Matrices organise numbers into rows and columns so that operations like addition and multiplication can be applied to entire arrays at once. This chapter’s exercises cover matrix algebra, transpose, and symmetric/skew-symmetric matrices, solved step by step below.
Last Updated: September 23, 2026
How to Approach This Chapter
Practice matrix operations (addition, multiplication, transpose) mechanically first, checking dimension compatibility before every multiplication — most errors here are arithmetic or dimension mismatches, not conceptual gaps. When finding an inverse via the adjoint method, compute the determinant first and confirm it’s non-zero before doing the more error-prone cofactor/adjoint work.
Exam Weightage: How Important Is This Chapter?
Matrices carries 5 marks, paired with Determinants (also 5 marks) to form the 10-mark Algebra unit in the CBSE Class 12 Maths board exam.
Exercise 3.1 Solutions (All 10 Questions)
1. In the matrix A = [[2,5,19,-7],[35,-2,5/2,12],[√3,1,-5,17]], write (i) the order, (ii) the number of elements, (iii) the elements a₁₃, a₂₁, a₃₃, a₂₄, a₂₃.
Ans: (i) Order 3×4. (ii) 12 elements. (iii) a₁₃=19, a₂₁=35, a₃₃=−5, a₂₄=12, a₂₃=5/2.
2. How many possible orders are there for a matrix with 24 elements? With 13 elements?
Ans: 24 = 1×24, 2×12, 3×8, 4×6, 6×4, 8×3, 12×2, 24×1 ⇒ 8 possible orders. For 13 (prime): 1×13, 13×1 ⇒ 2 possible orders.
3. How many possible orders are there for a matrix with 18 elements? With 5 elements?
Ans: 18 = 1×18, 2×9, 3×6, 6×3, 9×2, 18×1 ⇒ 6 possible orders. For 5 (prime): 1×5, 5×1 ⇒ 2 possible orders.
4. Construct a 2×2 matrix A=[aᵢᵇ] whose elements are given by: (i) aᵢᵇ=(i+j)²/2 (ii) aᵢᵇ=i/j (iii) aᵢᵇ=(i+2j)²/2.
Ans: (i) a₁₁=2, a₁₂=a₂₁=9/2, a₂₂=8, so A=[[2,9/2],[9/2,8]]. (ii) a₁₁=1, a₁₂=1/2, a₂₁=2, a₂₂=1, so A=[[1,1/2],[2,1]]. (iii) a₁₁=9/2, a₁₂=25/2, a₂₁=8, a₂₂=18, so A=[[9/2,25/2],[8,18]].
5. Construct a 3×4 matrix A=[aᵢᵇ] whose elements are given by: (i) aᵢᵇ=½|−3i+j| (ii) aᵢᵇ=2i−j.
Ans: (i) A=[[1,1/2,0,1/2],[5/2,2,3/2,1],[4,7/2,3,5/2]]. (ii) A=[[1,0,-1,-2],[3,2,1,0],[5,4,3,2]].
6. Find the values of x, y and z from the following equations: (i) [[4,3],[x,5]]=[[y,z],[1,5]] (ii) [[x+y,2],[5+z,xy]]=[[6,2],[5,8]] (iii) [[x+y+z],[x+z],[y+z]]=[[9],[5],[7]].
Ans: (i) Comparing entries: x=1, y=4, z=3. (ii) x+y=6 and xy=8, so x,y are roots of t²−6t+8=0, i.e. t=2,4 ⇒ (x,y)=(2,4) or (4,2); and 5+z=5 ⇒ z=0. (iii) x+z=5, y+z=7, x+y+z=9. Subtracting the first two from the third: (x+y+z)−(x+z)=9−5 ⇒ y=4; then z=7−y=3; then x=5−z=2. So x=2, y=4, z=3.
7. Find the values of a, b, c and d from the equation [[a-b,2a+c],[2a-b,3c+d]]=[[-1,5],[0,13]].
Ans: From a−b=−1 and 2a−b=0, subtracting gives a=1, so b=2. From 2a+c=5: c=3. From 3c+d=13: d=4. So a=1, b=2, c=3, d=4.
8. A=[aᵢᵇ]m×n is a square matrix, if: (A) m<n (B) m>n (C) m=n (D) None of these.
Ans: (C) m=n — a square matrix has an equal number of rows and columns.
9. Which of the given values of x and y make the following pair of matrices equal: [[3x+7,5],[y+1,2-3x]] = [[0,y-2],[8,4]]?
Ans: Comparing entries: 3x+7=0 ⇒ x=−7/3, but 2−3x=4 ⇒ x=−2/3 — these two required values of x conflict, so no single value of x satisfies both. (y−2=5 and y+1=8 both give y=7 consistently, but x cannot be found.) Hence it is not possible to find x and y satisfying the equality.
10. The number of all possible matrices of order 3×3 with each entry 0 or 1 is: (A) 27 (B) 18 (C) 81 (D) 512.
Ans: (D) 512 — a 3×3 matrix has 9 entries, and each entry independently has 2 choices (0 or 1), giving 2⁹=512 possible matrices.
Exercise 3.2 Solutions (All 22 Questions)
1. If A=[[2,4],[3,2]], B=[[1,3],[-2,5]], C=[[-2,5],[3,4]], find: (i) A+B (ii) A-B (iii) 3A-C (iv) AB (v) BA.
Ans: (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≠BA, confirming matrix multiplication is not commutative.)
2. Compute: (i) [[a,b],[-b,a]]+[[a,b],[b,a]] (ii) [[a²+b²,b²+c²],[a²+c²,a²+b²]]+[[2ab,2bc],[-2ac,-2ab]] (iii) two given 3×3 matrices [[-1,4,-6],[8,5,16],[2,8,5]]+[[12,7,6],[8,0,5],[3,2,4]] (iv) [[cos²x,sin²x],[sin²x,cos²x]]+[[sin²x,cos²x],[cos²x,sin²x]].
Ans: (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]], using sin²x+cos²x=1 in every entry.
3. Compute the indicated products: (i) [[a,b],[-b,a]]×[[a,-b],[b,a]] (ii) [1,2,3]T×[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]].
Ans: (i) (a²+b²)I = [[a²+b²,0],[0,a²+b²]]. (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. If A=[[1,2,-3],[5,0,2],[1,-1,1]], B=[[3,-1,2],[4,2,5],[2,0,3]], C=[[4,1,2],[0,3,2],[1,-2,3]], compute (A+B) and (B-C). Also verify A+(B-C)=(A+B)-C.
Ans: A+B=[[4,1,-1],[9,2,7],[3,-1,4]]. B-C=[[-1,-2,0],[4,-1,3],[1,2,0]]. Both A+(B-C) and (A+B)-C equal [[0,0,-3],[9,-1,5],[2,1,1]], confirming associativity of matrix subtraction/addition.
5. 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]], compute 3A-5B.
Ans: Every entry of 3A cancels exactly with the corresponding entry of 5B, so 3A-5B is the 3×3 zero matrix.
6. Simplify: cosθ[[cosθ,sinθ],[-sinθ,cosθ]] + sinθ[[sinθ,-cosθ],[cosθ,sinθ]].
Ans: [[cos²θ+sin²θ, cosθsinθ-cosθsinθ],[-sinθcosθ+sinθcosθ, sin²θ+cos²θ]] = [[1,0],[0,1]] = I, the 2×2 identity matrix.
7. 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]].
Ans: (i) Adding the equations: 2X=[[10,0],[2,8]] ⇒ X=[[5,0],[1,4]], and Y=(X+Y)-X=[[2,0],[1,1]]. (ii) Eliminating X (3×first − 2×second) gives Y=[[2/5,13/5],[14/5,-2]], and back-substituting gives X=[[2/5,-12/5],[-11/5,3]].
8. Find X, if Y=[[3,2],[1,4]] and 2X+Y=[[1,0],[-3,2]].
Ans: 2X=[[1,0],[-3,2]]-[[3,2],[1,4]]=[[-2,-2],[-4,-2]], so X=[[-1,-1],[-2,-1]].
9. Find x and y, if 2[[1,3],[0,x]]+[[y,0],[1,2]]=[[5,6],[1,8]].
Ans: Comparing entries: 2+y=5 ⇒ y=3; 2x+2=8 ⇒ x=3.
10. Solve for x, y, z and t, if 2[[x,z],[y,t]]+3[[1,-1],[0,2]]=3[[3,5],[4,6]].
Ans: Simplifying to [[2x+3,2z-3],[2y,2t+6]]=[[9,15],[12,18]] and comparing entries: x=3, z=9, y=6, t=6.
11. If x[[2],[3]]+y[[-1],[1]]=[[10],[5]], find x and y.
Ans: This gives 2x-y=10 and 3x+y=5. Adding: 5x=15 ⇒ x=3, and y=2x-10=-4.
12. Given 3[[x,y],[z,w]]=[[x,6],[-1,2w]]+[[4,x+y],[z+w,3]], find the values of x, y, z, w.
Ans: Comparing entries: 3x=x+4 ⇒ x=2; 3y=6+x+y ⇒ 2y=8 ⇒ y=4; 3w=2w+3 ⇒ w=3; 3z=-1+z+w ⇒ 2z=-1+3=2 ⇒ z=1.
13. 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).
Ans: Multiplying F(x) and F(y) and applying the angle-addition identities cos x cos y − sin x sin y=cos(x+y) and sin x cos y+cos x sin y=sin(x+y) in each entry gives exactly F(x+y). (The bottom-right 1 and the zero row/column are unaffected.)
14. Show that AB≠BA in each case: (i) A=[[5,-1],[6,7]], B=[[2,1],[3,4]] (ii) A=[[1,2,3],[0,1,0],[1,1,0]], B=[[-1,1,0],[0,-1,1],[2,3,4]].
Ans: (i) AB=[[7,1],[33,34]] but BA=[[16,5],[39,25]] — unequal. (ii) AB=[[5,8,14],[0,-1,1],[-1,0,1]] but BA=[[-1,-1,-3],[1,0,0],[6,11,6]] — unequal. Matrix multiplication is not commutative in general.
15. Find A²-5A+6I, if A=[[2,0,1],[2,1,3],[1,-1,0]].
Ans: A²-5A+6I = [[1,-1,-3],[-1,-1,-10],[-5,4,4]].
16. If A=[[1,0,2],[0,2,1],[2,0,3]], prove that A³-6A²+7A+2I=O.
Ans: Computing A² and A³ and combining with 7A and 2I, every entry cancels to give the 3×3 zero matrix O, verifying the identity.
17. Find k so that A²=kA-2I, if A=[[3,-2],[4,-2]].
Ans: A²=[[1,-2],[4,-4]]. Comparing with kA-2I entrywise gives k=1 consistently across all four entries.
18. If A=[[0,-tan(α/2)],[tan(α/2),0]] and I is the 2×2 identity matrix, show that I+A=(I-A)[[cosα,-sinα],[sinα,cosα]].
Ans: Using the half-angle identities cosα=(1-tan²(α/2))/(1+tan²(α/2)) and sinα=2tan(α/2)/(1+tan²(α/2)), expanding (I-A) times the rotation matrix simplifies to exactly [[1,-tan(α/2)],[tan(α/2),1]], which equals I+A.
19. A trust fund has Rs 30,000 to invest in two types of bonds: the first pays 5% interest per year, the second pays 7%. Using matrix multiplication, determine how to divide Rs 30,000 between the two bonds so the total annual interest is: (a) Rs 1,800 (b) Rs 2,000.
Ans: Let Rs x be invested at 5% and Rs(30000-x) at 7%: 0.05x+0.07(30000-x)=interest. (a) For Rs 1,800: x=15,000, so Rs 15,000 in each bond. (b) For Rs 2,000: x=5,000, so Rs 5,000 at 5% and Rs 25,000 at 7%.
20. A school bookshop has 10 dozen Chemistry books, 8 dozen Physics books, and 10 dozen Economics books, selling at Rs 80, Rs 60 and Rs 40 each respectively. Find the total amount the bookshop receives, using matrix algebra.
Ans: In individual units: 120 Chemistry, 96 Physics, 120 Economics books. Total revenue = 120(80)+96(60)+120(40) = 9,600+5,760+4,800 = Rs 20,160.
21. X, Y, Z, W and P are matrices of order 2×n, 3×k, 2×p, n×3 and p×k respectively. The restriction on n, k and p so that PY+WY is defined are: (A) k=3, p=n (B) k is arbitrary, p=2 (C) p is arbitrary, k=3 (D) k=2, p=3.
Ans: (A) k=3, p=n — P is p×k and Y is 3×k, so PY needs k=3; W is n×3 and Y is 3×k, so WY is n×k; for PY+WY to be defined, PY (order p×k) and WY (order n×k) must match, so p=n.
22. If n=p, then the order of the matrix 7X-5Z is: (A) p×2 (B) 2×n (C) n×3 (D) p×n.
Ans: (B) 2×n — X has order 2×n and Z has order 2×p; when n=p these match, so 7X-5Z is defined and has order 2×n.
Exercise 3.3 Solutions (All 12 Questions)
1. Find the transpose of each of the following matrices: (i) [[5],[1/2],[-1]] (ii) [[1,-1],[2,3]] (iii) [[-1,5,6],[√3,5,6],[2,3,-1]].
Ans: (i) [5, 1/2, -1] (a 1×3 row matrix). (ii) [[1,2],[-1,3]]. (iii) [[-1,√3,2],[5,5,3],[6,6,-1]].
2. If A=[[-1,2,3],[5,7,9],[-2,1,1]] and B=[[-4,1,-5],[1,2,0],[1,3,1]], verify: (i) (A+B)′=A′+B′ (ii) (A-B)′=A′-B′.
Ans: (i) Both sides equal [[-5,6,-1],[3,9,4],[-2,9,2]]. (ii) Both sides equal [[3,4,-3],[1,5,-2],[8,9,0]]. Both identities hold.
3. If A′=[[3,4],[-1,2],[0,1]] and B=[[-1,2,1],[1,2,3]], verify: (i) (A+B)′=A′+B′ (ii) (A-B)′=A′-B′.
Ans: Here A=[[3,-1,0],[4,2,1]] (the transpose of the given A′). (i) Both sides equal [[2,1],[5,4],[1,4]]. (ii) Both sides equal [[4,-3,-1],[3,0,-2]]. Both identities hold.
4. If A′=[[-2,3],[1,2]] and B=[[-1,0],[1,2]], find (A+2B)′.
Ans: A=[[-2,1],[3,2]] (transpose of given A′). A+2B=[[-4,1],[5,6]], so (A+2B)′=[[-4,5],[1,6]].
5. For matrices A and B, verify (AB)′=B′A′: (i) A=[[1],[-4],[3]], B=[[-1,2,1]] (ii) A=[[0],[1],[2]], B=[[1,5,7]].
Ans: (i) AB=[[-1,2,1],[4,-8,-4],[-3,6,3]], so (AB)′=[[-1,4,-3],[2,-8,6],[1,-4,3]], which equals B′A′. (ii) AB=[[0,0,0],[1,5,7],[2,10,14]], so (AB)′=[[0,1,2],[0,5,10],[0,7,14]], which equals B′A′. Both verified.
6. Verify A′A=I: (i) A=[[cosα,sinα],[-sinα,cosα]] (ii) A=[[sinα,cosα],[-cosα,sinα]].
Ans: In both cases, using sin²α+cos²α=1, A′A works out to [[1,0],[0,1]]=I. Verified for both.
7. (i) Show that the matrix A=[[1,-1,5],[-1,2,1],[5,1,3]] is symmetric. (ii) Show that the matrix A=[[0,1,-1],[-1,0,1],[1,-1,0]] is skew-symmetric.
Ans: (i) A′=A (every aᵢᵇ=aᵇᵢ), so A is symmetric. (ii) A′=-A (every diagonal entry is 0 and aᵢᵇ=-aᵇᵢ), so A is skew-symmetric.
8. For A=[[1,5],[6,7]], verify: (i) (A+A′) is a symmetric matrix (ii) (A-A′) is a skew-symmetric matrix.
Ans: (i) A+A′=[[2,11],[11,14]], which equals its own transpose — symmetric. (ii) A-A′=[[0,-1],[1,0]], whose transpose is its own negative — skew-symmetric.
9. Find ½(A+A′) and ½(A-A′), when A=[[0,a,b],[-a,0,c],[-b,-c,0]].
Ans: A is already skew-symmetric (A′=-A), so A+A′=O (the zero matrix), giving ½(A+A′)=O; and A-A′=2A, giving ½(A-A′)=A=[[0,a,b],[-a,0,c],[-b,-c,0]].
10. 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]].
Ans: Using P=½(A+A′) (symmetric) and Q=½(A-A′) (skew-symmetric): (i) P=[[3,3],[3,-1]], Q=[[0,2],[-2,0]]. (ii) The matrix is already symmetric, so P=[[6,-2,2],[-2,3,-1],[2,-1,3]] and Q=O (zero matrix). (iii) P=[[3,1/2,-5/2],[1/2,-2,-2],[-5/2,-2,2]], Q=[[0,5/2,3/2],[-5/2,0,3],[-3/2,-3,0]]. (iv) P=[[1,2],[2,2]], Q=[[0,3],[-3,0]].
11. If A, B are symmetric matrices of the same order, then AB-BA is a: (A) Skew-symmetric matrix (B) Symmetric matrix (C) Zero matrix (D) Identity matrix.
Ans: (A) Skew-symmetric matrix. Since A′=A and B′=B: (AB-BA)′=(AB)′-(BA)′=B′A′-A′B′=BA-AB=-(AB-BA), so AB-BA equals the negative of its own transpose — it is skew-symmetric.
12. If A=[[cosα,-sinα],[sinα,cosα]], and A+A′=I, then the value of α is: (A) π/6 (B) π/3 (C) π (D) 3π/2.
Ans: (B) π/3. A′=[[cosα,sinα],[-sinα,cosα]], so A+A′=[[2cosα,0],[0,2cosα]]. Setting this equal to I requires 2cosα=1, so cosα=1/2, giving α=π/3.
Exercise 3.4 Solutions (Question Retained After Syllabus Rationalisation)
18. Matrices A and B will be inverse of each other only if: (A) AB=BA (B) AB=BA=O (C) AB=O, BA=I (D) AB=BA=I.
Ans: (D) AB=BA=I — by definition, B is the inverse of A (and A is the inverse of B) precisely when both products AB and BA equal the identity matrix I.
Miscellaneous Exercise Solutions (All 11 Questions)
1. If A and B are symmetric matrices, prove that AB−BA is a skew-symmetric matrix.
Ans: Since A′=A and B′=B: (AB−BA)′=(AB)′−(BA)′=B′A′−A′B′=BA−AB=−(AB−BA). So AB−BA equals the negative of its own transpose — it is skew-symmetric.
2. Show that the matrix B′AB is symmetric or skew-symmetric according as A is symmetric or skew-symmetric.
Ans: (B′AB)′=B′A′(B′)′=B′A′B. If A is symmetric (A′=A), this gives (B′AB)′=B′AB, so B′AB is symmetric. If A is skew-symmetric (A′=−A), this gives (B′AB)′=B′(−A)B=−B′AB, so B′AB is skew-symmetric.
3. If A=[[0,2y,z],[x,y,-z],[x,-y,z]] satisfies A′A=I, find the values of x, y and z.
Ans: x=±1/√2, y=±1/√6, z=±1/√3. Matching the diagonal entries of A′A to 1 gives 2x²=1, 6y²=1 and 3z²=1; every off-diagonal entry of A′A works out to 0 automatically for any such x, y, z.
4. For what values of x: [1 2 1][[1,2,0],[2,0,1],[1,0,2]][[0],[2],[x]]=O?
Ans: x=−1. Multiplying the matrix and column vector first gives [4, x, 2x], and then [1 2 1]·[4,x,2x]=4+4x, which equals 0 only when x=−1.
5. If A=[[3,1],[-1,2]], show that A²-5A+7I=O.
Ans: A²=[[8,5],[-5,3]]. So A²-5A+7I=[[8-15+7,5-5+0],[-5+5+0,3-10+7]]=[[0,0],[0,0]]=O, as required.
6. Find x, if [x,-5,-1][[1,0,2],[0,2,1],[2,0,3]][[x],[4],[1]]=O.
Ans: x=±4√3. Multiplying the matrix and column vector first gives [x+2, 9, 2x+3], and then [x,-5,-1]·[x+2,9,2x+3]=x²+2x-45-2x-3=x²-48, which equals 0 only when x²=48, i.e. x=±4√3.
7. A manufacturer produces three products x, y, z which he sells in two markets. Annual sales are: Market I – 10,000 (x), 2,000 (y), 18,000 (z); Market II – 6,000 (x), 20,000 (y), 8,000 (z). (a) If unit sale prices of x, y and z are Rs 2.50, Rs 1.50 and Rs 1.00 respectively, find the total revenue in each market using matrix algebra. (b) If the unit costs of the above three commodities are Rs 2.00, Rs 1.00 and 50 paise respectively, find the gross profit.
Ans: (a) Revenue in Market I = 10000(2.50)+2000(1.50)+18000(1.00) = Rs 46,000. Revenue in Market II = 6000(2.50)+20000(1.50)+8000(1.00) = Rs 53,000. (b) Cost in Market I = 10000(2.00)+2000(1.00)+18000(0.50) = Rs 31,000, so profit = 46,000-31,000 = Rs 15,000. Cost in Market II = 6000(2.00)+20000(1.00)+8000(0.50) = Rs 36,000, so profit = 53,000-36,000 = Rs 17,000.
8. Find the matrix X so that X[[1,2,3],[4,5,6]]=[[-7,-8,-9],[2,4,6]].
Ans: X=[[1,-2],[2,0]]. (X must be 2×2; solving the resulting linear equations entrywise gives a=1, b=-2, c=2, d=0.)
9. If A=[[α,β],[γ,-α]] is such that A²=I, then: (A) 1+α²+βγ=0 (B) 1-α²+βγ=0 (C) 1-α²-βγ=0 (D) 1+α²-βγ=0.
Ans: (C) 1-α²-βγ=0. Computing A² gives (α²+βγ)I, and setting this equal to I requires α²+βγ=1, i.e. 1-α²-βγ=0.
10. If the matrix A is both symmetric and skew-symmetric, then: (A) A is a diagonal matrix (B) A is a zero matrix (C) A is a square matrix (D) None of these.
Ans: (B) A is a zero matrix. Symmetric means A′=A and skew-symmetric means A′=-A; together these force A=-A, i.e. 2A=O, so A must be the zero matrix.
11. If A is a square matrix such that A²=A, then (I+A)³-7A is equal to: (A) A (B) I-A (C) I (D) 3A.
Ans: (C) I. Since A²=A, also A³=A²·A=A·A=A²=A. Expanding (I+A)³=I+3A+3A²+A³=I+3A+3A+A=I+7A, so (I+A)³-7A=I.
Class 12 Mathematics Chapter 3 – Notes and Extra Questions
Along with these NCERT Solutions, students can also use the Class 12 Mathematics Chapter 3 Extra Questions and Class 12 Mathematics Chapter 3 Revision Notes for quick revision and extra practice.
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