric, then S₁ − S₂ = K₂ − K₁. The left side is symmetric, the right is skew-symmetric. A matrix that is both symmetric and skew-symmetric must be zero, so S₁ = S₂ and K₁ = K₂.

Final answer

A = (A+A′)/2 + (A−A′)/2 gives the unique decomposition into symmetric + skew-symmetric parts.

Step-by-step solution

  1. 1A symmetric: A′ = A.
  2. 2A skew-symmetric: A′ = −A.
  3. 3Therefore A = −A, which gives 2A = O, so A = O.

Final answer

A = A′ = −A ⟹ 2A = O ⟹ A = O.

Step-by-step solution

  1. 1(i) Let C = AB′ + BA′. C′ = (AB′)′ + (BA′)′ = (B′)′A′ + (A′)′B′ = BA′ + AB′ = C. Symmetric ✓.
  2. 2(ii) Let D = AB′ − BA′. D′ = (B′)′A′ − (A′)′B′ = BA′ − AB′ = −D. Skew-symmetric ✓.

Final answer

(i) AB′ + BA′ is symmetric since (AB′ + BA′)′ = BA′ + AB′. (ii) AB′ − BA′ is skew-symmetric since (AB′ − BA′)′ = BA′ − AB′ = −(AB′ − BA′).

Step-by-step solution

  1. 1A′ = [cos θ, −sin θ; sin θ, cos θ].
  2. 2For A to be symmetric, A′ = A, requiring sin θ = −sin θ, so sin θ = 0, hence θ = nπ.
  3. 3When θ = nπ: cos θ = ±1, and A = ±I.
  4. 4This is a scalar multiple of I with k = ±1.

Final answer

A symmetric requires sin θ = 0, giving θ = nπ and A = ±I (a scalar multiple of the identity).

Step-by-step solution

  1. 1Let B = A + A′.
  2. 2B′ = (A + A′)′ = A′ + (A′)′ = A′ + A = A + A′ = B.
  3. 3Since B′ = B, B is symmetric.
  4. 4This holds for any square matrix A, regardless of whether A itself is symmetric.

Final answer

(A + A′)′ = A′ + A = A + A′, so A + A′ is always symmetric.

Step-by-step solution

  1. 1B′ = [0, −a, −b; a, 0, −c; b, c, 0].
  2. 2−B = [0, −a, −b; a, 0, −c; b, c, 0].
  3. 3B′ = −B for all a, b, c.
  4. 4Diagonal entries are all 0, and the off-diagonal entries satisfy b′ᵢⱼ = −b′ⱼᵢ.
  5. 5Also note: det(B) = 0, so B is never invertible.

Final answer

B is skew-symmetric: B′ = −B for all a, b, c. The diagonal is all zeros and off-diagonal pairs satisfy b′ᵢⱼ = −b′ⱼᵢ.

Step-by-step solution

  1. 1A² = [1,−1; −1,1][1,−1; −1,1] = [1+1, −1−1; −1−1, 1+1] = [2, −2; −2, 2].
  2. 22A = [2, −2; −2, 2].
  3. 3A² = 2A ✓.

Final answer

A² = [2,−2; −2,2] = 2A, verified.

Step-by-step solution

  1. 1Let C = AB − BA.
  2. 2C′ = (AB − BA)′ = (AB)′ − (BA)′ = B′A′ − A′B′.
  3. 3Since A′ = A and B′ = B: C′ = BA − AB = −(AB − BA) = −C.
  4. 4Therefore C is skew-symmetric.

Final answer

(AB − BA)′ = B′A′ − A′B′ = BA − AB = −(AB − BA). Skew-symmetric.

05

Exercise 3.4 — Elementary Row and Column Operations

4Exercise questions

Step-by-step solution

  1. 1Type 1 (Row interchange): Rᵢ ↔ Rⱼ. Example: swapping rows 1 and 2.
  2. 2Type 2 (Row scaling): Rᵢ → kRᵢ for k ≠ 0. Example: R₂ → 3R₂.
  3. 3Type 3 (Row addition): Rᵢ → Rᵢ + kRⱼ. Example: R₁ → R₁ − 2R₂.

Final answer

The three elementary row operations are: (1) Interchange of two rows (Rᵢ ↔ Rⱼ); (2) Multiplication of a row by a non-zero scalar (Rᵢ → kRᵢ, k ≠ 0); (3) Addition of a scalar multiple of one row to another (Rᵢ → Rᵢ + kRⱼ).

Step-by-step solution

  1. 1Column operations are analogous to row operations, applied to columns instead of rows.
  2. 2Three types: (1) Cᵢ ↔ Cⱼ (column interchange), (2) Cᵢ → kCᵢ (column scaling), (3) Cᵢ → Cᵢ + kCⱼ (column addition).
  3. 3Every elementary column operation on A is equivalent to the corresponding elementary row operation on A′ (transpose).
  4. 4Column operations are right-multiplication by elementary matrices, while row operations are left-multiplication.

Final answer

Elementary column operations mirror row operations applied to columns: (1) Cᵢ ↔ Cⱼ, (2) Cᵢ → kCᵢ (k≠0), (3) Cᵢ → Cᵢ + kCⱼ. They correspond to right-multiplication by elementary matrices (versus left-multiplication for row operations).

Step-by-step solution

  1. 1Augment A with I: [1 3 | 1 0; 2 7 | 0 1].
  2. 2R₂ → R₂ − 2R₁: [1 3 | 1 0; 0 1 | −2 1].
  3. 3R₁ → R₁ − 3R₂: [1 0 | 7 −3; 0 1 | −2 1].
  4. 4The right half is A⁻¹. Check: A·A⁻¹ = [1,3; 2,7][7,−3; −2,1] = [1,0; 0,1] ✓.

Final answer

Step-by-step solution

  1. 1Augment A with I: [2 1 | 1 0; 7 4 | 0 1]. det A = 8 − 7 = 1 ≠ 0, so A is invertible.
  2. 2R₂ → R₂ − 3R₁: [2 1 | 1 0; 1 1 | −3 1].
  3. 3R₁ ↔ R₂: [1 1 | −3 1; 2 1 | 1 0].
  4. 4R₂ → R₂ − 2R₁: [1 1 | −3 1; 0 −1 | 7 −2].
  5. 5R₁ → R₁ + R₂: [1 0 | 4 −1; 0 −1 | 7 −2].
  6. 6R₂ → −R₂: [1 0 | 4 −1; 0 1 | −7 2]. So A⁻¹ = [4,−1; −7,2].
  7. 7Check: [2,1; 7,4][4,−1; −7,2] = [8−7, −2+2; 28−28, −7+8] = [1,0; 0,1] ✓.

Final answer

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Transpose of a product

Symmetric

Skew-symmetric

Sum of a skew-symmetric matrix

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

  • The order of a product cannot be swapped, and the transpose reverses it — (AB)ᵀ is BᵀAᵀ, never AᵀBᵀ.
  • A diagonal entry of a skew-symmetric matrix must be zero, so a non-zero a₁₁ is an immediate contradiction.

FAQ

Frequently asked questions

How many questions are in NCERT Class 12 Maths Chapter 3 (Matrices)?

There are 4 exercise questions in this chapter, numbered Ex 3.1 to Ex 3.4. Every one is solved step by step on this page in the official NCERT numbering.

Which formulas come up in Matrices Class 12 Maths?

The formulas this chapter's questions actually turn on are: Transpose of a product, Symmetric, Skew-symmetric, Sum of a skew-symmetric matrix. They are listed with their expressions in the key formulas section below, and the solved questions show where each one is used.

Is Matrices important for JEE Main?

Very important — matrix operations and transpose properties are a fixed board unit and a regular JEE Main topic, and they lead directly into determinants.

Next Chapters

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