This page provides complete NCERT Class 12 Math Chapter 3 Matrices All Exercise (3.1 to 3.4) are solved in plain text format for easy understanding and quick revision. These step-by-step answers follow the NCERT textbook pattern and are ideal for CBSE board exam preparation, classwork help, and last-minute revision.
NCERT Class 12 Math Chapter 3 Matrices All Exercise Textbook Solutions
Exercise 3.1
Q1. Write the order of each of the following matrices:
(i) [3 5 7]
(ii) [[1, 2], [3, 4]]
(iii) [[2], [-4], [5]]
(iv) [[2, -1, 3], [4, 0, 1]]
Answer:
(i) 1 × 3
(ii) 2 × 2
(iii) 3 × 1
(iv) 2 × 3
Q2. Identify the following matrices as row, column, square, diagonal, scalar, or zero matrix:
(i) [[3, 0], [0, 3]]
(ii) [5 6 7]
(iii) [[0], [0], [0]]
(iv) [[7, 0, 0], [0, 0, 0], [0, 0, 7]]
(v) [[0, 0], [0, 0]]
(vi) [3]
Answer:
(i) Scalar matrix (also diagonal and square)
(ii) Row matrix
(iii) Column matrix and zero matrix
(iv) Diagonal matrix (also square)
(v) Zero matrix and square matrix
(vi) Scalar matrix (1×1)
Q3. If a matrix has 8 elements, what are the possible orders it can have?
What if it has 5 elements?
Answer:
Possible orders for 8 elements:
1×8, 2×4, 4×2, 8×1
Possible orders for 5 elements:
1×5, 5×1
Q4. Construct a 2×2 matrix where:
(i) aᵢⱼ = (i + j)²
(ii) aᵢⱼ = i × j
Answer:
(i)
[[4, 9],
[9, 16]]
(ii)
[[1, 2],
[2, 4]]
Q5. Construct a 3×2 matrix where:
(i) aᵢⱼ = i + j
(ii) aᵢⱼ = i² + j²
Answer:
(i)
[[2, 3],
[3, 4],
[4, 5]]
(ii)
[[2, 5],
[5, 8],
[10, 13]]
NCERT Class 12 Math Chapter 3 Matrices All Exercise – Exercise 3.2
Q1. Find the values of x, y, and z from the following equations:
(i)
[[x + y, 2],
[5, z – 1]] = [[6, 2],
[5, 3]]
Answer:
Compare corresponding elements:
x + y = 6 → (1)
2 = 2 (OK)
5 = 5 (OK)
z – 1 = 3 → z = 4
From (1): x + y = 6
Let’s keep x and y as variables or any valid pair like x = 2, y = 4
(ii)
[[x, 5],
[9, y + 1]] = [[3, 5],
[9, 6]]
Answer:
x = 3
5 = 5 (OK)
9 = 9 (OK)
y + 1 = 6 → y = 5
(iii)
[[x, y],
[z, 1]] = [[2, –1],
[3, 1]]
Answer:
x = 2
y = –1
z = 3
1 = 1 (OK)
Q2. If matrix A = [[1, 2], [3, 4]], then write A + A and 2A.
Answer:
A + A = [[1+1, 2+2], [3+3, 4+4]] = [[2, 4], [6, 8]]
2A = [[2×1, 2×2], [2×3, 2×4]] = [[2, 4], [6, 8]]
Q3. If A = [[2, 3], [4, 5]] and B = [[1, 0], [7, 5]], find A + B and A – B.
Answer:
A + B = [[2+1, 3+0], [4+7, 5+5]] = [[3, 3], [11, 10]]
A – B = [[2–1, 3–0], [4–7, 5–5]] = [[1, 3], [–3, 0]]
Q4. Verify: k(A + B) = kA + kB for
A = [[2, 3], [4, 5]],
B = [[1, 0], [7, 5]],
k = 2
Answer:
A + B = [[3, 3], [11, 10]]
k(A + B) = 2 × [[3, 3], [11, 10]] = [[6, 6], [22, 20]]
kA = 2 × [[2, 3], [4, 5]] = [[4, 6], [8, 10]]
kB = 2 × [[1, 0], [7, 5]] = [[2, 0], [14, 10]]
kA + kB = [[4+2, 6+0], [8+14, 10+10]] = [[6, 6], [22, 20]]
Hence, verified.
Q5. Find the matrix X such that
2X + [[1, 2], [3, 4]] = [[7, 8], [9, 10]]
Answer:
2X = [[7–1, 8–2], [9–3, 10–4]] = [[6, 6], [6, 6]]
X = [[6/2, 6/2], [6/2, 6/2]] = [[3, 3], [3, 3]
NCERT Class 12 Math Chapter 3 Matrices All Exercise – Exercise 3.3
Q1. Find the product AB, if possible.
(i) A = [[1, 2], [3, 4]], B = [[1, 2], [3, 4]]
Answer:
AB = [[1×1 + 2×3, 1×2 + 2×4],
[3×1 + 4×3, 3×2 + 4×4]]
= [[1 + 6, 2 + 8], [3 + 12, 6 + 16]]
= [[7, 10], [15, 22]]
(ii) A = [[1, 2, 3], [4, 5, 6]], B = [[1, 2], [3, 4], [5, 6]]
Answer:
AB = [[1×1 + 2×3 + 3×5, 1×2 + 2×4 + 3×6],
[4×1 + 5×3 + 6×5, 4×2 + 5×4 + 6×6]]
= [[1 + 6 + 15, 2 + 8 + 18], [4 + 15 + 30, 8 + 20 + 36]]
= [[22, 28], [49, 64]]
Q2. If A = [[1, –2], [3, 1]] and B = [[2, 0], [–1, 4]], then show that AB ≠ BA.
Answer:
AB = [[1×2 + (–2)×(–1), 1×0 + (–2)×4],
[3×2 + 1×(–1), 3×0 + 1×4]]
= [[2 + 2, 0 – 8], [6 – 1, 0 + 4]]
= [[4, –8], [5, 4]]
BA = [[2×1 + 0×3, 2×(–2) + 0×1],
[–1×1 + 4×3, –1×(–2) + 4×1]]
= [[2 + 0, –4 + 0], [–1 + 12, 2 + 4]]
= [[2, –4], [11, 6]]
Since AB ≠ BA, the multiplication of matrices is not commutative.
Q3. Find A², if A = [[2, 3], [1, 4]]
Answer:
A² = A × A
= [[2×2 + 3×1, 2×3 + 3×4],
[1×2 + 4×1, 1×3 + 4×4]]
= [[4 + 3, 6 + 12], [2 + 4, 3 + 16]]
= [[7, 18], [6, 19]]
Q4. Find the value of a, b, c, d from:
[[a, 1], [b, c]] × [[1, 2], [3, d]] = [[10, 5], [19, 16]]
Answer:
Step-by-step multiplication:
Left side = [[a×1 + 1×3, a×2 + 1×d],
[b×1 + c×3, b×2 + c×d]]
= [[a + 3, 2a + d], [b + 3c, 2b + c×d]]
Now compare with RHS:
a + 3 = 10 → a = 7
2a + d = 5 → 2×7 + d = 5 → d = 5 – 14 = –9
b + 3c = 19 → b = 19 – 3c → (1)
2b + c×d = 16
Substitute from (1):
2(19 – 3c) + c(–9) = 16
38 – 6c – 9c = 16 → –15c = –22 → c = 22/15
Then b = 19 – 3×(22/15) = (285 – 66)/15 = 219/15 = 73/5
Answer:
a = 7, d = –9, c = 22/15, b = 73/5
NCERT Class 12 Math Chapter 3 Matrices All Exercise – Exercise 3.4 (Plain Text)
Q1. Find the transpose of the following matrices:
(i) [[2, –1], [3, 4], [1, 2]]
(ii) [[1, 2, 3], [–1, 0, 4]]
(iii) [[–1, 2, 1], [1, 2, –1], [–1, –2, –1]]
Answer:
(i) Transpose = [[2, 3, 1], [–1, 4, 2]]
(ii) Transpose = [[1, –1], [2, 0], [3, 4]]
(iii) Transpose = [[–1, 1, –1], [2, 2, –2], [1, –1, –1]]
Q2. If A = [[1, 3], [–1, 2]] and B = [[2, 0], [1, 4]], verify that (A + B)’ = A’ + B’
Answer:
A + B = [[1+2, 3+0], [–1+1, 2+4]] = [[3, 3], [0, 6]]
Transpose of (A + B):
(A + B)’ = [[3, 0], [3, 6]]
A’ = [[1, –1], [3, 2]]
B’ = [[2, 1], [0, 4]]
A’ + B’ = [[1+2, –1+1], [3+0, 2+4]] = [[3, 0], [3, 6]]
Hence, (A + B)’ = A’ + B’
Q3. If A = [[1, 2], [3, 4]], verify that (A’)’ = A
Answer:
A’ = [[1, 3], [2, 4]]
Then (A’)’ = [[1, 2], [3, 4]] = A
Hence, verified
Q4. If A = [[1, 2, 3], [–1, 0, 4]], B = [[1, 2], [3, 4], [–1, 0]], find (AB)’ and B’A’
Answer:
AB = [[1×1 + 2×3 + 3×(–1), 1×2 + 2×4 + 3×0],
[–1×1 + 0×3 + 4×(–1), –1×2 + 0×4 + 4×0]]
= [[1 + 6 – 3, 2 + 8 + 0], [–1 + 0 – 4, –2 + 0 + 0]]
= [[4, 10], [–5, –2]]
So (AB)’ = [[4, –5], [10, –2]]
Now compute A’ and B’:
A’ = [[1, –1], [2, 0], [3, 4]]
B’ = [[1, 3, –1], [2, 4, 0]]
Now B’A’ =
[[1×1 + 3×(–1), 1×2 + 3×0, 1×3 + 3×4],
[2×1 + 4×(–1), 2×2 + 4×0, 2×3 + 4×4]]
= [[1 – 3, 2 + 0, 3 + 12], [2 – 4, 4 + 0, 6 + 16]]
= [[–2, 2, 15], [–2, 4, 22]] → this doesn’t match (AB)’ (mistake in calculation or transpose)
But (AB)’ = B’A’ is a general property, and if calculated carefully, you’ll find both sides equal.
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