Class 12 Mathematics - NCERT Mathematics Part I
NCERT Solutions for Class 12 Maths Chapter 3: Matrices
Chapter 3 introduces matrices - rectangular arrays of numbers arranged in rows and columns. Students learn types of matrices, matrix operations (addition, scalar multiplication, multiplication), transpose, and properties of symmetric and skew-symmetric matrices. This chapter is part of the Algebra unit which carries 10 marks in the CBSE Class 12 board exam 2027 (combined with Chapter 4: Determinants). Matrices typically contribute about 5 marks.
Syllabus Update 2026-27
As per the CBSE 2026-27 syllabus, Chapter 3 Matrices has 4 exercises plus a Miscellaneous Exercise. Exercise 3.1 (10 Qs) covers types and order of matrices. Exercise 3.2 (22 Qs) covers matrix operations. Exercise 3.3 (12 Qs) covers transpose and symmetric matrices. Exercise 3.4 (18 Qs) covers elementary operations and invertible matrices. The Miscellaneous Exercise has 15 questions.
Table of Contents
3.1Exercise 3.1 - Types and Order of Matrices
Q1In the matrix A = [[1, 5, √3, 1], [−1/2, 3, 2, 4], [√3, 0, 5, 2/3]], write: (i) The order of the matrix (ii) The number of elements (iii) Write the elements a₁₃, a₂₁, a₃₃, a₂₄, a₂₃
The matrix A has 3 rows and 4 columns.
(i) Order of the matrix:
Number of rows = 3, Number of columns = 4
Order = 3 × 4
(ii) Number of elements:
Total elements = rows × columns = 3 × 4 = 12 elements
(iii) Elements:
aᵢⱼ means element in iᵗʰ row and jᵗʰ column.
• a₁₃ = element in 1st row, 3rd column = √3
• a₂₁ = element in 2nd row, 1st column = −1/2
• a₃₃ = element in 3rd row, 3rd column = 5
• a₂₄ = element in 2nd row, 4th column = 4
• a₂₃ = element in 2nd row, 3rd column = 2
Q2If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?
The order of a matrix is m × n where m × n = total number of elements.
For 24 elements:
Find all factor pairs of 24:
24 = 1×24 = 2×12 = 3×8 = 4×6 = 6×4 = 8×3 = 12×2 = 24×1
Possible orders: 1×24, 2×12, 3×8, 4×6, 6×4, 8×3, 12×2, 24×1
Total = 8 possible orders
For 13 elements:
13 is a prime number, so its only factor pairs are:
13 = 1×13 = 13×1
Possible orders: 1×13, 13×1
Total = 2 possible orders
Q3If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?
For 18 elements:
Factor pairs of 18:
18 = 1×18 = 2×9 = 3×6 = 6×3 = 9×2 = 18×1
Possible orders: 1×18, 2×9, 3×6, 6×3, 9×2, 18×1
Total = 6 possible orders
For 5 elements:
5 is a prime number.
5 = 1×5 = 5×1
Possible orders: 1×5, 5×1
Total = 2 possible orders
Q4Construct a 3 × 4 matrix, whose elements are given by: (i) aᵢⱼ = (1/2)|−3i + j| (ii) aᵢⱼ = 2i − j
(i) aᵢⱼ = (1/2)|−3i + j|
Compute each element:
i=1: a₁₁ = ½|−3+1| = ½(2) = 1, a₁₂ = ½|−3+2| = ½(1) = ½, a₁₃ = ½|−3+3| = 0, a₁₄ = ½|−3+4| = ½
i=2: a₂₁ = ½|−6+1| = ½(5) = 5/2, a₂₂ = ½|−6+2| = ½(4) = 2, a₂₃ = ½|−6+3| = 3/2, a₂₄ = ½|−6+4| = 1
i=3: a₃₁ = ½|−9+1| = ½(8) = 4, a₃₂ = ½|−9+2| = 7/2, a₃₃ = ½|−9+3| = 3, a₃₄ = ½|−9+4| = 5/2
A = [[1, 1/2, 0, 1/2], [5/2, 2, 3/2, 1], [4, 7/2, 3, 5/2]]
(ii) aᵢⱼ = 2i − j
i=1: a₁₁ = 2−1 = 1, a₁₂ = 2−2 = 0, a₁₃ = 2−3 = −1, a₁₄ = 2−4 = −2
i=2: a₂₁ = 4−1 = 3, a₂₂ = 4−2 = 2, a₂₃ = 4−3 = 1, a₂₄ = 4−4 = 0
i=3: a₃₁ = 6−1 = 5, a₃₂ = 6−2 = 4, a₃₃ = 6−3 = 3, a₃₄ = 6−4 = 2
A = [[1, 0, −1, −2], [3, 2, 1, 0], [5, 4, 3, 2]]
Q5Construct a 2 × 2 matrix, A = [aᵢⱼ], whose elements are given by: (i) aᵢⱼ = (i + j)²/2 (ii) aᵢⱼ = i/j (iii) aᵢⱼ = (i + 2j)²/2
(i) aᵢⱼ = (i + j)²/2
a₁₁ = (1+1)²/2 = 4/2 = 2
a₁₂ = (1+2)²/2 = 9/2
a₂₁ = (2+1)²/2 = 9/2
a₂₂ = (2+2)²/2 = 16/2 = 8
A = [[2, 9/2], [9/2, 8]]
(ii) aᵢⱼ = i/j
a₁₁ = 1/1 = 1, a₁₂ = 1/2, a₂₁ = 2/1 = 2, a₂₂ = 2/2 = 1
A = [[1, 1/2], [2, 1]]
(iii) aᵢⱼ = (i + 2j)²/2
a₁₁ = (1+2)²/2 = 9/2
a₁₂ = (1+4)²/2 = 25/2
a₂₁ = (2+2)²/2 = 16/2 = 8
a₂₂ = (2+4)²/2 = 36/2 = 18
A = [[9/2, 25/2], [8, 18]]
Q6Find 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]]
(i) [[4, 3], [x, 5]] = [[y, z], [1, 5]]
By equality of matrices (corresponding elements must be equal):
4 = y → y = 4
3 = z → z = 3
x = 1 → x = 1
5 = 5 ✓ (consistent)
(ii) [[x+y, 2], [5+z, xy]] = [[6, 2], [5, 8]]
x + y = 6 ... (1)
5 + z = 5 → z = 0
xy = 8 ... (2)
From (1) and (2): x and y are roots of t² − 6t + 8 = 0
(t − 2)(t − 4) = 0 → t = 2 or t = 4
x = 2, y = 4 (or x = 4, y = 2), z = 0
(iii) [[x+y+z], [x+z], [y+z]] = [[9], [5], [7]]
x + y + z = 9 ... (1)
x + z = 5 ... (2)
y + z = 7 ... (3)
From (1) − (2): y = 4
From (1) − (3): x = 2
From (2): z = 5 − 2 = 3
x = 2, y = 4, z = 3
Verification: 2 + 4 + 3 = 9 ✓, 2 + 3 = 5 ✓, 4 + 3 = 7 ✓
Q7Find the values of a, b, c and d from the equation: [[a−b, 2a+c], [2a−b, 3c+d]] = [[−1, 5], [0, 13]]
By equality of matrices:
a − b = −1 ... (1)
2a + c = 5 ... (2)
2a − b = 0 ... (3)
3c + d = 13 ... (4)
Step 1: From (3): b = 2a
Step 2: Substitute b = 2a in (1):
a − 2a = −1
−a = −1
a = 1
Step 3: b = 2(1) = 2
Step 4: From (2): 2(1) + c = 5 → c = 3, so c = 3
Step 5: From (4): 3(3) + d = 13 → 9 + d = 13 → d = 4
Answer: a = 1, b = 2, c = 3, d = 4
Q8A = [aᵢⱼ]ₘₓₙ is a square matrix, if (A) m < n (B) m > n (C) m = n (D) None of these
A matrix is square when the number of rows equals the number of columns.
For a matrix of order m × n to be square, we need m = n.
For example, a 3 × 3 matrix is square, but a 3 × 4 matrix is not.
Answer: (C) m = n
Q9Which 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]] (A) x = −1/3, y = 7 (B) Not possible to find (C) y = 7, x = −2/3 (D) x = −1/3, y = −2/3
By equality of corresponding elements:
3x + 7 = 0 → x = −7/3 ... (1)
5 = y − 2 → y = 7 ... (2)
y + 1 = 8 → y = 7 ... (3) ✓ (consistent with (2))
2 − 3x = 4 → −3x = 2 → x = −2/3 ... (4)
From (1): x = −7/3
From (4): x = −2/3
Since x = −7/3 and x = −2/3 are contradictory, no consistent values of x and y exist.
Answer: (B) Not possible to find
Q10The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is: (A) 27 (B) 18 (C) 81 (D) 512
A 3 × 3 matrix has 3 × 3 = 9 entries.
Each entry can be either 0 or 1, giving 2 choices per entry.
Total number of matrices = 2⁹ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 512
Answer: (D) 512
3.2Exercise 3.2 - Matrix Operations
12 key questions shown below covering addition, scalar multiplication, and multiplication. The remaining 10 questions follow similar patterns.
Q1Let 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
(i) A + B:
A + B = [[2+1, 4+3], [3+(-2), 2+5]] = [[3, 7], [1, 7]]
(ii) A - B:
A - B = [[2-1, 4-3], [3-(-2), 2-5]] = [[1, 1], [5, -3]]
(iii) 3A - C:
3A = [[6, 12], [9, 6]]
3A - C = [[6-(-2), 12-5], [9-3, 6-4]] = [[8, 7], [6, 2]]
(iv) AB:
AB = [[2(1)+4(-2), 2(3)+4(5)], [3(1)+2(-2), 3(3)+2(5)]]
= [[2-8, 6+20], [3-4, 9+10]]
= [[-6, 26], [-1, 19]]
(v) BA:
BA = [[1(2)+3(3), 1(4)+3(2)], [-2(2)+5(3), -2(4)+5(2)]]
= [[2+9, 4+6], [-4+15, -8+10]]
= [[11, 10], [11, 2]]
Note: AB ≠ BA. This demonstrates that matrix multiplication is not commutative.
Q2Compute the following: (i) [[a, b], [-b, a]] + [[a, b], [b, a]] (ii) [[a², ab], [ab, b²]] + [[b², -ab], [-ab, a²]]
(i) [[a+a, b+b], [-b+b, a+a]] = [[2a, 2b], [0, 2a]]
(ii) [[a²+b², ab+(-ab)], [ab+(-ab), b²+a²]] = [[a²+b², 0], [0, a²+b²]]
Note: The result in (ii) is (a²+b²)I, where I is the 2x2 identity matrix.
Q3Compute the indicated products: (i) [[a, b], [-b, a]] × [[a, -b], [b, a]] (ii) [[1], [2], [3]] × [[2, 3, 4]] (iii) [[2, 3, 4], [3, 4, 5], [4, 5, 6]] × [[1, -1, 1], [1, -1, 1], [1, -1, 1]]
(i) [[a, b], [-b, a]] x [[a, -b], [b, a]]
= [[a(a)+b(b), a(-b)+b(a)], [-b(a)+a(b), -b(-b)+a(a)]]
= [[a²+b², -ab+ab], [-ab+ab, b²+a²]]
= [[a²+b², 0], [0, a²+b²]] = (a²+b²)I
(ii) [[1], [2], [3]] x [[2, 3, 4]] (3x1 times 1x3 = 3x3)
= [[2, 3, 4], [4, 6, 8], [6, 9, 12]]
Each element (i,j) = (row i element) x (column j element).
(iii) [[2,3,4],[3,4,5],[4,5,6]] x [[1,-1,1],[1,-1,1],[1,-1,1]]
Row 1: [2(1)+3(1)+4(1), 2(-1)+3(-1)+4(-1), 2(1)+3(1)+4(1)] = [9, -9, 9]
Row 2: [3+4+5, -3-4-5, 3+4+5] = [12, -12, 12]
Row 3: [4+5+6, -4-5-6, 4+5+6] = [15, -15, 15]
Result = [[9, -9, 9], [12, -12, 12], [15, -15, 15]]
Q4If 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]], then compute (A+B) and (B-C). Also, verify that A + (B - C) = (A + B) - C.
A + B:
= [[1+3, 2+(-1), -3+2], [5+4, 0+2, 2+5], [1+2, -1+0, 1+3]]
= [[4, 1, -1], [9, 2, 7], [3, -1, 4]]
B - C:
= [[3-4, -1-1, 2-2], [4-0, 2-3, 5-2], [2-1, 0-(-2), 3-3]]
= [[-1, -2, 0], [4, -1, 3], [1, 2, 0]]
Verification: 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]]
(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 A + (B - C) = (A + B) - C, the associative property of matrix addition is verified. ✓
Q6Simplify: cos θ × [[cos θ, sin θ], [-sin θ, cos θ]] + sin θ × [[sin θ, -cos θ], [cos θ, sin θ]]
Step 1: Compute cos θ times the first matrix:
cos θ × [[cos θ, sin θ], [-sin θ, cos θ]] = [[cos²θ, cos θ sin θ], [-sin θ cos θ, cos²θ]]
Step 2: Compute sin θ times the second matrix:
sin θ × [[sin θ, -cos θ], [cos θ, sin θ]] = [[sin²θ, -sin θ cos θ], [sin θ cos θ, sin²θ]]
Step 3: Add the two results:
= [[cos²θ + sin²θ, cos θ sin θ - sin θ cos θ], [-sin θ cos θ + sin θ cos θ, cos²θ + sin²θ]]
= [[1, 0], [0, 1]]
= I (identity matrix)
Using the identity: sin²θ + cos²θ = 1.
Q7Find 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]]
(i) Adding the two equations:
2X = [[7+3, 0+0], [2+0, 5+3]] = [[10, 0], [2, 8]]
X = [[5, 0], [1, 4]]
Subtracting: 2Y = [[7-3, 0-0], [2-0, 5-3]] = [[4, 0], [2, 2]]
Y = [[2, 0], [1, 1]]
Verification: X + Y = [[5+2, 0+0], [1+1, 4+1]] = [[7, 0], [2, 5]] ✓
(ii) Multiply first equation by 3 and second by 2:
6X + 9Y = [[6, 9], [12, 0]]
6X + 4Y = [[4, -4], [-2, 10]]
Subtract: 5Y = [[6-4, 9-(-4)], [12-(-2), 0-10]] = [[2, 13], [14, -10]]
Y = [[2/5, 13/5], [14/5, -2]]
From 2X + 3Y = [[2, 3], [4, 0]]:
2X = [[2, 3], [4, 0]] - 3 × [[2/5, 13/5], [14/5, -2]]
= [[2-6/5, 3-39/5], [4-42/5, 0+6]]
= [[4/5, -24/5], [-22/5, 6]]
X = [[2/5, -12/5], [-11/5, 3]]
Q9Find x, y, z and w if 3[[x, y], [z, w]] = [[x, 6], [-1, 2w]] + [[4, x+y], [z+w, 3]]
[[3x, 3y], [3z, 3w]] = [[x+4, 6+x+y], [-1+z+w, 2w+3]]
Equating corresponding elements:
(1,1): 3x = x + 4 → 2x = 4 → x = 2
(1,2): 3y = 6 + x + y → 3y = 6 + 2 + y → 2y = 8 → y = 4
(2,1): 3z = -1 + z + w → 2z - w = -1 ... (i)
(2,2): 3w = 2w + 3 → w = 3
Substituting w = 3 in (i): 2z - 3 = -1 → 2z = 2 → z = 1
Answer: x = 2, y = 4, z = 1, w = 3
Q12If [[3, -2], [4, -2]] × A = [[-1, 0], [0, -1]] = -I, find the matrix A.
Let P = [[3, -2], [4, -2]]. We need PA = -I.
So A = P⁻¹ × (-I) = -P⁻¹.
First find P⁻¹:
det(P) = 3(-2) - (-2)(4) = -6 + 8 = 2.
P⁻¹ = (1/2) × [[-2, 2], [-4, 3]]
A = -P⁻¹ = -(1/2)[[-2, 2], [-4, 3]] = (1/2)[[2, -2], [4, -3]] = [[1, -1], [2, -3/2]]
Verification: PA = [[3,-2],[4,-2]] × [[1,-1],[2,-3/2]]
= [[3-4, -3+3], [4-4, -4+3]] = [[-1, 0], [0, -1]] = -I ✓
Q15Find A² - 5A + 6I if A = [[2, 0, 1], [2, 1, 3], [1, -1, 0]].
Step 1: Compute A²:
A² = A × A:
Row 1 of A² = [2(2)+0(2)+1(1), 2(0)+0(1)+1(-1), 2(1)+0(3)+1(0)] = [5, -1, 2]
Row 2 of A² = [2(2)+1(2)+3(1), 2(0)+1(1)+3(-1), 2(1)+1(3)+3(0)] = [9, -2, 5]
Row 3 of A² = [1(2)+(-1)(2)+0(1), 1(0)+(-1)(1)+0(-1), 1(1)+(-1)(3)+0(0)] = [0, -1, -2]
A² = [[5, -1, 2], [9, -2, 5], [0, -1, -2]]
Step 2: Compute 5A:
5A = [[10, 0, 5], [10, 5, 15], [5, -5, 0]]
Step 3: 6I:
6I = [[6, 0, 0], [0, 6, 0], [0, 0, 6]]
Step 4: A² - 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]]
Q17If A = [[0, -tan(α/2)], [tan(α/2), 0]] and I is the identity matrix of order 2, show that I + A = (I - A) × [[cos α, -sin α], [sin α, cos α]].
Let t = tan(α/2). Then A = [[0, -t], [t, 0]].
LHS = I + A:
= [[1, -t], [t, 1]]
RHS = (I - A) × [[cos α, -sin α], [sin α, cos α]]:
I - A = [[1, t], [-t, 1]]
Using half-angle formulas:
cos α = (1-t²)/(1+t²) and sin α = 2t/(1+t²)
(I - A) × [[cos α, -sin α], [sin α, cos α]]
= [[1, t], [-t, 1]] × [[(1-t²)/(1+t²), -2t/(1+t²)], [2t/(1+t²), (1-t²)/(1+t²)]]
Row 1: [(1-t²+2t²)/(1+t²), (-2t+t(1-t²))/(1+t²)] = [(1+t²)/(1+t²), (-2t+t-t³)/(1+t²)]
= [1, (-t-t³)/(1+t²)] = [1, -t(1+t²)/(1+t²)] = [1, -t]
Row 2: [(-t(1-t²)+2t)/(1+t²), (2t²+(1-t²))/(1+t²)]
= [(-t+t³+2t)/(1+t²), (1+t²)/(1+t²)] = [(t+t³)/(1+t²), 1] = [t(1+t²)/(1+t²), 1] = [t, 1]
RHS = [[1, -t], [t, 1]] = LHS ✓
Hence I + A = (I - A) × [[cos α, -sin α], [sin α, cos α]]. ∎
Q21If A is any square matrix, show that AA' and A'A are both symmetric matrices.
Show AA' is symmetric:
(AA')' = (A')'A' = AA' (using (XY)' = Y'X' and (A')' = A)
Since (AA')' = AA', AA' is symmetric. ✓
Show A'A is symmetric:
(A'A)' = A'(A')' = A'A (using (XY)' = Y'X' and (A')' = A)
Since (A'A)' = A'A, A'A is symmetric. ✓
Note: This result holds for any square matrix A, whether A itself is symmetric or not.
Q22If A and B are symmetric matrices of the same order, then show that AB is symmetric if and only if A and B commute, that is, AB = BA.
Given: A' = A, B' = B (both symmetric).
Forward direction (AB = BA implies AB is symmetric):
Assume AB = BA.
(AB)' = B'A' = BA = AB (given).
Since (AB)' = AB, AB is symmetric.
Reverse direction (AB is symmetric implies AB = BA):
Assume (AB)' = AB.
But (AB)' = B'A' = BA.
So BA = AB, i.e., A and B commute.
Therefore: AB is symmetric ⟺ AB = BA. ∎
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Ask AI TutorPrevious Year Board Exam Questions
If A = [[1, 2], [3, 4]], find A² − 5A + 6I.
A² = A × A = [[1,2],[3,4]] × [[1,2],[3,4]] = [[1+6, 2+8],[3+12, 6+16]] = [[7, 10],[15, 22]].
5A = [[5, 10],[15, 20]]. 6I = [[6, 0],[0, 6]].
A² − 5A + 6I = [[7−5+6, 10−10+0],[15−15+0, 22−20+6]] = [[8, 0],[0, 8]] = 8I.
Express the matrix [[3, 5], [1, −1]] as the sum of a symmetric and a skew-symmetric matrix.
Let A = [[3, 5],[1, −1]]. A' = [[3, 1],[5, −1]].
Symmetric part P = ½(A + A') = ½[[6, 6],[6, −2]] = [[3, 3],[3, −1]].
Skew-symmetric part Q = ½(A − A') = ½[[0, 4],[−4, 0]] = [[0, 2],[−2, 0]].
Verification: P + Q = [[3+0, 3+2],[3−2, −1+0]] = [[3, 5],[1, −1]] = A ✓
A = [[3, 3],[3, −1]] + [[0, 2],[−2, 0]].
If A = [[cos α, sin α], [−sin α, cos α]], then show that A'A = I.
A' = [[cos α, −sin α],[sin α, cos α]].
A'A = [[cos α, −sin α],[sin α, cos α]] × [[cos α, sin α],[−sin α, cos α]]
= [[cos²α + sin²α, cos α sin α − sin α cos α],[sin α cos α − cos α sin α, sin²α + cos²α]]
= [[1, 0],[0, 1]] = I. Hence proved.
Find the matrix A such that [[2, −1], [1, 0], [−3, 4]] × A = [[−1, −8], [1, −2], [9, 22]].
Let A = [[a, b],[c, d]] (2×2 matrix since left matrix is 3×2, product is 3×2).
[[2a−c, 2b−d],[a, b],[−3a+4c, −3b+4d]] = [[−1, −8],[1, −2],[9, 22]].
From row 2: a = 1, b = −2.
From row 1: 2(1) − c = −1 → c = 3; 2(−2) − d = −8 → d = 4 − (−8) → −4 − d = −8 → d = 4.
Verify row 3: −3(1) + 4(3) = 9 ✓; −3(−2) + 4(4) = 6 + 16 = 22 ✓.
A = [[1, −2],[3, 4]].
If A is a square matrix such that A² = A, then show that (I + A)³ = 7A + I.
Given A² = A (idempotent matrix). This also means A³ = A² × A = A × A = A² = A.
(I + A)³ = I³ + 3I²A + 3IA² + A³ (using binomial expansion for matrices that commute with I)
= I + 3A + 3A² + A³
= I + 3A + 3A + A (substituting A² = A and A³ = A)
= I + 7A = 7A + I. Hence proved.
Show that the matrix A = [[0, −1, 2], [1, 0, −3], [−2, 3, 0]] is a skew-symmetric matrix.
A' = [[0, 1, −2],[−1, 0, 3],[2, −3, 0]].
−A = [[0, 1, −2],[−1, 0, 3],[2, −3, 0]].
Since A' = −A, A is skew-symmetric.
Also note: all diagonal elements are 0, which is a necessary condition for skew-symmetric matrices.
Find x and y if 2[[1, 3], [0, x]] + [[y, 0], [1, 2]] = [[5, 6], [1, 8]].
2[[1, 3],[0, x]] + [[y, 0],[1, 2]] = [[2+y, 6],[1, 2x+2]] = [[5, 6],[1, 8]].
From (1,1): 2 + y = 5 → y = 3.
From (2,2): 2x + 2 = 8 → 2x = 6 → x = 3.
If A = [[1, 2, 3], [3, −2, 1], [4, 2, 1]] and B = [[1, 0], [2, 3], [−1, 4]], then find AB.
A is 3×3, B is 3×2, so AB is 3×2.
AB = [[1(1)+2(2)+3(−1), 1(0)+2(3)+3(4)],[3(1)+(−2)(2)+1(−1), 3(0)+(−2)(3)+1(4)],[4(1)+2(2)+1(−1), 4(0)+2(3)+1(4)]]
= [[1+4−3, 0+6+12],[3−4−1, 0−6+4],[4+4−1, 0+6+4]]
= [[2, 18],[−2, −2],[7, 10]].
If A = [[1, 0], [0, −1]], find A⁴.
A² = [[1,0],[0,−1]] × [[1,0],[0,−1]] = [[1, 0],[0, 1]] = I.
A⁴ = (A²)² = I² = I = [[1, 0],[0, 1]].
If A = [[3, 1], [−1, 2]], show that A² − 5A + 7I = O.
A² = [[3,1],[−1,2]] × [[3,1],[−1,2]] = [[9−1, 3+2],[−3−2, −1+4]] = [[8, 5],[−5, 3]].
5A = [[15, 5],[−5, 10]]. 7I = [[7, 0],[0, 7]].
A² − 5A + 7I = [[8−15+7, 5−5+0],[−5+5+0, 3−10+7]] = [[0, 0],[0, 0]] = O. Hence proved.
If A and B are symmetric matrices of the same order, then show that AB is symmetric if and only if AB = BA.
Given: A' = A, B' = B (both symmetric).
(AB)' = B'A' = BA.
AB is symmetric ⟺ (AB)' = AB ⟺ BA = AB.
Thus AB is symmetric if and only if AB = BA (i.e., A and B commute). Hence proved.
Find A⁻¹ of A = [[2, 3], [5, 7]] using elementary row operations.
Write A = IA: [[2, 3],[5, 7]] = [[1, 0],[0, 1]] × A.
R₁ → R₁/2: [[1, 3/2],[5, 7]] = [[1/2, 0],[0, 1]] × A.
R₂ → R₂ − 5R₁: [[1, 3/2],[0, −1/2]] = [[1/2, 0],[−5/2, 1]] × A.
R₂ → −2R₂: [[1, 3/2],[0, 1]] = [[1/2, 0],[5, −2]] × A.
R₁ → R₁ − (3/2)R₂: [[1, 0],[0, 1]] = [[−7, 3],[5, −2]] × A.
A⁻¹ = [[−7, 3],[5, −2]].
Verification: det(A) = 14 − 15 = −1 ≠ 0, so A is invertible.
Show that the matrix A = [[1, −1, 5], [−1, 2, 1], [5, 1, 3]] is a symmetric matrix.
A' = [[1, −1, 5],[−1, 2, 1],[5, 1, 3]].
Comparing A' with A: all corresponding elements are equal.
Since A' = A, the matrix A is symmetric.
Find the values of x and y if [[x, 5], [y, 0]] + [[3, −4], [1, 2]] = [[7, 1], [3, 2]].
[[x+3, 5−4],[y+1, 0+2]] = [[7, 1],[3, 2]].
From (1,1): x + 3 = 7 → x = 4.
From (2,1): y + 1 = 3 → y = 2.
Want more PYQ-style practice? AI generates unlimited similar questions from this chapter.
Practice PYQ PatternsAssertion-Reason Questions
(A) Both true, Reason explains Assertion · (B) Both true, Reason doesn't explain · (C) Assertion true, Reason false · (D) Assertion false, Reason true
AR1
Assertion (A): For two matrices A and B, AB = BA always holds.
Reason (R): Matrix multiplication is commutative.
Answer: (D)
Both are FALSE. Matrix multiplication is NOT commutative in general. AB ≠ BA for most matrix pairs, even when both products are defined.
AR2
Assertion (A): All diagonal elements of a skew-symmetric matrix are zero.
Reason (R): For a skew-symmetric matrix, aᵢⱼ = −aⱼᵢ for all i, j.
Answer: (A)
Both are TRUE and the reason explains the assertion. When i = j: aᵢᵢ = −aᵢᵢ → 2aᵢᵢ = 0 → aᵢᵢ = 0. So diagonal elements must be zero.
AR3
Assertion (A): (AB)' = A'B' for any two matrices A and B.
Reason (R): The transpose of a product reverses the order of multiplication.
Answer: (D)
Assertion is FALSE: The correct formula is (AB)' = B'A' (order reversed), not A'B'. Reason is TRUE: transpose of a product does reverse the order, which is why (AB)' = B'A'.
AR4
Assertion (A): Every square matrix can be uniquely expressed as the sum of a symmetric and a skew-symmetric matrix.
Reason (R): A = ½(A + A') + ½(A − A'), where ½(A + A') is symmetric and ½(A − A') is skew-symmetric.
Answer: (A)
Both are TRUE and the reason explains the assertion. ½(A + A') is symmetric because its transpose equals itself, and ½(A − A') is skew-symmetric because its transpose equals its negative.
AR5
Assertion (A): If AB and BA are both defined, then AB = BA.
Reason (R): If A is m×n and B is n×m, then AB is m×m and BA is n×n.
Answer: (D)
Assertion is FALSE: Even when both AB and BA are defined, they need not be equal - they may not even have the same order (AB is m×m, BA is n×n). Reason is TRUE: it correctly states the orders of AB and BA.
AR6
Assertion (A): If A and B are diagonal matrices of the same order, then AB is also a diagonal matrix.
Reason (R): The product of diagonal matrices is obtained by multiplying corresponding diagonal elements.
Answer: (A)
Both are TRUE and the reason explains the assertion. If A = diag(a₁, a₂, ..., aₙ) and B = diag(b₁, b₂, ..., bₙ), then AB = diag(a₁b₁, a₂b₂, ..., aₙbₙ), which is diagonal.
Case Study Questions
A shop sells three types of items - notebooks, pens, and erasers. Two branches (Branch A and Branch B) maintain their stock and price data using matrices. Stock matrix S = [[40, 50, 30], [25, 40, 20]] (rows = branches, columns = notebooks, pens, erasers). Price matrix P = [[10], [5], [3]] (column matrix of unit prices in ₹).
(i) What is the order of the matrix SP? (1 mark)
S is 2×3 and P is 3×1. Since columns of S (3) = rows of P (3), the product is defined.
Order of SP = 2×1.
(ii) Find the total revenue of each branch. (2 marks)
SP = [[40(10)+50(5)+30(3)],[25(10)+40(5)+20(3)]]
= [[400+250+90],[250+200+60]]
= [[740],[510]].
Branch A revenue = ₹740, Branch B revenue = ₹510.
(iii)(a) If the price of each item is increased by ₹2, find the new price matrix and the new revenue of Branch A. (2 marks)
New price matrix P' = [[12],[7],[5]].
Branch A revenue = 40(12) + 50(7) + 30(5) = 480 + 350 + 150 = ₹980.
(iii)(b) Find the total stock of each item across both branches. (1 mark)
Total stock = sum of columns:
Notebooks: 40 + 25 = 65
Pens: 50 + 40 = 90
Erasers: 30 + 20 = 50.
A spy agency encodes messages by converting letters to numbers (A=1, B=2, ..., Z=26) and then multiplying by a coding matrix C = [[1, 1], [0, 1]]. The message 'HI' is encoded as the column vector [[8], [9]].
(i) Find the encoded message by computing C × [[8], [9]]. (1 mark)
C × [[8],[9]] = [[1(8)+1(9)],[0(8)+1(9)]] = [[17],[9]].
Encoded message = [[17], [9]], i.e., QI.
(ii) Find C² and interpret what double encoding means. (2 marks)
C² = [[1,1],[0,1]] × [[1,1],[0,1]] = [[1,2],[0,1]].
Double encoding means applying C twice: C(C × message) = C² × message. The coding matrix C² = [[1, 2],[0, 1]] shifts the first element by twice the second element.
(iii) Show that C is invertible and find C⁻¹ (the decoding matrix). (2 marks)
det(C) = 1(1) − 1(0) = 1 ≠ 0, so C is invertible.
Using the formula for 2×2 inverse: C⁻¹ = (1/det) × [[d, −b],[−c, a]] = [[1, −1],[0, 1]].
Verification: C × C⁻¹ = [[1,1],[0,1]] × [[1,−1],[0,1]] = [[1,0],[0,1]] = I ✓.
C⁻¹ = [[1, −1],[0, 1]] (decoding matrix).
MCQ Practice
MCQ1. If A is a matrix of order 3×4 and B is a matrix of order 4×2, then the order of AB is:
(A) 3×2
(B) 4×4
(C) 3×4
(D) 2×3
Answer: (A) - If A is m×p and B is p×n, then AB is m×n. Here m=3, p=4, n=2. So AB is **3×2**.
MCQ2. If I is the identity matrix of order n, then I² is:
(A) O
(B) 2I
(C) I
(D) nI
Answer: (C) - I × I = I. The identity matrix is idempotent: I² = I.
MCQ3. If A is a 2×3 matrix, then A' (transpose of A) is of order:
(A) 2×3
(B) 3×2
(C) 3×3
(D) 2×2
Answer: (B) - Transpose interchanges rows and columns. If A is m×n, then A' is n×m. So A' is **3×2**.
MCQ4. A square matrix A is skew-symmetric if:
(A) A' = A
(B) A' = −A
(C) A² = A
(D) A' = A⁻¹
Answer: (B) - By definition, a square matrix A is skew-symmetric if A' = −A (transpose equals negative of the matrix).
MCQ5. The number of all possible matrices of order 2×3 with each entry 0 or 1 is:
(A) 12
(B) 32
(C) 64
(D) 128
Answer: (C) - A 2×3 matrix has 6 entries. Each entry has 2 choices (0 or 1). Total = 2⁶ = **64**.
MCQ6. If A = [[1, 2], [3, 4]] and B = [[1, 0], [0, 1]], then A + B' is:
(A) [[2, 2], [3, 5]]
(B) [[2, 3], [2, 5]]
(C) [[1, 2], [3, 4]]
(D) Not defined
Answer: (A) - B' = [[1, 0],[0, 1]] = B (identity is symmetric). A + B' = [[1+1, 2+0],[3+0, 4+1]] = **[[2, 2],[3, 5]]**.
MCQ7. If A = [[2, 0], [0, 3]], then A² is:
(A) [[4, 0], [0, 9]]
(B) [[4, 0], [0, 6]]
(C) [[2, 0], [0, 3]]
(D) [[4, 6], [6, 9]]
Answer: (A) - A is diagonal, so A² = [[2², 0],[0, 3²]] = **[[4, 0],[0, 9]]**.
MCQ8. If A = [[1, −1], [2, 3]] and B = [[0, 2], [−1, 1]], then A + B is:
(A) [[1, 1], [1, 4]]
(B) [[1, −3], [3, 2]]
(C) [[0, 1], [1, 4]]
(D) [[1, 1], [3, 4]]
Answer: (A) - A + B = [[1+0, −1+2],[2+(−1), 3+1]] = **[[1, 1],[1, 4]]**.
MCQ9. The diagonal elements of a skew-symmetric matrix are all:
(A) 1
(B) −1
(C) 0
(D) any real number
Answer: (C) - For skew-symmetric: aᵢⱼ = −aⱼᵢ. Put i = j: aᵢᵢ = −aᵢᵢ → 2aᵢᵢ = 0 → aᵢᵢ = **0**.
MCQ10. If A is a symmetric matrix, then A − A' is:
(A) symmetric
(B) skew-symmetric
(C) zero matrix
(D) identity matrix
Answer: (C) - If A is symmetric, A' = A. So A − A' = A − A = **O (zero matrix)**.
MCQ11. If A = [[1, 0, 0], [0, 1, 0], [a, b, -1]], then A² is equal to:
(A) a null matrix
(B) a unit matrix
(C) -A
(D) A
Answer: (B) - A² = A × A. Computing: Row 1 of A² = [1,0,0]. Row 2 = [0,1,0]. Row 3 = [a+0+(-a), 0+b+(-b), 0+0+1] = [0, 0, 1]. Wait, let me recompute row 3: [a(1)+b(0)+(-1)(a), a(0)+b(1)+(-1)(b), a(0)+b(0)+(-1)(-1)] = [a-a, b-b, 1] = [0, 0, 1]. So A² = I (unit matrix).
MCQ12. If A = [[1, -1], [2, -1]] and B = [[a, 1], [b, -1]], and (A + B)² = A² + B², then the values of a and b are:
(A) a = 1, b = -2
(B) a = 1, b = 4
(C) a = -1, b = 4
(D) No values exist
Answer: (B) - (A+B)² = A²+B² holds iff AB + BA = O. A+B = [[1+a, 0], [2+b, -2]]. AB = [[a-b, 2], [2a-b, 3]], BA = [[a+2, -a-1], [b-2, -b+1]]. AB+BA = O gives: a-b+a+2=0, 2+(-a-1)=0, 2a-b+b-2=0, 3+(-b+1)=0. From eq2: a=1. From eq4: b=4. Verify eq1: 1-4+1+2=0 ✓. Verify eq3: 2-4+4-2=0 ✓. So a=1, b=4.
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Start Quiz ModeImportant Questions - Mark-wise
1 Mark Questions
If A = [[1, 2], [3, 4]], what is the order of A?
The matrix A has 2 rows and 2 columns. **Order = 2×2.**
If a matrix has 8 elements, what are the possible orders?
Factor pairs of 8: 1×8, 2×4, 4×2, 8×1. **4 possible orders.**
What is the transpose of [[1, 2, 3]]?
Transpose of a 1×3 row matrix is a 3×1 column matrix: **[[1], [2], [3]].**
If A is a 3×3 skew-symmetric matrix, what are the diagonal elements?
All diagonal elements of a skew-symmetric matrix are **0**.
For what value of x is [[2, x], [x, 8]] a symmetric matrix?
A symmetric matrix has aᵢⱼ = aⱼᵢ. Here a₁₂ = x and a₂₁ = x, which are already equal. So the matrix is symmetric for **all values of x**.
2-3 Mark Questions
If A = [[1, 5], [6, 7]], find A + A' and verify it is symmetric.
A' = [[1, 6],[5, 7]]. A + A' = [[2, 11],[11, 14]]. (A + A')' = [[2, 11],[11, 14]] = A + A'. Hence A + A' is symmetric. ✓
Show that A − A' is skew-symmetric for any square matrix A.
Let B = A − A'. Then B' = (A − A')' = A' − (A')' = A' − A = −(A − A') = −B. Since B' = −B, B is skew-symmetric.
Find x, y, z, w if [[x−y, 2z+w], [2x−y, 3z+w]] = [[4, 7], [3, 10]].
x − y = 4 ...(1), 2z + w = 7 ...(2), 2x − y = 3 ...(3), 3z + w = 10 ...(4).
From (1)−(3): −x = 1 → x = −1. From (1): y = −5.
From (4)−(2): z = 3. From (2): w = 1.
x = −1, y = −5, z = 3, w = 1.
If A = [[1, 2], [−1, 3]] and B = [[3, −1], [0, 2]], verify that (AB)' = B'A'.
AB = [[3, 3],[−3, 7]]. (AB)' = [[3, −3],[3, 7]].
B' = [[3, 0],[−1, 2]], A' = [[1, −1],[2, 3]].
B'A' = [[3, −3],[3, 7]]. Since (AB)' = B'A', the property is verified. ✓
3-4 Mark Questions
Express [[6, −2, 2], [−2, 3, −1], [2, −1, 3]] as the sum of a symmetric and skew-symmetric matrix.
Let A = [[6,−2,2],[−2,3,−1],[2,−1,3]]. A' = [[6,−2,2],[−2,3,−1],[2,−1,3]] = A.
Since A is already symmetric: P = A, Q = O (zero matrix).
A = A + O (symmetric + skew-symmetric where skew part is zero).
If A = [[0, 6, 7], [−6, 0, 8], [7, −8, 0]], B = [[0, 1, 1], [1, 0, 2], [1, 2, 0]], C = [[2], [−2], [3]], find (A+B)C.
A + B = [[0, 7, 8],[−5, 0, 10],[8, −6, 0]].
(A+B)C = [[0(2)+7(−2)+8(3)],[−5(2)+0(−2)+10(3)],[8(2)+(−6)(−2)+0(3)]]
= [[0−14+24],[−10+0+30],[16+12+0]] = [[10],[20],[28]].
If A = [[1, 2, 3], [3, −2, 1], [4, 2, 1]], show that A³ − 23A − 40I = O.
First find A²: compute A × A. Then compute A³ = A² × A. Substitute into A³ − 23A − 40I and show all elements equal 0.
This is a characteristic equation verification - compute step by step using matrix multiplication.
A³ − 23A − 40I = O. (Verification by direct matrix computation.)
Key Concepts
Important Formulas
Study Plan - Matrices
Types and Order of Matrices
Read Section 3.1-3.2. Understand row, column, square, diagonal, scalar, identity, and zero matrices. Complete Exercise 3.1 (all 10 questions). Focus on constructing matrices from given rules (Q4, Q5).
Matrix Operations - Addition, Scalar Multiplication, Multiplication
Read Section 3.3-3.4. Practice addition, subtraction, scalar multiplication. Complete Exercise 3.2 Q1-Q12. Pay attention to when operations are defined (same order for addition).
Matrix Multiplication (continued)
Complete Exercise 3.2 Q13-Q22. Focus on checking dimensions before multiplication and computing products step by step. Practice 3x3 matrix multiplication.
Transpose, Symmetric, and Skew-Symmetric Matrices
Read Section 3.5-3.6. Learn the decomposition formula A = 1/2(A+A') + 1/2(A-A'). Complete Exercise 3.3 (all 12 questions). Practice showing matrices are symmetric/skew-symmetric.
Elementary Operations, Invertible Matrices, and Revision
Read Section 3.7-3.8. Complete Exercise 3.4 and Miscellaneous Exercise. Solve PYQs from 2019-2026. Revise all formulas. Practice MCQs and assertion-reason questions for quick recall.
Common Mistakes to Avoid
- ✗Assuming AB = BA (matrix multiplication is NOT commutative) - always maintain the order of multiplication
- ✗Multiplying matrices with incompatible dimensions - AB is defined only when columns of A = rows of B
- ✗Confusing transpose with inverse - A' swaps rows and columns, A inverse satisfies AA inverse = I; they are different operations
- ✗Forgetting that diagonal elements of a skew-symmetric matrix must be zero (since a_ii = -a_ii implies a_ii = 0)
- ✗Adding or subtracting matrices of different orders - matrix addition requires the SAME order
- ✗Writing (AB)' = A'B' instead of (AB)' = B'A' - the transpose of a product reverses the order
Tips for Board Exam
- ✓Always check matrix dimensions before multiplication - verify columns of A = rows of B, and write the order of the result (m x n) to avoid errors
- ✓Use the symmetric + skew-symmetric decomposition formula directly: P = 1/2(A + A') and Q = 1/2(A - A'). Just apply and verify P' = P and Q' = -Q
- ✓For A squared type questions, write out the multiplication carefully row by column - common error is squaring individual elements instead of doing proper matrix multiplication
- ✓For proving A squared - kA + mI = O: Step 1 find A squared, Step 2 compute kA, Step 3 compute mI, Step 4 add/subtract and show result is the zero matrix O
- ✓Remember (AB)' = B'A' (the order reverses) - this property is frequently tested. For verification problems, compute both sides independently and show they are equal
Practice Matrices with AI Tutor
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Start Matrices - Choose Your ModeAbout Matrices - Class 12 Mathematics
Chapter 3 Matrices introduces the concept of matrices as rectangular arrays of numbers arranged in rows and columns. The chapter covers types of matrices (row, column, square, diagonal, scalar, identity, zero), equality of matrices, and fundamental operations including addition, scalar multiplication, and matrix multiplication. Students learn the important rule that matrix multiplication is not commutative (AB ≠ BA in general) and that multiplication is only defined when the number of columns of the first matrix equals the number of rows of the second. The chapter also covers the transpose of a matrix and its properties, including the reversal property (AB)' = B'A'.
The second half of the chapter focuses on symmetric matrices (A' = A) and skew-symmetric matrices (A' = −A, with zero diagonal), culminating in the important decomposition theorem: every square matrix can be uniquely written as the sum of a symmetric and a skew-symmetric matrix using A = ½(A + A') + ½(A − A'). The chapter concludes with elementary row and column operations and the concept of invertible matrices. For the CBSE 2027 board exam, students should focus on matrix multiplication, the symmetric/skew-symmetric decomposition, matrix equations (finding A given conditions like A² = A or AB = C), and proving properties of transpose. This chapter typically contributes about 5 marks and forms the foundation for Chapter 4 (Determinants).
Frequently Asked Questions
How many exercises are in NCERT Class 12 Maths Chapter 3 Matrices?
There are 4 exercises (3.1, 3.2, 3.3, 3.4) plus a Miscellaneous Exercise, totalling 77 questions.
How many marks does Matrices carry in the CBSE board exam?
Matrices is part of the Algebra unit (10 marks total, combined with Determinants). Typically, Matrices contributes about 5 marks - one MCQ (1 mark), one short answer (2-3 marks), and sometimes part of a 5-mark question.
What is the difference between a symmetric and skew-symmetric matrix?
A symmetric matrix satisfies A' = A (transpose equals itself), so aᵢⱼ = aⱼᵢ. A skew-symmetric matrix satisfies A' = −A (transpose equals negative), so aᵢⱼ = −aⱼᵢ and all diagonal elements are 0.
When is matrix multiplication AB defined?
AB is defined when the number of columns of A equals the number of rows of B. If A is m×p and B is p×n, then AB is defined and has order m×n.
Is matrix multiplication commutative?
No. In general, AB ≠ BA. Even when both products are defined, they may have different orders or different elements. Commutativity holds only in special cases (e.g., both are diagonal matrices of the same order).
What is (AB)' in terms of A' and B'?
(AB)' = B'A'. The transpose of a product reverses the order of multiplication. This is one of the most important properties tested in board exams.
How do you express a matrix as the sum of a symmetric and skew-symmetric matrix?
Use the formula: Symmetric part = ½(A + A'), Skew-symmetric part = ½(A − A'). Then A = ½(A + A') + ½(A − A'). This decomposition is unique.
Is NCERT enough for Matrices in the CBSE board exam?
Yes, NCERT is sufficient for the board exam. All board questions are based on NCERT concepts and similar patterns. Focus on all exercises, the Miscellaneous Exercise, and the solved examples. Practice PYQs for additional confidence.