Exercise 3.2 — Matrices

1. 3.2.16 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

   Let A=[[2,4],[3,2]], B=[[1,3],[-2,5]], C=[[-2,5],[3,4]]. Find each of the following: (i) A+B (ii) A-B (iii) 3A-C (iv) AB (v) BA.

   Hint. Add/subtract entry by entry for (i)-(iii); for (iv)-(v) use row-times-column multiplication, remembering order matters.

   Since A, B, and C all share order 2x2, addition and subtraction work entry by entry. (i) A+B = [[2+1,4+3],[3-2,2+5]] = [[3,7],[1,7]]. (ii) A-B = [[2-1,4-3],[3+2,2-5]] = [[1,1],[5,-3]]. (iii) 3A-C = [[6-(-2),12-5],[9-3,6-4]] = [[8,7],[6,2]]. (iv) For AB, each entry is a row-of-A times a column-of-B: AB = [[21+4-2, 23+45],[31+2-2, 33+25]] = [[-6,26],[-1,19]]. (v) BA = [[12+33, 14+32],[-22+53, -24+52]] = [[11,10],[11,2]], which is different from AB, since matrix multiplication does not commute.

   ✦ (i) [[3,7],[1,7]] (ii) [[1,1],[5,-3]] (iii) [[8,7],[6,2]] (iv) AB=[[-6,26],[-1,19]] (v) BA=[[11,10],[11,2]] — note AB is not equal to BA.

2. 3.2.28 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

   Compute the following: (i) [[a,b],[-b,a]]+[[a,b],[b,a]] (ii) [[a^2+b^2,b^2+c^2],[a^2+c^2,a^2+b^2]]+[[2ab,2bc],[-2ac,-2ab]] (iii) [[-1,4,-6],[8,5,16],[2,8,5]]+[[12,7,6],[8,0,5],[3,2,4]] (iv) [[cos^2x,sin^2x],[sin^2x,cos^2x]]+[[sin^2x,cos^2x],[cos^2x,sin^2x]].

   Hint. Add entry by entry throughout; for (iv), each entry sums to sin^2x+cos^2x=1.

   Adding entry by entry throughout, since every pair here shares the same order. (i) [[2a,2b],[0,2a]]. (ii) [[a^2+b^2+2ab, b^2+c^2+2bc],[a^2+c^2-2ac, a^2+b^2-2ab]], which simplifies because each entry is a perfect-square expansion: [[(a+b)^2,(b+c)^2],[(a-c)^2,(a-b)^2]]. (iii) [[11,11,0],[16,5,21],[5,10,9]]. (iv) Every entry is sin^2x+cos^2x=1, since that is the Pythagorean identity, giving [[1,1],[1,1]].

   ✦ (i) [[2a,2b],[0,2a]] (ii) [[(a+b)^2,(b+c)^2],[(a-c)^2,(a-b)^2]] (iii) [[11,11,0],[16,5,21],[5,10,9]] (iv) [[1,1],[1,1]]

3. 3.2.310 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

   Compute the indicated products: (i) [[a,b],[-b,a]][[a,-b],[b,a]] (ii) [[1],[2],[3]][2,3,4] (iii) [[1,-2],[2,3]][[1,2,3],[2,3,1]] (iv) [[2,3,4],[3,4,5],[4,5,6]][[1,-3,5],[0,2,4],[3,0,5]] (v) [[2,1],[3,2],[-1,1]][[1,0,1],[-1,2,1]] (vi) [[3,-1,3],[-1,0,2]][[2,-3],[1,0],[3,1]].

   Hint. Apply row-times-column multiplication throughout, checking the resulting order matches (rows of first) x (columns of second).

   Since each pair of matrices has matching inner dimensions, every product below is defined; computing row-times-column throughout gives: (i) [[a^2+b^2,0],[0,a^2+b^2]]. (ii) [[2,3,4],[4,6,8],[6,9,12]]. (iii) [[-3,-4,1],[8,13,9]]. (iv) [[14,0,42],[18,-1,56],[22,-2,70]]. (v) [[1,2,3],[1,4,5],[-2,2,0]]. (vi) [[14,-6],[4,5]].

   ✦ (i) [[a^2+b^2,0],[0,a^2+b^2]] (ii) [[2,3,4],[4,6,8],[6,9,12]] (iii) [[-3,-4,1],[8,13,9]] (iv) [[14,0,42],[18,-1,56],[22,-2,70]] (v) [[1,2,3],[1,4,5],[-2,2,0]] (vi) [[14,-6],[4,5]]

4. 3.2.46 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

   If A=[[1,2,-3],[5,0,2],[1,-1,1]], B=[[3,-1,2],[4,2,5],[2,0,3]], and C=[[4,1,2],[0,3,2],[1,-2,3]], then compute (A+B) and (B-C). Also, verify that A+(B-C)=(A+B)-C.

   Hint. Compute A+B and B-C entry by entry, then compute both sides of the verification separately.

   A+B = [[4,1,-1],[9,2,7],[3,-1,4]]. B-C = [[-1,-2,0],[4,-1,3],[1,2,0]]. Computing the left side of the identity: A+(B-C) = [[1-1,2-2,-3+0],[5+4,0-1,2+3],[1+1,-1+2,1+0]] = [[0,0,-3],[9,-1,5],[2,1,1]]. Computing the right side independently: (A+B)-C = [[4-4,1-1,-1-2],[9-0,2-3,7-2],[3-1,-1+2,4-3]] = [[0,0,-3],[9,-1,5],[2,1,1]]. Since both sides give the identical matrix, the associativity-style identity is confirmed.

   ✦ A+B=[[4,1,-1],[9,2,7],[3,-1,4]], B-C=[[-1,-2,0],[4,-1,3],[1,2,0]]; A+(B-C)=(A+B)-C=[[0,0,-3],[9,-1,5],[2,1,1]], verified equal.

5. 3.2.55 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

   If A=[[2/3,1,5/3],[1/3,2/3,4/3],[7/3,2,2/3]] and B=[[2/5,3/5,1],[1/5,2/5,4/5],[7/5,6/5,2/5]], then compute 3A-5B.

   Hint. Scale each matrix first (3A and 5B), then subtract entry by entry, converting to a common form.

   3A = [[2,3,5],[1,2,4],[7,6,2]]. 5B = [[2,3,5],[1,2,4],[7,6,2]]. Since 3A and 5B are identical (each entry of A is exactly 5/3 times the corresponding entry of B), 3A-5B is the zero matrix.

   ✦ 3A-5B = the 3x3 zero matrix.

6. 3.2.63 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

   Simplify cos(theta)[[cos theta,sin theta],[-sin theta,cos theta]] + sin(theta)[[sin theta,-cos theta],[cos theta,sin theta]].

   Hint. Distribute the scalars into each matrix first, then add entry by entry using sin^2+cos^2=1.

   First matrix scaled: [[cos^2 theta, sin theta cos theta],[-sin theta cos theta, cos^2 theta]]. Second matrix scaled: [[sin^2 theta, -sin theta cos theta],[sin theta cos theta, sin^2 theta]]. Adding, the diagonal entries become cos^2 theta+sin^2 theta=1 each, since that is the Pythagorean identity, while the off-diagonal entries cancel to 0 because sin theta cos theta and -sin theta cos theta are negatives of each other.

   ✦ The 2x2 identity matrix [[1,0],[0,1]].

7. 3.2.76 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

   Find X and Y, if (i) X+Y=[[7,0],[2,5]] and X-Y=[[3,0],[0,3]] (ii) 2X+3Y=[[2,3],[4,0]] and 3X+2Y=[[2,-2],[-1,5]].

   Hint. For (i), add and subtract the two given equations directly. For (ii), solve the 2x2 linear system for each entry pair.

   (i) Adding the two equations: 2X=[[10,0],[2,8]], so X=[[5,0],[1,4]]. Subtracting: 2Y=[[4,0],[2,2]], so Y=[[2,0],[1,1]]. (ii) From 2X+3Y=E1 and 3X+2Y=E2: multiply the first by 3 and the second by 2: 6X+9Y=3E1, 6X+4Y=2E2. Subtracting: 5Y=3E1-2E2, so Y=(3E1-2E2)/5. Then X=(E1-3Y)/2. Computing entry by entry: Y=[[0.4,2.6],[2.8,-2]], X=[[0.4,-2.4],[-2.2,3]].

   ✦ (i) X=[[5,0],[1,4]], Y=[[2,0],[1,1]]. (ii) X=[[0.4,-2.4],[-2.2,3]], Y=[[0.4,2.6],[2.8,-2]].

8. 3.2.83 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

   Find X, if Y=[[3,2],[1,4]] and 2X+Y=[[1,0],[-3,2]].

   Hint. Isolate X by subtracting Y and dividing by 2.

   2X = [[1,0],[-3,2]] - [[3,2],[1,4]] = [[-2,-2],[-4,-2]]. So X = [[-1,-1],[-2,-1]].

   ✦ X = [[-1,-1],[-2,-1]]

9. 3.2.93 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

   Find x and y, if 2[[1,3],[0,x]]+[[y,0],[1,2]]=[[5,6],[1,8]].

   Hint. Scale the first matrix, then match each entry to build simple equations in x and y.

   2[[1,3],[0,x]] = [[2,6],[0,2x]]. Adding [[y,0],[1,2]]: [[2+y,6],[1,2x+2]] = [[5,6],[1,8]]. Matching: 2+y=5 gives y=3; 2x+2=8 gives x=3.

   ✦ x=3, y=3

10. 3.2.104 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

    Solve the equation for x, y, z and t, if 2[[x,z],[y,t]]+3[[1,-1],[0,2]]=3[[3,5],[4,6]].

    Hint. Simplify the right side first, then isolate the unknown matrix.

    Right side: 3[[3,5],[4,6]]=[[9,15],[12,18]]. Subtracting 3[[1,-1],[0,2]]=[[3,-3],[0,6]]: 2[[x,z],[y,t]] = [[6,18],[12,12]]. Dividing by 2: [[x,z],[y,t]]=[[3,9],[6,6]].

    ✦ x=3, z=9, y=6, t=6

11. 3.2.113 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

    If x[[2],[3]]+y[[-1],[1]]=[[10],[5]], find the values of x and y.

    Hint. This gives two simultaneous linear equations in x and y, one per row.

    Row 1: 2x-y=10. Row 2: 3x+y=5. Adding the two equations: 5x=15, so x=3. Substituting into row 2: 9+y=5, so y=-4.

    ✦ x=3, y=-4

12. 3.2.125 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

    Given 3[[x,y],[z,w]]=[[x,6],[-1,2w]]+[[4,x+y],[z+w,3]], find the values of x, y, z and w.

    Hint. Match each entry to get four equations, solving for x and w first since their equations are self-contained.

    Entry (1,1): 3x=x+4, so 2x=4, x=2. Entry (1,2): 3y=6+x+y, so 2y=6+2=8, y=4. Entry (2,2): 3w=2w+3, so w=3. Entry (2,1): 3z=-1+z+w=-1+z+3, so 2z=2, z=1.

    ✦ x=2, y=4, z=1, w=3

13. 3.2.134 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

    If F(x)=[[cos x,-sin x,0],[sin x,cos x,0],[0,0,1]], show that F(x)F(y)=F(x+y).

    Hint. Multiply the two matrices directly and simplify each entry using the cosine and sine addition formulas.

    F(x)F(y) has (1,1) entry cos x cos y - sin x sin y = cos(x+y); (1,2) entry -cos x sin y - sin x cos y = -sin(x+y); (2,1) entry sin x cos y + cos x sin y = sin(x+y); (2,2) entry -sin x sin y + cos x cos y = cos(x+y). The bottom row and right column stay [0,0,1] since only the top-left 2x2 block has nonzero off-diagonal terms. So F(x)F(y) = [[cos(x+y),-sin(x+y),0],[sin(x+y),cos(x+y),0],[0,0,1]] = F(x+y).

    ✦ Identity proved: F(x)F(y)=F(x+y), using the cosine and sine addition formulas.

14. 3.2.146 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

    Show that (i) [[5,-1],[6,7]][[2,1],[3,4]] is not equal to [[2,1],[3,4]][[5,-1],[6,7]] (ii) [[1,2,3],[0,1,0],[1,1,0]][[-1,1,0],[0,-1,1],[2,3,4]] is not equal to [[-1,1,0],[0,-1,1],[2,3,4]][[1,2,3],[0,1,0],[1,1,0]].

    Hint. Compute both products fully in each case and compare entries directly.

    (i) Left product: [[52+-13, 51+-14],[62+73, 61+74]] = [[7,1],[33,34]]. Right product: [[25+16, 2*-1+17],[35+46, 3-1+4*7]] = [[16,5],[39,25]]. These are different matrices. (ii) Computing both 3x3 products similarly gives AB=[[5,8,14],[0,-1,1],[-1,0,1]] and BA=[[-1,-1,-3],[1,0,0],[6,11,6]], which are different.

    ✦ Both pairs of products are verified unequal by direct computation, confirming matrix multiplication is not commutative in either case.

15. 3.2.155 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

    Find A^2-5A+6I, if A=[[2,0,1],[2,1,3],[1,-1,0]].

    Hint. Compute A^2 first via matrix multiplication, then combine with -5A and 6I entry by entry.

    Since the expression needs A^2 first, compute A.A = [[22+02+11, 20+01+1-1, 21+03+10],[22+12+31, 20+11+3*-1, 21+13+30],[12+-12+01, 10+-11+0*-1, 11+-13+0*0]] = [[5,-1,2],[9,-2,5],[0,-1,-2]]. Then, combining with -5A and 6I entry by entry, A^2-5A+6I = [[5-10+6, -1-0+0, 2-5+0],[9-10+0, -2-5+6, 5-15+0],[0-5+0, -1+5+0, -2-0+6]] = [[1,-1,-3],[-1,-1,-10],[-5,4,4]].

    ✦ A^2-5A+6I = [[1,-1,-3],[-1,-1,-10],[-5,4,4]]

16. 3.2.166 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

    If A=[[1,0,2],[0,2,1],[2,0,3]], prove that A^3-6A^2+7A+2I=0.

    Hint. Compute A^2, then A^3=A^2.A, then combine all four terms entry by entry.

    A^2 = [[5,0,8],[2,4,5],[8,0,13]]. A^3 = A^2.A = [[21,0,34],[12,8,23],[34,0,55]]. Combining: A^3-6A^2+7A+2I gives, for example, entry (1,1): 21-30+7+2=0; entry (2,2): 8-24+14+2=0; entry (3,3): 55-78+21+2=0; and every off-diagonal entry cancels similarly.

    ✦ A^3-6A^2+7A+2I is the zero matrix, as required.

17. 3.2.174 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

    If A=[[3,-2],[4,-2]] and I=[[1,0],[0,1]], find k so that A^2=kA-2I.

    Hint. Compute A^2 first, then compare A^2+2I against A entry by entry to read off k.

    A^2 = [[33+-24, 3*-2+-2*-2],[43+-24, 4*-2+-2*-2]] = [[1,-2],[4,-4]]. So A^2+2I = [[3,-2],[4,-2]], which is exactly A itself. Since A^2+2I=A means A^2=A-2I, comparing to A^2=kA-2I gives k=1.

    ✦ k=1

18. 3.2.186 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

    If A=[[0,-tan(alpha/2)],[tan(alpha/2),0]] and I is the identity matrix of order 2, show that I+A=(I-A)[[cos alpha,-sin alpha],[sin alpha,cos alpha]].

    Hint. Use the half-angle substitution t=tan(alpha/2) and the identities cos(alpha)=(1-t^2)/(1+t^2), sin(alpha)=2t/(1+t^2).

    Let t=tan(alpha/2), so I+A=[[1,-t],[t,1]] and I-A=[[1,t],[-t,1]]. Since cos(alpha)=(1-t^2)/(1+t^2) and sin(alpha)=2t/(1+t^2) by the standard tangent half-angle identities, computing (I-A) times the rotation matrix and multiplying through by (1+t^2) simplifies the product to exactly [[1,-t],[t,1]], which matches I+A.

    ✦ Identity verified: I+A = (I-A)[[cos alpha,-sin alpha],[sin alpha,cos alpha]], using the tangent half-angle substitution.

19. 3.2.195 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

    A trust fund has Rs 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs 30,000 among the two types of bonds if the trust fund must obtain an annual total interest of (a) Rs 1800 (b) Rs 2000.

    Hint. Let x be the amount in the first bond; set up the interest equation 0.05x+0.07(30000-x)=target, and represent it as a matrix product.

    Let x = amount invested at 5%, so 30000-x is invested at 7%. Total interest, as a matrix product [x, 30000-x] times [0.05, 0.07]', equals 0.05x+0.07(30000-x) = 2100-0.02x. (a) 2100-0.02x=1800, so 0.02x=300, x=15000. So Rs 15,000 at 5% and Rs 15,000 at 7%. (b) 2100-0.02x=2000, so 0.02x=100, x=5000. So Rs 5,000 at 5% and Rs 25,000 at 7%.

    ✦ (a) Rs 15,000 at 5% and Rs 15,000 at 7%. (b) Rs 5,000 at 5% and Rs 25,000 at 7%.

20. 3.2.204 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

    The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are Rs 80, Rs 60 and Rs 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

    Hint. Convert dozens to actual book counts, then represent quantities and prices as matrices and multiply.

    Since prices are given per book, first convert dozens to book counts: 10 dozen=120 chemistry, 8 dozen=96 physics, 10 dozen=120 economics. As a row matrix [120,96,120] times the column matrix of prices [80,60,40]', the total is 120(80)+96(60)+120(40) = 9600+5760+4800 = 20160.

    ✦ Rs 20,160

21. 3.2.212 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

    Assume X, Y, Z, W and P are matrices of order 2xn, 3xk, 2xp, nx3 and pxk, respectively. The restriction on n, k and p so that PY+WY will be defined are: (A) k=3, p=n (B) k is arbitrary, p=2 (C) p is arbitrary, k=3 (D) k=2, p=3.

    Hint. PY needs P's columns to match Y's rows; WY needs W's columns to match Y's rows; and PY, WY must have matching orders to be added.

    P is pxk, Y is 3xk... actually Y is 3xk means for PY to be defined, P's columns (k) must equal Y's rows (3), so k=3. For WY to be defined, W is nx3, and Y is 3xk, so this is automatically consistent since Y's rows are 3. For PY+WY to be defined, PY (order px k=p x3... with k=3, PY is p x k's column count from Y, i.e. p x k) and WY (order n x k) must have the same order, so p=n.

    ✦ (A) k=3, p=n

22. 3.2.222 marksNCERT Class 12 Mathematics, Matrices, Reprint 2026-27

    If n=p, then the order of the matrix 7X-5Z is: (A) px2 (B) 2xn (C) nx3 (D) pxn.

    Hint. X and Z must have the same order for 7X-5Z to be defined; recall X is 2xn and Z is 2xp.

    X has order 2xn and Z has order 2xp. Since n=p, both matrices have order 2xn (equivalently 2xp), so 7X-5Z is also of order 2xn.

    ✦ (B) 2xn

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.