Exercise 3.3 — Matrices

1. 3.3.13 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

   Find the transpose of each of the following matrices: (i) [[5],[1/2],[-1]] (ii) [[1,-1],[2,3]] (iii) [[-1,5,6],[sqrt3,5,6],[2,3,-1]].

   Hint. Interchange rows and columns in each case.

   Interchanging rows and columns in each case: (i) since a column matrix's rows become a single row, the transpose of a column matrix is a row matrix: [5, 1/2, -1]. (ii) [[1,2],[-1,3]]. (iii) [[-1,sqrt3,2],[5,5,3],[6,6,-1]].

   ✦ (i) [5, 0.5, -1] (ii) [[1,2],[-1,3]] (iii) [[-1,sqrt3,2],[5,5,3],[6,6,-1]]

2. 3.3.24 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

   If A=[[-1,2,3],[5,7,9],[-2,1,1]] and B=[[-4,1,-5],[1,2,0],[1,3,1]], then verify that (i) (A+B)'=A'+B' (ii) (A-B)'=A'-B'.

   Hint. Compute both sides of each identity independently and confirm they match.

   (i) A+B = [[-5,3,-2],[6,9,9],[-1,4,2]], so (A+B)' = [[-5,6,-1],[3,9,4],[-2,9,2]]. A' = [[-1,5,-2],[2,7,1],[3,9,1]], B' = [[-4,1,1],[1,2,3],[-5,0,1]], so A'+B' = [[-5,6,-1],[3,9,4],[-2,9,2]], matching. (ii) A-B = [[3,1,8],[4,5,9],[-3,-2,0]], so (A-B)' = [[3,4,-3],[1,5,-2],[8,9,0]]. A'-B' = [[-1-(-4),5-1,-2-1],[2-1,7-2,1-3],[3-(-5),9-0,1-1]] = [[3,4,-3],[1,5,-2],[8,9,0]], matching.

   ✦ Both identities verified: (A+B)'=A'+B' and (A-B)'=A'-B'.

3. 3.3.34 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

   If A'=[[3,4],[-1,2],[0,1]] and B=[[-1,2,1],[1,2,3]], then verify that (i) (A+B)'=A'+B' (ii) (A-B)'=A'-B'.

   Hint. First recover A from A' by transposing back, then proceed as before.

   Since A'=[[3,4],[-1,2],[0,1]], A=[[3,-1,0],[4,2,1]]. (i) A+B = [[2,1,1],[5,4,4]], so (A+B)'=[[2,5],[1,4],[1,4]]. A'+B' = [[3,4],[-1,2],[0,1]] + [[-1,1],[2,2],[1,3]] = [[2,5],[1,4],[1,4]], matching. (ii) A-B = [[4,-3,-1],[3,0,-2]], so (A-B)'=[[4,3],[-3,0],[-1,-2]]. A'-B' = [[3-(-1),4-1],[-1-2,2-2],[0-1,1-3]] = [[4,3],[-3,0],[-1,-2]], matching.

   ✦ Both identities verified: (A+B)'=A'+B' and (A-B)'=A'-B'.

4. 3.3.43 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

   If A'=[[-2,3],[1,2]] and B=[[-1,0],[1,2]], then find (A+2B)'.

   Hint. Recover A from A', compute A+2B directly, then transpose the result.

   A = [[-2,1],[3,2]]. 2B = [[-2,0],[2,4]]. A+2B = [[-4,1],[5,6]]. Transposing: (A+2B)' = [[-4,5],[1,6]].

   ✦ (A+2B)' = [[-4,5],[1,6]]

5. 3.3.54 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

   For the matrices A and B, verify that (AB)'=B'A', where (i) A=[[1],[-4],[3]], B=[-1,2,1] (ii) A=[[0],[1],[2]], B=[1,5,7].

   Hint. AB is a 3x3 matrix here since A is 3x1 and B is 1x3; compute both AB and B'A' independently.

   Since A is 3x1 and B is 1x3, AB is a 3x3 matrix. (i) AB = [[1*-1,12,11],[-4*-1,-42,-41],[3*-1,32,31]] = [[-1,2,1],[4,-8,-4],[-3,6,3]]. (AB)' = [[-1,4,-3],[2,-8,6],[1,-4,3]]. B'A' = [[-1],[2],[1]] * [1,-4,3] = [[-1,4,-3],[2,-8,6],[1,-4,3]], matching. (ii) AB = [[0,0,0],[1,5,7],[2,10,14]]. (AB)' = [[0,1,2],[0,5,10],[0,7,14]]. B'A' = [[1],[5],[7]]*[0,1,2] = [[0,1,2],[0,5,10],[0,7,14]], matching.

   ✦ Both cases verified: (AB)'=B'A'.

6. 3.3.64 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

   If (i) A=[[cos alpha,sin alpha],[-sin alpha,cos alpha]], then verify that A'A=I (ii) If A=[[sin alpha,cos alpha],[-cos alpha,sin alpha]], then verify that A'A=I.

   Hint. Compute A' first, then multiply A'A and simplify using sin^2+cos^2=1.

   (i) A' = [[cos alpha,-sin alpha],[sin alpha,cos alpha]]. A'A = [[cos^2 alpha+sin^2 alpha, cos alpha sin alpha - sin alpha cos alpha],[sin alpha cos alpha - cos alpha sin alpha, sin^2 alpha+cos^2 alpha]], which simplifies to [[1,0],[0,1]] = I, since each diagonal entry is the Pythagorean identity and each off-diagonal entry is a term minus itself. (ii) A' = [[sin alpha,-cos alpha],[cos alpha,sin alpha]]. A'A = [[sin^2 alpha+cos^2 alpha, sin alpha cos alpha-cos alpha sin alpha],[cos alpha sin alpha - sin alpha cos alpha, cos^2 alpha+sin^2 alpha]] = [[1,0],[0,1]] = I, by the same reasoning.

   ✦ Both verified: A'A=I in each case.

7. 3.3.74 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

   (i) Show that the matrix A=[[1,-1,5],[-1,2,1],[5,1,3]] is a symmetric matrix. (ii) Show that the matrix A=[[0,1,-1],[-1,0,1],[1,-1,0]] is a skew symmetric matrix.

   Hint. Compute A' in each case and compare to A (symmetric) or -A (skew-symmetric).

   (i) A' = [[1,-1,5],[-1,2,1],[5,1,3]], which is identical to A, so A is symmetric. (ii) A' = [[0,-1,1],[1,0,-1],[-1,1,0]] = -A (every entry is negated), so A is skew-symmetric.

   ✦ (i) A'=A, confirming A is symmetric. (ii) A'=-A, confirming A is skew-symmetric.

8. 3.3.84 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

   For the matrix A=[[1,5],[6,7]], verify that (i) (A+A') is a symmetric matrix (ii) (A-A') is a skew symmetric matrix.

   Hint. Compute A' first, then form both sums and check the defining property of each.

   A' = [[1,6],[5,7]]. (i) A+A' = [[2,11],[11,14]]. Its transpose is [[2,11],[11,14]], identical to itself, so it is symmetric. (ii) A-A' = [[0,-1],[1,0]]. Its transpose is [[0,1],[-1,0]], which is the negative of A-A', so it is skew-symmetric.

   ✦ (i) A+A'=[[2,11],[11,14]], symmetric. (ii) A-A'=[[0,-1],[1,0]], skew-symmetric.

9. 3.3.94 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

   Find (1/2)(A+A') and (1/2)(A-A'), when A=[[0,a,b],[-a,0,c],[-b,-c,0]].

   Hint. Compute A' (which is -A here, since A is already skew-symmetric), then form both halves.

   A' = [[0,-a,-b],[a,0,-c],[b,c,0]] = -A. So A+A' is the zero matrix, giving (1/2)(A+A') = the zero matrix. And A-A' = A-(-A) = 2A, giving (1/2)(A-A') = A = [[0,a,b],[-a,0,c],[-b,-c,0]].

   ✦ (1/2)(A+A') = the zero matrix. (1/2)(A-A') = A = [[0,a,b],[-a,0,c],[-b,-c,0]].

10. 3.3.108 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

    Express the following matrices as the sum of a symmetric and a skew symmetric matrix: (i) [[3,5],[1,-1]] (ii) [[6,-2,2],[-2,3,-1],[2,-1,3]] (iii) [[3,3,-1],[-2,-2,1],[-4,-5,2]] (iv) [[1,5],[-1,2]].

    Hint. For each matrix, compute A', then form (1/2)(A+A') and (1/2)(A-A') as the symmetric and skew-symmetric parts.

    (i) A'=[[3,1],[5,-1]]. Symmetric part (1/2)(A+A')=[[3,3],[3,-1]]. Skew-symmetric part (1/2)(A-A')=[[0,2],[-2,0]]. (ii) A'=[[6,-2,2],[-2,3,-1],[2,-1,3]]=A, so the symmetric part is A itself and the skew-symmetric part is the zero matrix. (iii) A'=[[3,-2,-4],[3,-2,-5],[-1,1,2]]. Symmetric part=[[3,0.5,-2.5],[0.5,-2,-2],[-2.5,-2,2]]. Skew-symmetric part=[[0,2.5,1.5],[-2.5,0,3],[-1.5,-3,0]]. (iv) A'=[[1,-1],[5,2]]. Symmetric part=[[1,2],[2,2]]. Skew-symmetric part=[[0,3],[-3,0]].

    ✦ (i) [[3,3],[3,-1]] + [[0,2],[-2,0]] (ii) A itself (symmetric) + zero matrix (iii) [[3,0.5,-2.5],[0.5,-2,-2],[-2.5,-2,2]] + [[0,2.5,1.5],[-2.5,0,3],[-1.5,-3,0]] (iv) [[1,2],[2,2]] + [[0,3],[-3,0]]

11. 3.3.112 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

    If A, B are symmetric matrices of same order, then AB-BA is a (A) Skew symmetric matrix (B) Symmetric matrix (C) Zero matrix (D) Identity matrix.

    Hint. Use A'=A, B'=B and the transpose-of-a-product rule (AB)'=B'A' to find (AB-BA)'.

    (AB-BA)' = (AB)'-(BA)' = B'A'-A'B' = BA-AB (using A'=A, B'=B) = -(AB-BA). So (AB-BA)' = -(AB-BA), which is exactly the defining property of a skew-symmetric matrix.

    ✦ (A) Skew symmetric matrix

12. 3.3.123 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

    If A=[[cos alpha,-sin alpha],[sin alpha,cos alpha]], and A+A'=I, then the value of alpha is (A) pi/6 (B) pi/3 (C) pi (D) 3pi/2.

    Hint. Compute A+A' directly and match it to the identity matrix.

    A' = [[cos alpha,sin alpha],[-sin alpha,cos alpha]]. A+A' = [[2cos alpha,0],[0,2cos alpha]]. Setting this equal to I=[[1,0],[0,1]]: 2cos alpha=1, so cos alpha=1/2, giving alpha=pi/3 (the standard angle in the usual range).

    ✦ (B) pi/3

Solutions written by the tuition.in editorial team and checked against the NCERT Class 12 Mathematics textbook, Reprint 2026-27 (lemh103.pdf) — Exercise 3.1 (10 questions), Exercise 3.2 (22 questions), Exercise 3.3 (12 questions), Exercise 3.4 (1 question), plus the chapter's Miscellaneous Exercise (11 questions), 56 questions total. Exercise pages were rendered as 300dpi images throughout to read matrix bracket notation accurately, since raw text extraction badly garbled every matrix in the chapter. The old stub taught 'finding the inverse using elementary row/column operations' as a core section, but that method has been fully removed from the current edition — Section 3.7 only proves existence and uniqueness of an inverse, and Exercise 3.4 has shrunk from what older editions had down to a single leftover MCQ, confirmed by reading straight through the section with nothing about row-reducing a matrix anywhere in it. The old stub also collapsed all four real exercises into one invented 30-question group with no solutions file behind it. Every matrix computation in this file (all 56 questions) was independently verified with a Python script using numpy for numeric matrices and sympy for the symbolic ones (F(x)F(y)=F(x+y), the A'A=I system, and the two symbolic quadratic-in-x miscellaneous questions), catching zero discrepancies against the hand-derived working.. Questions are referenced from the NCERT textbook for identification.