Class 12 : Maths (English) – Chapter 3: Matrices

Posted by

On June 25, 2025

EXPLANATION & SUMMARY

🔵 Explanation
1️⃣ Introduction
A matrix is a rectangular arrangement of numbers in rows and columns.
If a matrix has m rows and n columns, it is called an m × n matrix.
Matrices help in solving linear equations, geometric transformations, statistics, economics, and computer graphics.
Notation:
A = [aᵢⱼ] means element aᵢⱼ is in the i-th row and j-th column.
Example:
A = [[a₁₁, a₁₂], [a₂₁, a₂₂]]

2️⃣ Types of Matrices
Row matrix – one row: [3 5 7]
Column matrix – one column:
[2]
[4]
[6]
Square matrix – rows = columns (n × n)
Zero / Null matrix – all elements 0
Diagonal matrix – all non-diagonal elements 0
Scalar matrix – diagonal elements equal
Identity matrix Iₙ – diagonal elements 1, others 0
Equal matrices – same order and equal corresponding elements
Order = rows × columns.

3️⃣ Operations on Matrices
(a) Equality
A = B ⇔ same order and aᵢⱼ = bᵢⱼ for every i, j.
(b) Addition
If A and B are m × n,
A + B = [aᵢⱼ + bᵢⱼ]
Properties:
Commutative A + B = B + A
Associative (A + B) + C = A + (B + C)
Identity A + 0 = A
Inverse A + (−A) = 0
(c) Scalar Multiplication
For scalar k,
kA = [k·aᵢⱼ]
Properties:
k(A + B) = kA + kB
(k + l)A = kA + lA
(kl)A = k(lA)
(d) Subtraction
A − B = A + (−1)B
(e) Multiplication
If A is m × n and B is n × p, then AB is m × p.
Each element cᵢⱼ = Σ (aᵢₖ · bₖⱼ)
Condition: columns(A) = rows(B)
Properties:
Associative (AB)C = A(BC)
Distributive A(B + C) = AB + AC
Not commutative generally (AB ≠ BA)
Identity AI = IA = A
Null A·0 = 0

4️⃣ Transpose of a Matrix
Transpose (Aᵀ) is formed by interchanging rows and columns.
Example:
A = [[1, 2], [3, 4]] ⇒ Aᵀ = [[1, 3], [2, 4]]
Properties:
(Aᵀ)ᵀ = A
(A + B)ᵀ = Aᵀ + Bᵀ
(kA)ᵀ = kAᵀ
(AB)ᵀ = BᵀAᵀ

5️⃣ Symmetric and Skew-Symmetric
Symmetric: Aᵀ = A
Skew-symmetric: Aᵀ = −A
Every square matrix A can be written as
A = ½(A + Aᵀ) + ½(A − Aᵀ)

6️⃣ Invertible Matrix
A square matrix A is invertible if a matrix A⁻¹ exists such that
A·A⁻¹ = A⁻¹·A = I
Condition: determinant |A| ≠ 0
For 2 × 2:
A = [[a, b], [c, d]]
A⁻¹ = (1 / (ad − bc)) × [[d, −b], [−c, a]]

7️⃣ Elementary Operations
Used to simplify matrices:
Row: interchange rows, multiply a row by non-zero scalar, add multiple of one row to another
Column: same types
Applied in finding inverses or solving equations.

8️⃣ Matrix Equation AX = B
System
a₁₁x + a₁₂y = b₁
a₂₁x + a₂₂y = b₂
Matrix form:
A = [[a₁₁, a₁₂], [a₂₁, a₂₂]]
X = [[x], [y]]
B = [[b₁], [b₂]]
If |A| ≠ 0 ⇒ solution X = A⁻¹B

9️⃣ Applications
Solving linear equations (matrix method)
Computer graphics transformations
Statistics and economics models
Engineering computations

🔟 Key Properties
Addition/scalar multiplication: same order
Multiplication possible only if cols(A) = rows(B)
(AB)ᵀ = BᵀAᵀ
Inverse exists only when |A| ≠ 0
I acts as multiplicative identity

⚠️ Common Mistakes
Trying to multiply incompatible matrices
Assuming AB = BA
Ignoring determinant condition for inverse
Mixing transpose with inverse

✳️ Example
Solve
2x + 3y = 5
4x + 1y = 6
A = [[2, 3], [4, 1]] X = [[x], [y]] B = [[5], [6]]
Solution: X = A⁻¹B (provided |A| ≠ 0)

SUMMARY (~300 words)
A matrix is a rectangular array; order = rows × columns.
Types: row, column, square, diagonal, scalar, identity, zero.
Operations: addition, subtraction, scalar multiplication, matrix multiplication.
Transpose interchanges rows and columns.
Symmetric: Aᵀ = A; Skew-symmetric: Aᵀ = −A.
Inverse A⁻¹ exists if |A| ≠ 0; for 2×2, A⁻¹ = (1/|A|) adj(A).
Matrix equation AX = B ⇒ X = A⁻¹B.
Applications include solving linear systems, graphics, statistics, economics, engineering.
Elementary operations help in finding inverses.
Matrix multiplication is generally not commutative.

QUICK RECAP
✔️ Matrix order m × n
✔️ Operations: +, −, scalar, ×
✔️ Transpose Aᵀ
✔️ Symmetric Aᵀ = A, Skew Aᵀ = −A
✔️ Inverse A⁻¹ if |A| ≠ 0
✔️ Solve AX = B ⇒ X = A⁻¹B

————————————————————————————————————————————————————————————————————————————

QUESTIONS FROM TEXTBOOK

Exercise 3.1

🔵 Question 1:
In the matrix A =
[ 2 5 19 −7 ]
[ 35 −2 5/2 12 ]
[ √3 1 −5 17 ], write:
(i) The order of the matrix, (ii) The number of elements, (iii) Write the elements a₁₃, a₂₁, a₃₃, a₂₄, a₂₃.

🟢 Answer:
➡️ Step 1 (Order): Rows = 3, Columns = 4 ⇒ Order = 3 × 4.
➡️ Step 2 (Number of elements): 3 × 4 = 12.
➡️ Step 3 (Specified elements):
• a₁₃ = element (row 1, column 3) = 19.
• a₂₁ = element (row 2, column 1) = 35.
• a₃₃ = element (row 3, column 3) = −5.
• a₂₄ = element (row 2, column 4) = 12.
• a₂₃ = element (row 2, column 3) = 5/2.
✔️ Final: Order = 3 × 4; No. of elements = 12; a₁₃ = 19, a₂₁ = 35, a₃₃ = −5, a₂₄ = 12, a₂₃ = 5/2.

🔵 Question 2:
If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?

🟢 Answer:
💡 Concept: If a matrix has m × n elements, then m × n equals the total number of elements.
➡️ Step 1 (24 elements): m × n = 24.
Possible orders (m, n): (1,24), (2,12), (3,8), (4,6), (6,4), (8,3), (12,2), (24,1).
➡️ Step 2 (13 elements): 13 is prime ⇒ factor pairs (1,13), (13,1).
✔️ Final: For 24 elements: all factor pairs of 24; for 13 elements: (1,13) or (13,1).

🔵 Question 3:
If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?

🟢 Answer:
➡️ Step 1 (18 elements): m × n = 18.
Possible orders: (1,18), (2,9), (3,6), (6,3), (9,2), (18,1).
➡️ Step 2 (5 elements): 5 is prime ⇒ (1,5), (5,1).
✔️ Final: Orders as listed above.

🔵 Question 4:
Construct 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.

🟢 Answer:
💡 Take i = 1,2 and j = 1,2.

(i) aᵢⱼ = ((i + j)²)/2
➡️ a₁₁ = (2²)/2 = 2; a₁₂ = (3²)/2 = 9/2; a₂₁ = (3²)/2 = 9/2; a₂₂ = (4²)/2 = 8.
✔️ A = [ 2 9/2 ; 9/2 8 ].

(ii) aᵢⱼ = i/j
➡️ a₁₁ = 1; a₁₂ = 1/2; a₂₁ = 2; a₂₂ = 1.
✔️ A = [ 1 1/2 ; 2 1 ].

(iii) aᵢⱼ = ((i + 2j)²)/2
➡️ a₁₁ = (3²)/2 = 9/2; a₁₂ = (5²)/2 = 25/2; a₂₁ = (4²)/2 = 8; a₂₂ = (6²)/2 = 18.
✔️ A = [ 9/2 25/2 ; 8 18 ].

🔵 Question 5:
Construct a 3 × 4 matrix, whose elements are given by:
(i) aᵢⱼ = (1/2) | −3i + j |, (ii) aᵢⱼ = 2i − j.

🟢 Answer:
💡 Take i = 1,2,3 and j = 1,2,3,4.

(i) aᵢⱼ = (1/2) | −3i + j |
➡️ Row i = 1 (j = 1..4): (1/2)|−2| = 1; (1/2)|−1| = 1/2; (1/2)|0| = 0; (1/2)|1| = 1/2.
➡️ Row i = 2: (1/2)|−5| = 5/2; (1/2)|−4| = 2; (1/2)|−3| = 3/2; (1/2)|−2| = 1.
➡️ Row i = 3: (1/2)|−8| = 4; (1/2)|−7| = 7/2; (1/2)|−6| = 3; (1/2)|−5| = 5/2.
✔️ Required matrix =
[ 1 1/2 0 1/2 ]
[ 5/2 2 3/2 1 ]
[ 4 7/2 3 5/2 ].

(ii) aᵢⱼ = 2i − j
➡️ Row i = 1: 1, 0, −1, −2.
➡️ Row i = 2: 3, 2, 1, 0.
➡️ Row i = 3: 5, 4, 3, 2.
✔️ Required matrix =
[ 1 0 −1 −2 ]
[ 3 2 1 0 ]
[ 5 4 3 2 ].

🔵 Question 6(i):
Find the values of x, y and z from the equation
[ 4 3 ] = [ y z ]
[ x 5 ] [ 1 5 ]

🟢 Answer:
✨ Step 1 (Equality of corresponding entries):
➡️ 4 = y, 3 = z, x = 1, 5 = 5 (checks out)
✔️ Final: x = 1, y = 4, z = 3

🔵 Question 6(ii):
Find the values of x, y and z from the equation
[ x + y 2 ] = [ 6 2 ]
[ 5 + z x y ] [ 5 8 ]

🟢 Answer:
✨ Step 1 (Match entries):
➡️ x + y = 6
➡️ 2 = 2 (already equal)
➡️ 5 + z = 5 ⇒ z = 0
➡️ x y = 8

🌿 Step 2 (Solve for x and y using sum and product):
➡️ t^2 − (x + y)t + x y = 0 ⇒ t^2 − 6t + 8 = 0
➡️ (t − 2)(t − 4) = 0 ⇒ t = 2 or t = 4

✔️ Final: z = 0 and (x, y) = (2, 4) or (4, 2)

🔵 Question 6(iii):
Find the values of x, y and z from the equation
[ x + y + z ] [ 9 ]
[ x + z ] = [ 5 ]
[ y + z ] [ 7 ]

🟢 Answer:
✨ Step 1 (Write equations):
➡️ x + y + z = 9
➡️ x + z = 5
➡️ y + z = 7

🌿 Step 2 (Express x, y in terms of z):
➡️ x = 5 − z
➡️ y = 7 − z

⚡ Step 3 (Substitute into first equation):
➡️ (5 − z) + (7 − z) + z = 9 ⇒ 12 − z = 9 ⇒ z = 3

🧠 Step 4 (Back-substitute):
➡️ x = 5 − 3 = 2
➡️ y = 7 − 3 = 4

✔️ Final: x = 2, y = 4, z = 3

🔵 Question 7:
Find the value of a, b, c and d from the equation
[ a − b 2a + c ] = [ −1 5 ]
[ 2a − b 3c + d ] [ 0 13 ]

🟢 Answer:
✨ Step 1 (Equate corresponding entries):
➡️ a − b = −1 … (1)
➡️ 2a + c = 5 … (2)
➡️ 2a − b = 0 … (3)
➡️ 3c + d = 13 … (4)

🌿 Step 2 (Solve for a, b):
➡️ (3) − (1): (2a − b) − (a − b) = 0 − (−1) ⇒ a = 1
➡️ From (3): 2(1) − b = 0 ⇒ b = 2

⚡ Step 3 (Solve for c):
➡️ From (2): 2(1) + c = 5 ⇒ c = 3

🧠 Step 4 (Solve for d):
➡️ From (4): 3(3) + d = 13 ⇒ d = 4

✔️ Final: a = 1, b = 2, c = 3, d = 4

🔵 Question 8:
A = [aᵢⱼ]ₘₓₙ is a square matrix, if
(A) m < n
(B) m > n
(C) m = n
(D) None of these
🟢 Answer:
💡 A matrix is called a square matrix when the number of rows equals the number of columns.
➡️ That is, m = n.
✔️ Correct Option: (C) m = n

🔵 Question 9:
Which of the given values of x and y make the following pair of matrices equal:
[ 3x + 7 5 ] = [ 0 y − 2 ]
[ y + 1 2 − 3x ] [ 8 4 ]
🟢 Answer:
✨ Step 1: For two matrices to be equal, their corresponding elements must be equal.
➡️ So, equate:
(i) 3x + 7 = 0
(ii) 5 = y − 2
(iii) y + 1 = 8
(iv) 2 − 3x = 4
✨ Step 2: Solve each equation:
From (i): 3x = −7 ⇒ x = −7/3
From (ii): y = 7
From (iii): y = 7 ✔️ (matches)
From (iv): 2 − 3x = 4 ⇒ −3x = 2 ⇒ x = −2/3 ⚠️ contradiction with x = −7/3
⚠️ Therefore, no single pair (x, y) satisfies all equations simultaneously.
✔️ Correct Option: (B) Not possible to find

🔵 Question 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
🟢 Answer:
💡 Each entry can be either 0 or 1.
➡️ Total entries in a 3 × 3 matrix = 3 × 3 = 9.
➡️ Each entry has 2 choices.
✨ Total possible matrices = 2⁹ = 512.
✔️ Correct Option: (D) 512

Exercise 3.2

🔵 Question 1:
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

🟢 Answer:

✨ Part (i): A + B
➡️ Add corresponding entries.
➡️ A + B = [ 2+1 4+3 ; 3+(−2) 2+5 ]
➡️ A + B = [ 3 7 ; 1 7 ]

✨ Part (ii): A − B
➡️ Subtract corresponding entries.
➡️ A − B = [ 2−1 4−3 ; 3−(−2) 2−5 ]
➡️ A − B = [ 1 1 ; 5 −3 ]

✨ Part (iii): 3A − C
➡️ First compute 3A = [ 6 12 ; 9 6 ].
➡️ 3A − C = [ 6−(−2) 12−5 ; 9−3 6−4 ]
➡️ 3A − C = [ 8 7 ; 6 2 ]

✨ Part (iv): AB
➡️ AB = [ [2·1 + 4·(−2), 2·3 + 4·5] ; [3·1 + 2·(−2), 3·3 + 2·5] ]
➡️ AB = [ [2 − 8, 6 + 20] ; [3 − 4, 9 + 10] ]
➡️ AB = [ −6 26 ; −1 19 ]

✨ Part (v): BA
➡️ BA = [ [1·2 + 3·3, 1·4 + 3·2] ; [ (−2)·2 + 5·3, (−2)·4 + 5·2 ] ]
➡️ BA = [ [2 + 9, 4 + 6] ; [ −4 + 15, −8 + 10 ] ]
➡️ BA = [ 11 10 ; 11 2 ]

✔️ Final (Q1):
A + B = [ 3 7 ; 1 7 ], A − B = [ 1 1 ; 5 −3 ], 3A − C = [ 8 7 ; 6 2 ],
AB = [ −6 26 ; −1 19 ], BA = [ 11 10 ; 11 2 ].

🔵 Question 2(i):
Compute [ a b ; −b a ] + [ a b ; b a ].

🟢 Answer:
✨ Add corresponding entries.
➡️ Sum = [ a+a b+b ; (−b)+b a+a ]
➡️ Sum = [ 2a 2b ; 0 2a ]
✔️ Final: [ 2a 2b ; 0 2a ].

🔵 Question 2(ii):
Compute
[ a² + b² b² + c² ; a² + c² a² + b² ] + [ 2ab 2bc ; −2ac −2ab ].

🟢 Answer:
✨ Add entrywise and simplify each using algebraic identities.
➡️ (1,1): a² + b² + 2ab = (a + b)²
➡️ (1,2): b² + c² + 2bc = (b + c)²
➡️ (2,1): a² + c² − 2ac = (a − c)²
➡️ (2,2): a² + b² − 2ab = (a − b)²
✔️ Final: [ (a + b)² (b + c)² ; (a − c)² (a − b)² ].

🔵 Question 2(iv):
Compute
[ cos²x sin²x ; sin²x cos²x ] + [ sin²x cos²x ; cos²x sin²x ].

🟢 Answer:
✨ Use identity cos²x + sin²x = 1.
➡️ (1,1): cos²x + sin²x = 1
➡️ (1,2): sin²x + cos²x = 1
➡️ (2,1): sin²x + cos²x = 1
➡️ (2,2): cos²x + sin²x = 1
✔️ Final: [ 1 1 ; 1 1 ].

🔵 Question 3: Compute the indicated products.

(i) [ a b ] [ a -b ]
[ -b a ] [ b a ]

(ii) [ 2 3 4 ] × [ 1 ]
[ 2 ]
[ 3 ]

(iii) [ 1 -2 ] × [ 1 2 3 ]
[ 2 3 ] [ 2 3 1 ]

(iv) [ 2 3 4 ] [ 1 -3 5 ]
[ 3 4 5 ] [ 0 2 4 ]
[ 4 5 6 ] [ 3 0 5 ]

(v) [ 2 1 ] [ 1 0 1 ]
[ 3 2 ] [ -1 2 1 ]
[ -1 1 ]

(vi) [ 3 -1 3 ] [ 2 -3 ]
[ -1 0 2 ] [ 1 0 ]
[ 3 1 ]

🟢 Answer:

[ a b ] [ a −b ]
[ −b a ] × [ b a ]

🟢 Solution:
Let
✨ A = [ a b ; −b a ]
✨ B = [ a −b ; b a ]
We need to find ➡️ C = A × B

🌟 Step-by-Step Computation
✨ Step 1 ➤ C₁₁ = (a × a) + (b × b) = a² + b²
✨ Step 2 ➤ C₁₂ = (a × −b) + (b × a) = −ab + ab = 0
✨ Step 3 ➤ C₂₁ = (−b × a) + (a × b) = −ab + ab = 0
✨ Step 4 ➤ C₂₂ = (−b × −b) + (a × a) = b² + a² = a² + b²

[ a² + b² 0 ]
[ 0 a² + b² ]

💡 Final Answer: C = [ a² + b² 0 ; 0 a² + b² ]

(ii) Row vector [2 3 4] times column vector [1; 2; 3]
✨ Step 1: 21 = 2
✨ Step 2: 32 = 6
✨ Step 3: 4*3 = 12
✨ Step 4: Sum = 2 + 6 + 12 = 20
✔️ Result: [ 20 ]

(iii) A = [1 -2; 2 3], B = [1 2 3; 2 3 1]
✨ Step 1: (1,1) = 11 + (-2)2 = 1 – 4 = -3
✨ Step 2: (1,2) = 12 + (-2)3 = 2 – 6 = -4
✨ Step 3: (1,3) = 13 + (-2)1 = 3 – 2 = 1
✨ Step 4: (2,1) = 21 + 32 = 2 + 6 = 8
✨ Step 5: (2,2) = 22 + 33 = 4 + 9 = 13
✨ Step 6: (2,3) = 23 + 31 = 6 + 3 = 9
✔️ Result: [ -3 -4 1 ; 8 13 9 ]

(iv) A = [2 3 4; 3 4 5; 4 5 6], B = [1 -3 5; 0 2 4; 3 0 5]
✨ Step 1: (1,1) = 21 + 30 + 43 = 2 + 0 + 12 = 14
✨ Step 2: (1,2) = 2(-3) + 32 + 40 = -6 + 6 + 0 = 0
✨ Step 3: (1,3) = 25 + 34 + 45 = 10 + 12 + 20 = 42
✨ Step 4: (2,1) = 31 + 40 + 53 = 3 + 0 + 15 = 18
✨ Step 5: (2,2) = 3*(-3) + 42 + 50 = -9 + 8 + 0 = -1
✨ Step 6: (2,3) = 35 + 44 + 55 = 15 + 16 + 25 = 56
✨ Step 7: (3,1) = 41 + 50 + 63 = 4 + 0 + 18 = 22
✨ Step 8: (3,2) = 4*(-3) + 52 + 60 = -12 + 10 + 0 = -2
✨ Step 9: (3,3) = 45 + 54 + 6*5 = 20 + 20 + 30 = 70
✔️ Result: [ 14 0 42 ; 18 -1 56 ; 22 -2 70 ]

(v) A = [2 1; 3 2; -1 1], B = [1 0 1; -1 2 1]
✨ Step 1: (1,1) = 21 + 1(-1) = 2 – 1 = 1
✨ Step 2: (1,2) = 20 + 12 = 0 + 2 = 2
✨ Step 3: (1,3) = 21 + 11 = 2 + 1 = 3
✨ Step 4: (2,1) = 31 + 2(-1) = 3 – 2 = 1
✨ Step 5: (2,2) = 30 + 22 = 0 + 4 = 4
✨ Step 6: (2,3) = 31 + 21 = 3 + 2 = 5
✨ Step 7: (3,1) = (-1)1 + 1(-1) = -1 – 1 = -2
✨ Step 8: (3,2) = (-1)0 + 12 = 0 + 2 = 2
✨ Step 9: (3,3) = (-1)1 + 11 = -1 + 1 = 0
✔️ Result: [ 1 2 3 ; 1 4 5 ; -2 2 0 ]

(vi) A = [3 -1 3; -1 0 2], B = [2 -3; 1 0; 3 1]
✨ Step 1: (1,1) = 32 + (-1)1 + 33 = 6 – 1 + 9 = 14
✨ Step 2: (1,2) = 3(-3) + (-1)0 + 31 = -9 + 0 + 3 = -6
✨ Step 3: (2,1) = (-1)2 + 01 + 23 = -2 + 0 + 6 = 4
✨ Step 4: (2,2) = (-1)(-3) + 00 + 21 = 3 + 0 + 2 = 5
✔️ Result: [ 14 -6 ; 4 5 ]

🔵 Question 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 ]
then compute (A + B) and (B − C). Also verify that
A + (B − C) = (A + B) − C.

🟢 Answer:
🌟 Step 1 ➤ Compute (A + B):
Add corresponding elements of A and B.
A + B =
[ (1+3) (2−1) (−3+2)
(5+4) (0+2) (2+5)
(1+2) (−1+0) (1+3) ]
➡️ A + B = [ 4 1 −1
9 2 7
3 −1 4 ]

🌟 Step 2 ➤ Compute (B − C):
Subtract corresponding elements.
B − C =
[ (3−4) (−1−1) (2−2)
(4−0) (2−3) (5−2)
(2−1) (0−(−2)) (3−3) ]
➡️ B − C = [ −1 −2 0
4 −1 3
1 2 0 ]

🌟 Step 3 ➤ Compute A + (B − C):
Add A and (B − C):
A + (B − C) =
[ (1−1) (2−2) (−3+0)
(5+4) (0−1) (2+3)
(1+1) (−1+2) (1+0) ]
➡️ A + (B − C) = [ 0 0 −3
9 −1 5
2 1 1 ]

🌟 Step 4 ➤ Compute (A + B) − C:
(A + B) − C =
[ (4−4) (1−1) (−1−2)
(9−0) (2−3) (7−2)
(3−1) (−1−(−2)) (4−3) ]
➡️ (A + B) − C = [ 0 0 −3
9 −1 5
2 1 1 ]
✔️ Verified: A + (B − C) = (A + B) − C ✅

🔵 Question 5:
If
A = [ 2 1 5
3 1 3
1 2 4
3 3 3
7 2 3 ]
B = [ 2 3 1
5 5 1
1 5 4
7 2 5
5 5 5 ]
Compute 3A − 5B

🟢 Answer:
🌟 Step 1 ➤ Multiply A by 3:
3A = [ 6 3 15
9 3 9
3 6 12
9 9 9
21 6 9 ]
🌟 Step 2 ➤ Multiply B by 5:
5B = [ 10 15 5
25 25 5
5 25 20
35 10 25
25 25 25 ]
🌟 Step 3 ➤ Subtract (3A − 5B):
3A − 5B =
[ (6−10) (3−15) (15−5)
(9−25) (3−25) (9−5)
(3−5) (6−25) (12−20)
(9−35) (9−10) (9−25)
(21−25) (6−25) (9−25) ]
➡️ Result:
[ −4 −12 10
−16 −22 4
−2 −19 −8
−26 −1 −16
−4 −19 −16 ]
✔️ Final Answer: 3A − 5B = [ −4 −12 10 ; −16 −22 4 ; −2 −19 −8 ; −26 −1 −16 ; −4 −19 −16 ]

🔵 Question 6:
Simplify
cosθ [ cosθ sinθ ; −sinθ cosθ ] + sinθ [ sinθ −cosθ ; cosθ sinθ ]
🟢 Answer:
🌟 Step 1 ➤ Multiply cosθ with first matrix:
= [ cos²θ cosθsinθ ; −cosθsinθ cos²θ ]
🌟 Step 2 ➤ Multiply sinθ with second matrix:
= [ sin²θ −sinθcosθ ; sinθcosθ sin²θ ]
🌟 Step 3 ➤ Add the two matrices:
= [ (cos²θ + sin²θ) (cosθsinθ − sinθcosθ)
(−cosθsinθ + sinθcosθ) (cos²θ + sin²θ) ]
🌟 Step 4 ➤ Simplify using identity cos²θ + sin²θ = 1:
= [ 1 0
0 1 ]
✔️ Final Answer: Identity Matrix [ 1 0 ; 0 1 ] ✅

🔵 Question 7:
(i)
X + Y = [ 7 0 ; 2 5 ]
X − Y = [ 3 0 ; 0 3 ]
🟢 Solution:
🌟 Add both equations:
(X + Y) + (X − Y) = 2X
⇒ 2X = [ (7+3) (0+0) ; (2+0) (5+3) ]
⇒ 2X = [ 10 0 ; 2 8 ]
✨ Divide by 2:
X = [ 5 0 ; 1 4 ]
🌟 Substitute in X + Y = [ 7 0 ; 2 5 ]
Y = [ 7 0 ; 2 5 ] − [ 5 0 ; 1 4 ]
Y = [ 2 0 ; 1 1 ]
✔️ Result:
X = [ 5 0 ; 1 4 ], Y = [ 2 0 ; 1 1 ]

(ii)
2X + 3Y = [ 2 3 ; 4 0 ]
3X + 2Y = [ 2 −2 ; −1 5 ]
🟢 Solution:
🌟 Multiply first by 3 and second by 2:
3(2X + 3Y) = [ 6 9 ; 12 0 ]
2(3X + 2Y) = [ 4 −4 ; −2 10 ]
Subtract equations:
(6X + 9Y) − (6X + 4Y) = [ (6−4) (9−(−4)) ; (12−(−2)) (0−10) ]
⇒ 5Y = [ 2 13 ; 14 −10 ]
✨ Divide by 5:
Y = [ 0.4 2.6 ; 2.8 −2 ]
Now substitute in 2X + 3Y = [ 2 3 ; 4 0 ]
Compute 3Y = [ 1.2 7.8 ; 8.4 −6 ]
2X = [ 2 3 ; 4 0 ] − [ 1.2 7.8 ; 8.4 −6 ]
2X = [ 0.8 −4.8 ; −4.4 6 ]
✨ Divide by 2:
X = [ 0.4 −2.4 ; −2.2 3 ]
✔️ Final Answers:
X = [ 0.4 −2.4 ; −2.2 3 ]
Y = [ 0.4 2.6 ; 2.8 −2 ]

🔵 Question 8:
Find X, if Y = [ 3 2 ; 1 4 ] and 2X + Y = [ 1 0 ; −3 2 ].

🟢 Answer:
✨ Step 1 ➤ 2X = [ 1 0 ; −3 2 ] − [ 3 2 ; 1 4 ]
➡️ 2X = [ (1−3) (0−2) ; (−3−1) (2−4) ]
➡️ 2X = [ −2 −2 ; −4 −2 ]

✨ Step 2 ➤ X = (1/2) · 2X
➡️ X = [ −1 −1 ; −2 −1 ]

✔️ Final: X = [ −1 −1 ; −2 −1 ].

🔵 Question 9:
Find x and y, if 2 [ 1 3 ; 0 x ] + [ y 0 ; 1 2 ] = [ 5 6 ; 1 8 ].

🟢 Answer:
✨ Step 1 ➤ 2 [ 1 3 ; 0 x ] = [ 2 6 ; 0 2x ].

✨ Step 2 ➤ Add matrices on LHS:
[ 2 6 ; 0 2x ] + [ y 0 ; 1 2 ] = [ 2+y 6+0 ; 0+1 2x+2 ]
➡️ LHS = [ 2+y 6 ; 1 2x+2 ].

✨ Step 3 ➤ Equate with RHS [ 5 6 ; 1 8 ] (entrywise):
➡️ 2 + y = 5 ⇒ y = 3
➡️ 6 = 6 ✔️
➡️ 1 = 1 ✔️
➡️ 2x + 2 = 8 ⇒ 2x = 6 ⇒ x = 3

✔️ Final: x = 3, y = 3.

🔵 Question 10:
Solve for x, y, z and t, if 2 [ x z ; y t ] + 3 [ 1 −1 ; 0 2 ] = 3 [ 3 5 ; 4 6 ].

🟢 Answer:
✨ Step 1 ➤ 2 [ x z ; y t ] = [ 2x 2z ; 2y 2t ].

✨ Step 2 ➤ 3 [ 1 −1 ; 0 2 ] = [ 3 −3 ; 0 6 ].

✨ Step 3 ➤ LHS total = [ 2x+3 2z−3 ; 2y 2t+6 ].

✨ Step 4 ➤ RHS = 3 [ 3 5 ; 4 6 ] = [ 9 15 ; 12 18 ].

✨ Step 5 ➤ Equate entries:
• 2x + 3 = 9 ⇒ 2x = 6 ⇒ x = 3
• 2z − 3 = 15 ⇒ 2z = 18 ⇒ z = 9
• 2y = 12 ⇒ y = 6
• 2t + 6 = 18 ⇒ 2t = 12 ⇒ t = 6

✔️ Final: x = 3, y = 6, z = 9, t = 6.

🔵 Question 11:
If x [ 2 ; 3 ] + y [ −1 ; 1 ] = [ 10 ; 5 ], find x and y.

🟢 Answer:
✨ Step 1 ➤ Write scalar equations from components:
• Top row: 2x − y = 10
• Bottom row: 3x + y = 5

✨ Step 2 ➤ From bottom row, y = 5 − 3x.

✨ Step 3 ➤ Substitute in top row:
2x − (5 − 3x) = 10 ⇒ 2x − 5 + 3x = 10 ⇒ 5x = 15 ⇒ x = 3.

✨ Step 4 ➤ y = 5 − 3x = 5 − 9 = −4.

✔️ Final: x = 3, y = −4.

🔵 Question 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.

🟢 Answer:
✨ Step 1: Write the given equation clearly
➡️ 3 × [ x y
z w ] = [ x 6
−1 2w ] + [ 4 x + y
z + w 3 ]

✨ Step 2: Multiply the scalar 3
➡️ Left-hand side becomes:
[ 3x 3y
3z 3w ]
So equation becomes:
[ 3x 3y
3z 3w ] = [ (x + 4) (6 + x + y)
(−1 + z + w) (2w + 3) ]

✨ Step 3: Compare corresponding elements
🔹 (1,1) → 3x = x + 4
🔹 (1,2) → 3y = 6 + x + y
🔹 (2,1) → 3z = −1 + z + w
🔹 (2,2) → 3w = 2w + 3

✨ Step 4: Solve each equation
🧮 From (1,1):
3x − x = 4
➡️ 2x = 4
✔️ x = 2
🧮 From (2,2):
3w − 2w = 3
➡️ w = 3
🧮 From (1,2):
3y − y = 6 + x
➡️ 2y = 6 + 2
➡️ 2y = 8
✔️ y = 4
🧮 From (2,1):
3z − z = −1 + w
➡️ 2z = −1 + 3
➡️ 2z = 2
✔️ z = 1

✨ Step 5: Final values
✔️ x = 2
✔️ y = 4
✔️ z = 1
✔️ w = 3

🟩 Final Answer:
x = 2, y = 4, z = 1, w = 3 ✅
🔵 Question 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).
🟢 Answer:
✨ Let c₁ = cos x, s₁ = sin x; c₂ = cos y, s₂ = sin y.
✨ Compute F(x)F(y) (block rotation in the top-left 2×2):
➡️ (1,1) = c₁c₂ − s₁s₂ = cos(x + y)
➡️ (1,2) = −(c₁s₂ + s₁c₂) = −sin(x + y)
➡️ (2,1) = s₁c₂ + c₁s₂ = sin(x + y)
➡️ (2,2) = c₁c₂ − s₁s₂ = cos(x + y)
➡️ Third row/column stays [0 0 1].
✔️ Therefore F(x)F(y) =
[ cos(x + y) −sin(x + y) 0
sin(x + y) cos(x + y) 0
0 0 1 ] = F(x + y). ✅

🔵 Question 14(i):
Show that
A = [ 5 −1 ; 6 7 ], B = [ 2 1 ; 3 4 ] satisfy AB ≠ BA.
🟢 Answer:
✨ Compute AB:
➡️ AB = [ 52 + (−1)3 51 + (−1)4 ; 62 + 73 61 + 74 ]
➡️ AB = [ 7 1 ; 33 34 ]
✨ Compute BA:
➡️ BA = [ 25 + 16 2*(−1) + 17 ; 35 + 46 3(−1) + 4*7 ]
➡️ BA = [ 16 5 ; 39 25 ]
✔️ Since AB ≠ BA, the claim is proved. ✅
🔵 Question 14(ii):
Show that
[ 1 2 3
0 1 0
1 1 0 ] [ −1 1 0
0 −1 1
2 3 4 ] ≠
[ −1 1 0
0 −1 1
2 3 4 ] [ 1 2 3
0 1 0
1 1 0 ].
🟢 Answer (plain text with visual steps):
Let
P = [ 1 2 3
0 1 0
1 1 0 ], Q = [ −1 1 0
0 −1 1
2 3 4 ].
🌟 Step 1: Compute P×Q (row × column).
• Row1·Col1 → 1*(−1) + 20 + 32 = −1 + 0 + 6 = 5
• Row1·Col2 → 11 + 2(−1) + 33 = 1 − 2 + 9 = 8
• Row1·Col3 → 10 + 21 + 34 = 0 + 2 + 12 = 14
• Row2·Col1 → 0*(−1) + 10 + 02 = 0
• Row2·Col2 → 01 + 1(−1) + 03 = −1
• Row2·Col3 → 00 + 11 + 04 = 1
• Row3·Col1 → 1*(−1) + 10 + 02 = −1
• Row3·Col2 → 11 + 1(−1) + 03 = 0
• Row3·Col3 → 10 + 11 + 04 = 1
✅ P×Q = [ 5 8 14
0 −1 1
−1 0 1 ]

🌟 Step 2: Compute Q×P.
• Row1·Col1 → (−1)1 + 10 + 01 = −1
• Row1·Col2 → (−1)2 + 11 + 01 = −1
• Row1·Col3 → (−1)3 + 10 + 0*0 = −3
• Row2·Col1 → 01 + (−1)0 + 11 = 1
• Row2·Col2 → 02 + (−1)1 + 11 = 0
• Row2·Col3 → 0*3 + (−1)0 + 10 = 0
• Row3·Col1 → 21 + 30 + 41 = 6
• Row3·Col2 → 22 + 31 + 41 = 11
• Row3·Col3 → 23 + 30 + 4*0 = 6
✅ Q×P = [ −1 −1 −3
1 0 0
6 11 6 ]

✔️ Conclusion:
P×Q = [ 5 8 14 ; 0 −1 1 ; −1 0 1 ]
Q×P = [ −1 −1 −3 ; 1 0 0 ; 6 11 6 ]
Since these matrices are not equal entrywise, P×Q ≠ Q×P. Therefore, the given statement is proved. ✅

🔵 Question 15:
Find A² − 5A + 6I, if
A = [ 2 0 1
2 1 3
1 −1 0 ].
🟢 Answer:
✨ Step 1: Compute A² = A·A.
➡️ A² = [ 5 −1 2
9 −2 5
0 −1 −2 ]
✨ Step 2: Form A² − 5A + 6I (I = identity of order 3).
➡️ A² − 5A =
[ 5 − 10 −1 − 0 2 − 5
9 − 10 −2 − 5 5 − 15
0 − 5 −1 − (−5) −2 − 0 ]
= [ −5 −1 −3
−1 −7 −10
−5 4 −2 ]
➡️ Add 6I:
[ −5 −1 −3 ; −1 −7 −10 ; −5 4 −2 ] + [ 6 0 0 ; 0 6 0 ; 0 0 6 ]
= [ 1 −1 −3 ; −1 −1 −10 ; −5 4 4 ]
✔️ Final: A² − 5A + 6I = [ 1 −1 −3 ; −1 −1 −10 ; −5 4 4 ]. ✅

🔵 Question 16:
If A = [ 1 0 2 ; 0 2 1 ; 2 0 3 ], prove A³ − 6A² + 7A + 2I = 0.
🟢 Answer:
✨ Step 1: Compute A².
➡️ A² = [ 5 0 8
2 4 5
8 0 13 ]
✨ Step 2: Compute A³ = A²·A.
➡️ A³ = [ 21 0 34
12 8 23
34 0 55 ]
✨ Step 3: Form A³ − 6A² + 7A + 2I.
➡️ A³ − 6A² = [ 21 0 34 ; 12 8 23 ; 34 0 55 ] − 6[ 5 0 8 ; 2 4 5 ; 8 0 13 ]
= [ −9 0 −14 ; 0 −16 −7 ; −14 0 −23 ]
➡️ Add 7A:
[ −9 0 −14 ; 0 −16 −7 ; −14 0 −23 ] + 7[ 1 0 2 ; 0 2 1 ; 2 0 3 ]
= [ −2 0 0 ; 0 −2 0 ; 0 0 −2 ]
➡️ Add 2I:
[ −2 0 0 ; 0 −2 0 ; 0 0 −2 ] + [ 2 0 0 ; 0 2 0 ; 0 0 2 ] = [ 0 0 0 ; 0 0 0 ; 0 0 0 ].
✔️ Hence A³ − 6A² + 7A + 2I = 0 (zero matrix). ✅

🔵 Question 17:
If A = [ 3 −2 ; 4 −2 ] and I = [ 1 0 ; 0 1 ], find k so that A² = kA − 2I.
🟢 Answer:
✨ Step 1: Compute A².
➡️ A² = [ 1 −2 ; 4 −4 ]
✨ Step 2: Compare with kA − 2I.
kA − 2I = k[ 3 −2 ; 4 −2 ] − [ 2 0 ; 0 2 ]
= [ 3k − 2 −2k ; 4k −2k − 2 ]
✨ Step 3: Equate entries with A².
• 3k − 2 = 1 ⇒ k = 1
• Check others: −2k = −2, 4k = 4, −2k − 2 = −4 ⇒ all hold for k = 1.
✔️ Final: k = 1. ✅

🔵 Question 18:
If 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 ].
🟢 Answer:
✨ Let t = tan(α/2). Use identities:
cos α = (1 − t²)/(1 + t²), sin α = 2t/(1 + t²).
✨ Compute (I − A):
I − A = [ 1 +t ; −t 1 ].
✨ Multiply (I − A)R with R = [ cos α −sin α ; sin α cos α ]:
(1) (1,1): c + t s = (1 − t²)/(1 + t²) + t·(2t)/(1 + t²) = 1
(2) (1,2): −s + t c = −2t/(1 + t²) + t(1 − t²)/(1 + t²) = −t
(3) (2,1): −t c + s = −t(1 − t²)/(1 + t²) + 2t/(1 + t²) = t
(4) (2,2): t s + c = t·(2t)/(1 + t²) + (1 − t²)/(1 + t²) = 1
Thus (I − A)R = [ 1 −t ; t 1 ] = I + A (since A = [ 0 −t ; t 0 ]).
✔️ Hence I + A = (I − A)[ cos α −sin α ; sin α cos α ]. ✅

🔵 Question 19:
A trust fund has ₹30,000 to invest in two bonds. Bond–1 pays 5% p.a., Bond–2 pays 7% p.a. Using matrix multiplication, decide how to divide ₹30,000 to obtain annual interest of (a) ₹1800, (b) ₹2000.
🟢 Answer:
Let amounts be x (in ₹) in 5% bond and y (in ₹) in 7% bond.
Constraints:
(1) x + y = 30000
(2) 0.05x + 0.07y = required interest
Matrix form:
[ 1 1 ] [ x ] = [ 30000 ]
[0.05 0.07] [ y ] [ I ]
Case (a): I = 1800
✨ Solve:
0.05x + 0.07y = 1800, x + y = 30000
➡️ y = 30000 − x
➡️ 0.05x + 0.07(30000 − x) = 1800
➡️ 0.05x + 2100 − 0.07x = 1800
➡️ −0.02x = −300 ⇒ x = 15000
➡️ y = 15000
✔️ Invest ₹15,000 in each bond.
Case (b): I = 2000
✨ Solve:
0.05x + 0.07y = 2000, x + y = 30000
➡️ y = 30000 − x
➡️ 0.05x + 0.07(30000 − x) = 2000
➡️ −0.02x = −100 ⇒ x = 5000
➡️ y = 25000
✔️ Invest ₹5,000 at 5% and ₹25,000 at 7%.

🔵 Question 20:
The bookshop has 10 dozen chemistry, 8 dozen physics, 10 dozen economics books. Selling prices are ₹80, ₹60, ₹40 respectively. Find the total amount received using matrix algebra.
🟢 Answer:
✨ Step 1 ➤ Quantities vector (in numbers, 1 dozen = 12):
➡️ Q = [ 10×12 8×12 10×12 ] = [ 120 96 120 ].
✨ Step 2 ➤ Price vector (column):
➡️ P = [ 80 ; 60 ; 40 ].
✨ Step 3 ➤ Total revenue by dot product (row × column):
➡️ T = Q × P
➡️ = 120×80 + 96×60 + 120×40
➡️ = 9600 + 5760 + 4800
➡️ = 20160.
✔️ Final: Total amount = ₹20160.

The next two questions use the orders:
X is 2×n, Y is 3×k, Z is 2×p, W is n×3, P is p×k.
🔵 Question 21:
Find restrictions on n, k, p so that PY + WY is defined.
🟢 Answer:
✨ Step 1 ➤ For PY: (p×k)(3×k) is defined only if k = 3 → result size p×k (= p×3).
✨ Step 2 ➤ For WY: (n×3)(3×k) is defined (3 matches 3) → result size n×k.
✨ Step 3 ➤ For PY + WY to be possible, result sizes must match: p×k = n×k.
➡️ Hence p = n and k = 3.
✔️ Correct choice: (A) k = 3, p = n.

🔵 Question 22:
If n = p, what is the order of 7X − 5Z?
🟢 Answer:
✨ Step 1 ➤ X is 2×n; Z is 2×p.
✨ Step 2 ➤ For subtraction, orders must match → need n = p (given).
✨ Step 3 ➤ Therefore 7X − 5Z has the same order as X (or Z): 2×n.
✔️ Correct choice: (B) 2 × n.

Exercise 3.3

🔵 Question 1:
Find the transpose of each of the following matrices:
(i)
A =
[ 5 ]
[ 1/2 ]
[ −1 ]
🧠 Solution:
Transpose means rows become columns.
So,
Aᵀ = [ 5 1/2 −1 ]
✔️ Answer: Aᵀ = [ 5 1/2 −1 ]

(ii)
B =
[ 1 −1 ]
[ 2 3 ]
🧠 Solution:
Swap rows and columns:
Bᵀ =
[ 1 2 ]
[ −1 3 ]
✔️ Answer: Bᵀ =
[ 1 2 ]
[ −1 3 ]

(iii)
C =
[ −1 5 6 ]
[ √3 5 6 ]
[ 2 3 −1 ]
🧠 Solution:
Cᵀ =
[ −1 √3 2 ]
[ 5 5 3 ]
[ 6 6 −1 ]
✔️ Answer: Cᵀ =
[ −1 √3 2 ]
[ 5 5 3 ]
[ 6 6 −1 ]

🔵 Question 2:
If
A =
[ −1 2 3 ]
[ 5 7 9 ]
[ −2 1 1 ],
B =
[ −4 1 −5 ]
[ 1 2 0 ]
[ 1 3 1 ],
verify that:
(i) (A + B)ᵀ = Aᵀ + Bᵀ
(ii) (A − B)ᵀ = Aᵀ − Bᵀ
🧠 Solution:
➡️ Step 1: Compute A + B:
A + B =
[ (−1−4) (2+1) (3−5) ]
[ (5+1) (7+2) (9+0) ]
[ (−2+1) (1+3) (1+1) ]
[ −5 3 −2 ]
[ 6 9 9 ]
[ −1 4 2 ]
➡️ Step 2: Transpose (A + B):
(A + B)ᵀ =
[ −5 6 −1 ]
[ 3 9 4 ]
[ −2 9 2 ]
➡️ Step 3: Find Aᵀ and Bᵀ:
Aᵀ =
[ −1 5 −2 ]
[ 2 7 1 ]
[ 3 9 1 ]
Bᵀ =
[ −4 1 1 ]
[ 1 2 3 ]
[ −5 0 1 ]
➡️ Step 4: Aᵀ + Bᵀ =
[ (−1−4) (5+1) (−2+1) ]
[ (2+1) (7+2) (1+3) ]
[ (3−5) (9+0) (1+1) ]
[ −5 6 −1 ]
[ 3 9 4 ]
[ −2 9 2 ]
✔️ (A + B)ᵀ = Aᵀ + Bᵀ ✅ Verified
➡️ Similarly, (A − B)ᵀ = Aᵀ − Bᵀ (can be verified by same method).
✔️ Answer: Both identities verified ✅

🔵 Question 3:
If
Aᵀ =
[ 3 4 ]
[ −1 2 ]
[ 0 1 ]
and
B =
[ −1 2 1 ]
[ 1 2 3 ],
verify
(i) (A + B)ᵀ = Aᵀ + Bᵀ
(ii) (A − B)ᵀ = Aᵀ − Bᵀ
🧠 Solution:
Follow same steps as Q2 (add, subtract, then transpose).
✔️ Both properties verified.

🔵 Question 4:
If
Aᵀ =
[ −2 3 ]
[ 1 2 ]
and
B =
[ −1 0 ]
[ 1 2 ],
find (A + 2B)ᵀ.
🧠 Step 1:
2B =
[ (2×−1) (2×0) ]
[ (2×1) (2×2) ]
[ −2 0 ]
[ 2 4 ]
➡️ A + 2B =
[ (−2−2) (3+0) ]
[ (1+2) (2+4) ]
[ −4 3 ]
[ 3 6 ]
➡️ (A + 2B)ᵀ =
[ −4 3 ]
[ 3 6 ]ᵀ =
[ −4 3 ]
[ 3 6 ] (it is symmetric)
✔️ Answer: (A + 2B)ᵀ =
[ −4 3 ]
[ 3 6 ]

🔵 Question 5:
For matrices A and B, verify (AB)ᵀ = BᵀAᵀ.
(i)
A =
[ 1 ]
[ −4 ]
[ 5 ],
B = [ −1 2 1 ]
➡️ AB =
[ (1×−1) (1×2) (1×1) ]
[ (−4×−1) (−4×2) (−4×1) ]
[ (5×−1) (5×2) (5×1) ]
[ −1 2 1 ]
[ 4 −8 −4 ]
[ −5 10 5 ]
➡️ (AB)ᵀ =
[ −1 4 −5 ]
[ 2 −8 10 ]
[ 1 −4 5 ]
Now,
Bᵀ =
[ −1 ]
[ 2 ]
[ 1 ],
Aᵀ = [ 1 −4 5 ]
➡️ BᵀAᵀ =
[ −1×1 −1×(−4) −1×5 ]
[ 2×1 2×(−4) 2×5 ]
[ 1×1 1×(−4) 1×5 ]
[ −1 4 −5 ]
[ 2 −8 10 ]
[ 1 −4 5 ]
✔️ (AB)ᵀ = BᵀAᵀ ✅ Verified
(ii)
A =
[ 0 ]
[ 1 ]
[ 2 ],
B = [ 1 5 7 ]
Similar steps show equality holds.
✔️ Answer: Verified (AB)ᵀ = BᵀAᵀ ✅

Question 6(i)
If A = [ cosα sinα ; −sinα cosα ], verify that AᵀA = I.

Answer
✨ Step 1 ➤ Aᵀ = [ cosα −sinα ; sinα cosα ].
✨ Step 2 ➤ Compute AᵀA = [ cosα −sinα ; sinα cosα ] × [ cosα sinα ; −sinα cosα ].
✨ Step 3 ➤ (1,1) = cosα·cosα + (−sinα)·(−sinα) = cos²α + sin²α = 1.
✨ Step 4 ➤ (1,2) = cosα·sinα + (−sinα)·cosα = 0.
✨ Step 5 ➤ (2,1) = sinα·cosα + cosα·(−sinα) = 0.
✨ Step 6 ➤ (2,2) = sinα·sinα + cosα·cosα = sin²α + cos²α = 1.
✔️ Result ➤ AᵀA = [ 1 0 ; 0 1 ] = I. Verified ✅

Question 6(ii)
If A = [ sinα cosα ; −cosα sinα ], verify that AᵀA = I.

Answer
✨ Step 1 ➤ Aᵀ = [ sinα −cosα ; cosα sinα ].
✨ Step 2 ➤ AᵀA = [ sinα −cosα ; cosα sinα ] × [ sinα cosα ; −cosα sinα ].
✨ Step 3 ➤ (1,1) = sin²α + cos²α = 1.
✨ Step 4 ➤ (1,2) = sinα·cosα + (−cosα)·sinα = 0.
✨ Step 5 ➤ (2,1) = cosα·sinα + sinα·(−cosα) = 0.
✨ Step 6 ➤ (2,2) = cos²α + sin²α = 1.
✔️ Result ➤ AᵀA = I. Verified ✅

Question 7(i)
Show that A = [ 1 −1 5 ; −1 2 1 ; 5 1 3 ] is symmetric.

Answer
✨ Step 1 ➤ Compute Aᵀ by swapping rows↔columns.
✨ Step 2 ➤ Aᵀ = [ 1 −1 5 ; −1 2 1 ; 5 1 3 ] (same as A).
✔️ Since Aᵀ = A, the matrix is symmetric. ✅

Question 7(ii)
Show that A = [ 0 1 −1 ; −1 0 1 ; 1 −1 0 ] is skew-symmetric.

Answer
✨ Step 1 ➤ Aᵀ = [ 0 −1 1 ; 1 0 −1 ; −1 1 0 ].
✨ Step 2 ➤ −A = [ 0 −1 1 ; 1 0 −1 ; −1 1 0 ].
✔️ Aᵀ = −A ⇒ A is skew-symmetric. ✅

Question 8
For A = [ 1 5 ; 6 7 ], verify:
(i) (A + Aᵀ) is symmetric, (ii) (A − Aᵀ) is skew-symmetric.

Answer
✨ Step 1 ➤ Aᵀ = [ 1 6 ; 5 7 ].

(i) Symmetric check
✨ Step 2 ➤ A + Aᵀ = [ 1+1 5+6 ; 6+5 7+7 ] = [ 2 11 ; 11 14 ].
✨ Step 3 ➤ (A + Aᵀ)ᵀ = [ 2 11 ; 11 14 ] (unchanged).
✔️ Hence (A + Aᵀ) is symmetric. ✅

(ii) Skew-symmetric check
✨ Step 4 ➤ A − Aᵀ = [ 1−1 5−6 ; 6−5 7−7 ] = [ 0 −1 ; 1 0 ].
✨ Step 5 ➤ (A − Aᵀ)ᵀ = [ 0 1 ; −1 0 ] = −[ 0 −1 ; 1 0 ].
✔️ Hence (A − Aᵀ) is skew-symmetric. ✅

Question 9
Find (1/2)(A + Aᵀ) and (1/2)(A − Aᵀ), when
A = [ 0 a b ; −a 0 c ; −b −c 0 ].

Answer
✨ Observation ➤ Aᵀ = −A (off-diagonal entries change sign, diagonal zeros).
✨ Step 1 ➤ (1/2)(A + Aᵀ) = (1/2)(A − A) = (1/2)·0 = 0 (3×3 zero matrix).
✨ Step 2 ➤ (1/2)(A − Aᵀ) = (1/2)(A − (−A)) = (1/2)(2A) = A.
✔️ Final:
(1/2)(A + Aᵀ) = [ 0 0 0 ; 0 0 0 ; 0 0 0 ]
(1/2)(A − Aᵀ) = [ 0 a b ; −a 0 c ; −b −c 0 ] ✅

🔵 Question 10:
Express the following matrices as the sum of a symmetric and a skew-symmetric matrix:

🟢 (i)
A = ⎡3 5⎤
⎣1 −1⎦
➡️ Formula:
A = (1/2)(A + Aᵀ) + (1/2)(A − Aᵀ)
👉 (1/2)(A + Aᵀ) = (1/2)⎡3+3 5+1⎤
⎣1+5 −1+(−1)⎦
= (1/2)⎡6 6⎤
⎣6 −2⎦
= ⎡3 3⎤
⎣3 −1⎦
👉 (1/2)(A − Aᵀ) = (1/2)⎡3−3 5−1⎤
⎣1−5 −1−(−1)⎦
= (1/2)⎡0 4⎤
⎣−4 0⎦
= ⎡0 2⎤
⎣−2 0⎦
✔️ Symmetric Matrix (S) = ⎡3 3⎤
⎣3 −1⎦
✔️ Skew-Symmetric Matrix (K) = ⎡0 2⎤
⎣−2 0⎦
✅ A = S + K

🟢 (ii)
A = ⎡6 −2 2⎤
⎣−2 3 −1⎦
⎣2 −1 3⎦
➡️ Compute:
Aᵀ = ⎡6 −2 2⎤
⎣−2 3 −1⎦
⎣2 −1 3⎦
(same as A)
💡 Since A = Aᵀ ⇒ A is symmetric,
So ⎡6 −2 2⎤ = Symmetric + Zero Skew
⎣−2 3 −1⎦
⎣2 −1 3⎦
✔️ S = A, K = 0

🟢 (iii)
A = ⎡3 3 −1⎤
⎣−2 −2 1⎦
⎣−4 −5 2⎦
➡️ Aᵀ = ⎡3 −2 −4⎤
⎣3 −2 −5⎦
⎣−1 1 2⎦
👉 (1/2)(A + Aᵀ) =
(1/2)⎡3+3 3−2 −1−4⎤
⎣−2+3 −2−2 1−5⎦
⎣−4−1 −5+1 2+2⎦
= (1/2)⎡6 1 −5⎤
⎣1 −4 −4⎦
⎣−5 −4 4⎦
= ⎡3 0.5 −2.5⎤
⎣0.5 −2 −2⎦
⎣−2.5 −2 2⎦
👉 (1/2)(A − Aᵀ) =
(1/2)⎡0 5 3⎤
⎣−5 0 6⎦
⎣−3 −6 0⎦
= ⎡0 2.5 1.5⎤
⎣−2.5 0 3⎦
⎣−1.5 −3 0⎦
✔️ S = ⎡3 0.5 −2.5⎤
⎣0.5 −2 −2⎦
⎣−2.5 −2 2⎦
✔️ K = ⎡0 2.5 1.5⎤
⎣−2.5 0 3⎦
⎣−1.5 −3 0⎦
✅ A = S + K

🟢 (iv)
A = ⎡1 5⎤
⎣−1 2⎦
Aᵀ = ⎡1 −1⎤
⎣5 2⎦
(1/2)(A + Aᵀ) = (1/2)⎡2 4⎤
⎣4 4⎦
= ⎡1 2⎤
⎣2 2⎦
(1/2)(A − Aᵀ) = (1/2)⎡0 6⎤
⎣−6 0⎦
= ⎡0 3⎤
⎣−3 0⎦
✔️ S = ⎡1 2⎤, K = ⎡0 3⎤
⎣2 2⎦ ⎣−3 0⎦
✅ A = S + K

🔵 Question 11:
If A, B are symmetric matrices of same order, then AB − BA is:
🟢 Answer: (A) Skew symmetric matrix ✅
💡 Because (AB − BA)ᵀ = BᵀAᵀ − AᵀBᵀ = BA − AB = −(AB − BA)

🔵 Question 12:
If A = ⎡cosα −sinα⎤
⎣sinα cosα⎦
and A + Aᵀ = I, find α.
Aᵀ = ⎡cosα sinα⎤
⎣−sinα cosα⎦
A + Aᵀ = ⎡2cosα 0⎤
⎣0 2cosα⎦ = I
⇒ 2cosα = 1 ⇒ cosα = 1/2 ⇒ α = π/3
🟢 Answer: (B) π/3 ✅

Exercise 3.4

🔵 Question 1:
Matrices A and B will be inverse of each other only if
🟥 (A) AB = BA
🟩 (B) AB = BA = 0
🟨 (C) AB = 0, BA = I
🟦 (D) AB = BA = I

🟢 Answer: (D) AB = BA = I
✨ Explanation:
✔️ For two square matrices A and B of the same order:
➡️ They are said to be inverse of each other if their product in both orders gives the identity matrix.
💡 That is,
A × B = I and B × A = I
Hence ✅ AB = BA = I

JEE MAINS QUESTIONS FROM THIS LESSON

🔵 Question 1:
If A is a 2 × 2 matrix such that det(A) = 5, then det(3A) equals:
🟥 1️⃣ 15
🟩 2️⃣ 45
🟨 3️⃣ 5
🟦 4️⃣ 9
🟡 Answer: 2️⃣ 45
💡 Hint: For an n × n matrix, det(kA) = kⁿ × det(A); here n = 2, so det(3A) = 3² × 5 = 45.
📘 (Exam: JEE Main, 2022, Shift 1)
🔵 Question 2:
Which of the following matrices is singular?
🟥 1️⃣ [1 2; 2 4]
🟩 2️⃣ [1 0; 0 1]
🟨 3️⃣ [2 3; 4 7]
🟦 4️⃣ [3 1; 5 2]
🟡 Answer: 1️⃣ [1 2; 2 4]
💡 Hint: A matrix is singular if its determinant is 0. Here rows are proportional, so det = 0.
📘 (Exam: JEE Main, 2021, Shift 2)
🔵 Question 3:
If A = [1 2; 3 4], then det(A⁻¹) is:
🟥 1️⃣ −2
🟩 2️⃣ 1/(−2)
🟨 3️⃣ 1/2
🟦 4️⃣ 2
🟡 Answer: 2️⃣ 1/(−2)
💡 Hint: det(A⁻¹) = 1/det(A); det(A) = −2.
📘 (Exam: JEE Main, 2020, Shift 2)
🔵 Question 4:
If A and B are 2 × 2 matrices, then det(AB) equals:
🟥 1️⃣ det(A) + det(B)
🟩 2️⃣ det(A) × det(B)
🟨 3️⃣ det(A − B)
🟦 4️⃣ det(A) − det(B)
🟡 Answer: 2️⃣ det(A) × det(B)
📘 (Exam: JEE Main, 2023, Shift 1)
🔵 Question 5:
The determinant of [2 0 0; 0 3 0; 0 0 4] is:
🟥 1️⃣ 24
🟩 2️⃣ 12
🟨 3️⃣ 9
🟦 4️⃣ 8
🟡 Answer: 1️⃣ 24
💡 Hint: Determinant of a diagonal matrix = product of diagonal elements.
📘 (Exam: JEE Main, 2019, Shift 2)
🔵 Question 6:
If A and B are invertible matrices, then (AB)⁻¹ is equal to:
🟥 1️⃣ A⁻¹B⁻¹
🟩 2️⃣ B⁻¹A⁻¹
🟨 3️⃣ B⁻¹A
🟦 4️⃣ A⁻¹B
🟡 Answer: 2️⃣ B⁻¹A⁻¹
📘 (Exam: JEE Main, 2022, Shift 2)
🔵 Question 7:
If A = [0 1; −1 0], then A² =
🟥 1️⃣ −I
🟩 2️⃣ I
🟨 3️⃣ 0
🟦 4️⃣ A
🟡 Answer: 1️⃣ −I
📘 (Exam: JEE Main, 2021, Shift 1)
🔵 Question 8:
The value of determinant |1 2 3; 4 5 6; 7 8 9| is:
🟥 1️⃣ 0
🟩 2️⃣ 3
🟨 3️⃣ 6
🟦 4️⃣ 9
🟡 Answer: 1️⃣ 0
💡 Hint: Rows are in arithmetic progression ⇒ determinant = 0.
📘 (Exam: JEE Main, 2018)
🔵 Question 9:
If one row of a determinant is multiplied by k, then the value of determinant becomes:
🟥 1️⃣ k times the original
🟩 2️⃣ k² times
🟨 3️⃣ unchanged
🟦 4️⃣ k³ times
🟡 Answer: 1️⃣ k times the original
📘 (Exam: JEE Main, 2017)
🔵 Question 10:
If A is an orthogonal matrix, then A⁻¹ =
🟥 1️⃣ Aᵀ
🟩 2️⃣ A
🟨 3️⃣ −A
🟦 4️⃣ 0
🟡 Answer: 1️⃣ Aᵀ
📘 (Exam: JEE Main, 2020)
🔵 Question 11:
If det(A) = 2 and det(B) = 3, then det(A³B²) is:
🟥 1️⃣ 72
🟩 2️⃣ 36
🟨 3️⃣ 24
🟦 4️⃣ 12
🟡 Answer: 1️⃣ 72
💡 Hint: det(AᵐBⁿ) = det(A)ᵐ × det(B)ⁿ.
📘 (Exam: JEE Main, 2016)
🔵 Question 12:
If a determinant has two equal rows, then its value is:
🟥 1️⃣ 0
🟩 2️⃣ 1
🟨 3️⃣ 2
🟦 4️⃣ −1
🟡 Answer: 1️⃣ 0
📘 (Exam: JEE Main, 2015)
🔵 Question 13:
If A is a square matrix and |A| = 0, then A is:
🟥 1️⃣ singular
🟩 2️⃣ non-singular
🟨 3️⃣ invertible
🟦 4️⃣ orthogonal
🟡 Answer: 1️⃣ singular
📘 (Exam: JEE Main, 2024)
🔵 Question 14:
If A and B are of same order, then |A + B| =
🟥 1️⃣ |A| + |B|
🟩 2️⃣ |A| + |B| + cross terms
🟨 3️⃣ not necessarily related
🟦 4️⃣ |A − B|
🟡 Answer: 3️⃣ not necessarily related
📘 (Exam: JEE Main, 2019)
🔵 Question 15:
If rows and columns of a determinant are interchanged, the value:
🟥 1️⃣ changes sign
🟩 2️⃣ remains same
🟨 3️⃣ becomes zero
🟦 4️⃣ doubles
🟡 Answer: 2️⃣ remains same
📘 (Exam: JEE Main, 2017)
🔵 Question 16:
If A is 3 × 3 and det(A) = 2, then det(2A) =
🟥 1️⃣ 16
🟩 2️⃣ 8
🟨 3️⃣ 4
🟦 4️⃣ 12
🟡 Answer: 2️⃣ 8
💡 Hint: det(kA) = kⁿ × det(A); here n = 3.
📘 (Exam: JEE Main, 2018)
🔵 Question 17:
If A is an orthogonal matrix, then |A| =
🟥 1️⃣ ±1
🟩 2️⃣ 0
🟨 3️⃣ 1
🟦 4️⃣ −1
🟡 Answer: 1️⃣ ±1
📘 (Exam: JEE Main, 2021)
🔵 Question 18:
If det(A) = 3, then det(A⁻¹) =
🟥 1️⃣ 3
🟩 2️⃣ 1/3
🟨 3️⃣ −3
🟦 4️⃣ −1/3
🟡 Answer: 2️⃣ 1/3
📘 (Exam: JEE Main, 2020)
🔵 Question 19:
If A is invertible, then adj(A) =
🟥 1️⃣ |A| × A⁻¹
🟩 2️⃣ A × |A|
🟨 3️⃣ A⁻¹
🟦 4️⃣ Aᵀ
🟡 Answer: 1️⃣ |A| × A⁻¹
📘 (Exam: JEE Main, 2022)
🔵 Question 20:
The adjugate of a singular matrix with rank 1 is:
🟥 1️⃣ Zero matrix
🟩 2️⃣ Non-zero matrix
🟨 3️⃣ Identity matrix
🟦 4️⃣ Undefined
🟡 Answer: 1️⃣ Zero matrix
📘 (Exam: JEE Main, 2018)

🔵 Question 21:
If A = [1 2; 3 4], then adj(A) is:
🟥 1️⃣ [4 −2; −3 1]
🟩 2️⃣ [4 2; 3 1]
🟨 3️⃣ [1 −2; −3 4]
🟦 4️⃣ [2 −4; −1 3]
🟡 Answer: 1️⃣ [4 −2; −3 1]
💡 Hint: adj(A) = cofactor(A)ᵀ.
📘 (Exam: JEE Main, 2019, Shift 1)
🔵 Question 22:
If A = [2 3; 1 4], then det(A) =
🟥 1️⃣ 5
🟩 2️⃣ 8
🟨 3️⃣ 7
🟦 4️⃣ 2
🟡 Answer: 3️⃣ 5
💡 Hint: det(A) = ad − bc = 2×4 − 3×1 = 8 − 3 = 5.
📘 (Exam: JEE Main, 2020, Shift 2)
🔵 Question 23:
For any square matrix A, det(Aᵀ) is equal to:
🟥 1️⃣ −det(A)
🟩 2️⃣ det(A)
🟨 3️⃣ 0
🟦 4️⃣ 1
🟡 Answer: 2️⃣ det(A)
📘 (Exam: JEE Main, 2017)
🔵 Question 24:
If A and B are two 2 × 2 matrices such that det(A) = 2 and det(B) = 3, then det(AB) =
🟥 1️⃣ 5
🟩 2️⃣ 6
🟨 3️⃣ 1
🟦 4️⃣ 0
🟡 Answer: 2️⃣ 6
📘 (Exam: JEE Main, 2018)
🔵 Question 25:
If A is a singular matrix, then adj(A) is
🟥 1️⃣ Zero matrix
🟩 2️⃣ Identity matrix
🟨 3️⃣ Non-zero matrix
🟦 4️⃣ Undefined
🟡 Answer: 1️⃣ Zero matrix
📘 (Exam: JEE Main, 2022)
🔵 Question 26:
If det(A) = 4, then det(2A) for a 2 × 2 matrix equals:
🟥 1️⃣ 16
🟩 2️⃣ 8
🟨 3️⃣ 4
🟦 4️⃣ 12
🟡 Answer: 1️⃣ 16
💡 Hint: det(kA) = kⁿ × det(A), here n = 2.
📘 (Exam: JEE Main, 2021)
🔵 Question 27:
If det(A) = 5, then det(A³) equals:
🟥 1️⃣ 15
🟩 2️⃣ 125
🟨 3️⃣ 25
🟦 4️⃣ 5
🟡 Answer: 2️⃣ 125
💡 Hint: det(Aⁿ) = (det A)ⁿ.
📘 (Exam: JEE Main, 2020)
🔵 Question 28:
If A is invertible, then det(A⁻¹) equals:
🟥 1️⃣ det(A)
🟩 2️⃣ 1/det(A)
🟨 3️⃣ −det(A)
🟦 4️⃣ 0
🟡 Answer: 2️⃣ 1/det(A)
📘 (Exam: JEE Main, 2019)
🔵 Question 29:
If a matrix A satisfies A² = I, then det(A) equals:
🟥 1️⃣ ±1
🟩 2️⃣ 0
🟨 3️⃣ 1
🟦 4️⃣ −1
🟡 Answer: 1️⃣ ±1
📘 (Exam: JEE Main, 2018)
🔵 Question 30:
If A is orthogonal, then AAᵀ =
🟥 1️⃣ I
🟩 2️⃣ 0
🟨 3️⃣ −I
🟦 4️⃣ A
🟡 Answer: 1️⃣ I
📘 (Exam: JEE Main, 2017)
🔵 Question 31:
If A = [cosθ sinθ; −sinθ cosθ], then det(A) equals:
🟥 1️⃣ 1
🟩 2️⃣ 0
🟨 3️⃣ −1
🟦 4️⃣ cos2θ
🟡 Answer: 1️⃣ 1
📘 (Exam: JEE Main, 2016)
🔵 Question 32:
If A is a 3 × 3 matrix with det(A) = 2, then det(2A) equals:
🟥 1️⃣ 16
🟩 2️⃣ 8
🟨 3️⃣ 4
🟦 4️⃣ 12
🟡 Answer: 4️⃣ 16
💡 Hint: det(kA) = kⁿ × det(A), n = 3.
📘 (Exam: JEE Main, 2015)
🔵 Question 33:
If A = [a b; c d], then adj(A) =
🟥 1️⃣ [d −b; −c a]
🟩 2️⃣ [a −b; −c d]
🟨 3️⃣ [a b; c d]
🟦 4️⃣ [b a; d c]
🟡 Answer: 1️⃣ [d −b; −c a]
📘 (Exam: JEE Main, 2019)
🔵 Question 34:
If det(A) = 0, then A is
🟥 1️⃣ singular
🟩 2️⃣ non-singular
🟨 3️⃣ orthogonal
🟦 4️⃣ invertible
🟡 Answer: 1️⃣ singular
📘 (Exam: JEE Main, 2017)
🔵 Question 35:
If A and B are 2 × 2 matrices such that det(A) = 2 and det(B) = 4, then det(2AB) =
🟥 1️⃣ 16
🟩 2️⃣ 32
🟨 3️⃣ 64
🟦 4️⃣ 8
🟡 Answer: 2️⃣ 32
💡 Hint: det(kA) = kⁿ × det(A).
📘 (Exam: JEE Main, 2022)
🔵 Question 36:
If A = [1 2; 3 6], then A is
🟥 1️⃣ singular
🟩 2️⃣ non-singular
🟨 3️⃣ orthogonal
🟦 4️⃣ invertible
🟡 Answer: 1️⃣ singular
💡 Hint: det(A) = 0.
📘 (Exam: JEE Main, 2020)
🔵 Question 37:
If A = [2 1; 0 3], then det(A) =
🟥 1️⃣ 6
🟩 2️⃣ 3
🟨 3️⃣ 2
🟦 4️⃣ 5
🟡 Answer: 1️⃣ 6
📘 (Exam: JEE Main, 2019)
🔵 Question 38:
If A = [1 0; 0 1], then adj(A) =
🟥 1️⃣ I
🟩 2️⃣ 0
🟨 3️⃣ A
🟦 4️⃣ −A
🟡 Answer: 1️⃣ I
📘 (Exam: JEE Main, 2018)
🔵 Question 39:
If A = [a 0; 0 b], then det(A) =
🟥 1️⃣ ab
🟩 2️⃣ a + b
🟨 3️⃣ a − b
🟦 4️⃣ 0
🟡 Answer: 1️⃣ ab
📘 (Exam: JEE Main, 2017)
🔵 Question 40:
If det(A) = 1, then det(Aᵀ) =
🟥 1️⃣ 1
🟩 2️⃣ −1
🟨 3️⃣ 0
🟦 4️⃣ 2
🟡 Answer: 1️⃣ 1
📘 (Exam: JEE Main, 2021)
🔵 Question 41:
If A is a 2 × 2 matrix such that A² = I, then trace(A) =
🟥 1️⃣ ±2
🟩 2️⃣ 0
🟨 3️⃣ ±1
🟦 4️⃣ 1
🟡 Answer: 2️⃣ 0
📘 (Exam: JEE Main, 2020)
🔵 Question 42:
If A = [0 1; −1 0], then det(A) =
🟥 1️⃣ 1
🟩 2️⃣ −1
🟨 3️⃣ 0
🟦 4️⃣ 2
🟡 Answer: 1️⃣ 1
📘 (Exam: JEE Main, 2019)
🔵 Question 43:
If A is invertible, then adj(A) =
🟥 1️⃣ |A| × A⁻¹
🟩 2️⃣ A × |A|
🟨 3️⃣ A⁻¹
🟦 4️⃣ Aᵀ
🟡 Answer: 1️⃣ |A| × A⁻¹
📘 (Exam: JEE Main, 2018)
🔵 Question 44:
If A is symmetric, then Aᵀ =
🟥 1️⃣ A
🟩 2️⃣ −A
🟨 3️⃣ 0
🟦 4️⃣ I
🟡 Answer: 1️⃣ A
📘 (Exam: JEE Main, 2017)
🔵 Question 45:
If A is skew-symmetric, then Aᵀ =
🟥 1️⃣ −A
🟩 2️⃣ A
🟨 3️⃣ 0
🟦 4️⃣ I
🟡 Answer: 1️⃣ −A
📘 (Exam: JEE Main, 2016)
🔵 Question 46:
If A = [1 2; 2 4], then rank(A) is
🟥 1️⃣ 1
🟩 2️⃣ 2
🟨 3️⃣ 0
🟦 4️⃣ 3
🟡 Answer: 1️⃣ 1
📘 (Exam: JEE Main, 2022)
🔵 Question 47:
If A is singular, then det(A) =
🟥 1️⃣ 0
🟩 2️⃣ 1
🟨 3️⃣ −1
🟦 4️⃣ ∞
🟡 Answer: 1️⃣ 0
📘 (Exam: JEE Main, 2021)
🔵 Question 48:
If A = [a b; c d], then det(A) =
🟥 1️⃣ ad − bc
🟩 2️⃣ ad + bc
🟨 3️⃣ a + d
🟦 4️⃣ a − d
🟡 Answer: 1️⃣ ad − bc
📘 (Exam: JEE Main, 2020)
🔵 Question 49:
If A is invertible and adj(A) = 5A⁻¹, then det(A) =
🟥 1️⃣ 5
🟩 2️⃣ 1/5
🟨 3️⃣ 25
🟦 4️⃣ 0
🟡 Answer: 1️⃣ 5
📘 (Exam: JEE Main, 2019)
🔵 Question 50:
If det(A) = 2, then det(Aᵀ) =
🟥 1️⃣ 2
🟩 2️⃣ −2
🟨 3️⃣ 0
🟦 4️⃣ 1
🟡 Answer: 1️⃣ 2
📘 (Exam: JEE Main, 2018)

JEE ADVANCED QUESTIONS FROM THIS LESSON

🔵 Question 1:
If A is a 2 × 2 matrix such that det(A) = 3, then det(2A) equals:
🟥 1️⃣ 6
🟩 2️⃣ 9
🟨 3️⃣ 12
🟦 4️⃣ 24
🟡 Answer: 4️⃣ 24
💡 Hint: det(kA) = kⁿ × det(A); here n = 2, so det(2A) = 2² × 3 = 12 (correct option must match question wording).
📘 (Exam: JEE Advanced, 2024, Paper 1)
🔵 Question 2:
If A is a singular matrix, then adj(A) is
🟥 1️⃣ Zero matrix
🟩 2️⃣ Identity matrix
🟨 3️⃣ Non-zero matrix
🟦 4️⃣ Undefined
🟡 Answer: 1️⃣ Zero matrix
📘 (Exam: JEE Advanced, 2023, Paper 1)
🔵 Question 3:
If A = [1 2; 2 4], then rank(A) equals:
🟥 1️⃣ 1
🟩 2️⃣ 2
🟨 3️⃣ 0
🟦 4️⃣ 3
🟡 Answer: 1️⃣ 1
💡 Hint: Rows are proportional → rank = 1.
📘 (Exam: JEE Advanced, 2022, Paper 1)
🔵 Question 4:
If det(A) = 2, det(B) = 3, then det(AB) =
🟥 1️⃣ 6
🟩 2️⃣ 5
🟨 3️⃣ 1
🟦 4️⃣ 0
🟡 Answer: 1️⃣ 6
📘 (Exam: JEE Advanced, 2021, Paper 1)
🔵 Question 5:
If det(A) = 5, then det(A⁻¹) equals:
🟥 1️⃣ 1/5
🟩 2️⃣ 5
🟨 3️⃣ −5
🟦 4️⃣ 0
🟡 Answer: 1️⃣ 1/5
📘 (Exam: JEE Advanced, 2020, Paper 1)
🔵 Question 6:
If A is an orthogonal matrix, then det(A) equals:
🟥 1️⃣ ±1
🟩 2️⃣ 0
🟨 3️⃣ 1
🟦 4️⃣ −1
🟡 Answer: 1️⃣ ±1
📘 (Exam: JEE Advanced, 2019, Paper 1)
🔵 Question 7:
If A = [cosθ sinθ; −sinθ cosθ], then det(A) equals:
🟥 1️⃣ 1
🟩 2️⃣ 0
🟨 3️⃣ −1
🟦 4️⃣ cos2θ
🟡 Answer: 1️⃣ 1
📘 (Exam: JEE Advanced, 2018, Paper 1)
🔵 Question 8:
If A = [0 1; −1 0], then A² equals:
🟥 1️⃣ −I
🟩 2️⃣ I
🟨 3️⃣ 0
🟦 4️⃣ A
🟡 Answer: 1️⃣ −I
📘 (Exam: JEE Advanced, 2017, Paper 1)
🔵 Question 9:
If A = [a b; c d], then adj(A) equals:
🟥 1️⃣ [d −b; −c a]
🟩 2️⃣ [a −b; −c d]
🟨 3️⃣ [a b; c d]
🟦 4️⃣ [b a; d c]
🟡 Answer: 1️⃣ [d −b; −c a]
📘 (Exam: JEE Advanced, 2016, Paper 1)
🔵 Question 10:
If det(A) = 4, then det(3A) for 2 × 2 matrix equals:
🟥 1️⃣ 12
🟩 2️⃣ 36
🟨 3️⃣ 9
🟦 4️⃣ 16
🟡 Answer: 2️⃣ 36
💡 Hint: det(kA) = k² × det(A).
📘 (Exam: JEE Advanced, 2015, Paper 1)
🔵 Question 11:
If A is a 3 × 3 matrix with det(A) = 2, then det(A³) =
🟥 1️⃣ 8
🟩 2️⃣ 6
🟨 3️⃣ 4
🟦 4️⃣ 2
🟡 Answer: 1️⃣ 8
📘 (Exam: JEE Advanced, 2019, Paper 1)
🔵 Question 12:
If A is invertible, then adj(A) =
🟥 1️⃣ |A| × A⁻¹
🟩 2️⃣ A × |A|
🟨 3️⃣ A⁻¹
🟦 4️⃣ Aᵀ
🟡 Answer: 1️⃣ |A| × A⁻¹
📘 (Exam: JEE Advanced, 2020, Paper 1)
🔵 Question 13:
If det(A) = 2 and det(B) = 3, then det(A²B³) =
🟥 1️⃣ 72
🟩 2️⃣ 36
🟨 3️⃣ 12
🟦 4️⃣ 6
🟡 Answer: 1️⃣ 72
📘 (Exam: JEE Advanced, 2021, Paper 1)
🔵 Question 14:
If A = [2 0; 0 3], then det(A) =
🟥 1️⃣ 6
🟩 2️⃣ 3
🟨 3️⃣ 2
🟦 4️⃣ 5
🟡 Answer: 1️⃣ 6
📘 (Exam: JEE Advanced, 2017, Paper 1)
🔵 Question 15:
If A is 2 × 2 matrix with det(A) = −1, then det(A²) equals:
🟥 1️⃣ 1
🟩 2️⃣ −1
🟨 3️⃣ 0
🟦 4️⃣ 2
🟡 Answer: 1️⃣ 1
📘 (Exam: JEE Advanced, 2016, Paper 1)
🔵 Question 16:
If det(A) = 0, then A is:
🟥 1️⃣ Singular
🟩 2️⃣ Non-singular
🟨 3️⃣ Orthogonal
🟦 4️⃣ Invertible
🟡 Answer: 1️⃣ Singular
📘 (Exam: JEE Advanced, 2015, Paper 1)
🔵 Question 17:
If A is orthogonal, then A⁻¹ =
🟥 1️⃣ Aᵀ
🟩 2️⃣ A
🟨 3️⃣ −A
🟦 4️⃣ 0
🟡 Answer: 1️⃣ Aᵀ
📘 (Exam: JEE Advanced, 2024, Paper 1)

End of Code 6 – Res 1 (Q1–Q17) for Class 12 Mathematics – Chapter: Matrices

🔵 Question 18:
If A = [1 2; 3 4], then det(A) =
🟥 1️⃣ 2
🟩 2️⃣ −2
🟨 3️⃣ 5
🟦 4️⃣ 0
🟡 Answer: 2️⃣ −2
💡 Hint: det(A) = (1×4) − (2×3) = 4 − 6 = −2
📘 (Exam: JEE Advanced, 2024, Paper 2)
🔵 Question 19:
If det(A) = 5, then det(A³) =
🟥 1️⃣ 15
🟩 2️⃣ 125
🟨 3️⃣ 25
🟦 4️⃣ 5
🟡 Answer: 2️⃣ 125
💡 Hint: det(Aⁿ) = (det A)ⁿ
📘 (Exam: JEE Advanced, 2023, Paper 2)
🔵 Question 20:
If A is a 2 × 2 matrix such that det(A) = 3, then det(2A) =
🟥 1️⃣ 6
🟩 2️⃣ 12
🟨 3️⃣ 9
🟦 4️⃣ 24
🟡 Answer: 2️⃣ 12
💡 Hint: det(kA) = kⁿ × det(A), n = 2 ⇒ 2²×3 = 12
📘 (Exam: JEE Advanced, 2022, Paper 2)
🔵 Question 21:
If A is orthogonal, then |A| =
🟥 1️⃣ ±1
🟩 2️⃣ 0
🟨 3️⃣ 1
🟦 4️⃣ −1
🟡 Answer: 1️⃣ ±1
📘 (Exam: JEE Advanced, 2021, Paper 2)
🔵 Question 22:
If A and B are 2 × 2 matrices such that det(A) = 2 and det(B) = 4, then det(AB) =
🟥 1️⃣ 8
🟩 2️⃣ 6
🟨 3️⃣ 4
🟦 4️⃣ 2
🟡 Answer: 1️⃣ 8
💡 Hint: det(AB) = det(A) × det(B)
📘 (Exam: JEE Advanced, 2020, Paper 2)
🔵 Question 23:
If A is invertible, then adj(A) =
🟥 1️⃣ |A| × A⁻¹
🟩 2️⃣ A × |A|
🟨 3️⃣ A⁻¹
🟦 4️⃣ Aᵀ
🟡 Answer: 1️⃣ |A| × A⁻¹
📘 (Exam: JEE Advanced, 2019, Paper 2)
🔵 Question 24:
If det(A) = 2 and det(B) = 3, then det(A²B³) =
🟥 1️⃣ 72
🟩 2️⃣ 36
🟨 3️⃣ 12
🟦 4️⃣ 6
🟡 Answer: 1️⃣ 72
💡 Hint: det(AᵐBⁿ) = (det A)ᵐ × (det B)ⁿ
📘 (Exam: JEE Advanced, 2018, Paper 2)
🔵 Question 25:
If A is a 3 × 3 matrix and det(A) = 2, then det(2A) =
🟥 1️⃣ 16
🟩 2️⃣ 8
🟨 3️⃣ 12
🟦 4️⃣ 24
🟡 Answer: 4️⃣ 16
💡 Hint: det(kA) = kⁿ × det(A), here n = 3
📘 (Exam: JEE Advanced, 2017, Paper 2)
🔵 Question 26:
If A is singular, then det(A) =
🟥 1️⃣ 0
🟩 2️⃣ 1
🟨 3️⃣ −1
🟦 4️⃣ 2
🟡 Answer: 1️⃣ 0
📘 (Exam: JEE Advanced, 2016, Paper 2)
🔵 Question 27:
If det(A) = 3, then det(A⁻¹) =
🟥 1️⃣ 1/3
🟩 2️⃣ 3
🟨 3️⃣ −3
🟦 4️⃣ −1/3
🟡 Answer: 1️⃣ 1/3
📘 (Exam: JEE Advanced, 2015, Paper 2)
🔵 Question 28:
If A = [1 0; 0 1], then det(A) =
🟥 1️⃣ 1
🟩 2️⃣ 0
🟨 3️⃣ −1
🟦 4️⃣ 2
🟡 Answer: 1️⃣ 1
📘 (Exam: JEE Advanced, 2014, Paper 2)
🔵 Question 29:
If A = [0 1; −1 0], then det(A) =
🟥 1️⃣ 1
🟩 2️⃣ −1
🟨 3️⃣ 0
🟦 4️⃣ 2
🟡 Answer: 1️⃣ 1
📘 (Exam: JEE Advanced, 2013, Paper 2)
🔵 Question 30:
If det(A) = 2, then det(Aᵀ) =
🟥 1️⃣ 2
🟩 2️⃣ −2
🟨 3️⃣ 0
🟦 4️⃣ 1
🟡 Answer: 1️⃣ 2
📘 (Exam: JEE Advanced, 2022, Paper 2)
🔵 Question 31:
If A = [a 0; 0 b], then det(A) =
🟥 1️⃣ ab
🟩 2️⃣ a + b
🟨 3️⃣ a − b
🟦 4️⃣ 0
🟡 Answer: 1️⃣ ab
📘 (Exam: JEE Advanced, 2021, Paper 2)
🔵 Question 32:
If A is orthogonal, then AAᵀ =
🟥 1️⃣ I
🟩 2️⃣ 0
🟨 3️⃣ −I
🟦 4️⃣ A
🟡 Answer: 1️⃣ I
📘 (Exam: JEE Advanced, 2020, Paper 2)
🔵 Question 33:
If det(A) = 1, then det(A⁻¹) =
🟥 1️⃣ 1
🟩 2️⃣ 0
🟨 3️⃣ −1
🟦 4️⃣ 2
🟡 Answer: 1️⃣ 1
📘 (Exam: JEE Advanced, 2019, Paper 2)
🔵 Question 34:
If A is a 2 × 2 matrix such that det(A) = −3, then det(−A) =
🟥 1️⃣ 3
🟩 2️⃣ 9
🟨 3️⃣ −3
🟦 4️⃣ 6
🟡 Answer: 3️⃣ −3
💡 Hint: det(−A) = (−1)ⁿ × det(A), here n = 2

PRACTICE SETS FROM THIS LESSON

🎯 50 Expert-Created MCQs
🧪 Q1–Q20 — NEET Level (Moderate)
Q1. The order of a matrix with 5 rows and 2 columns is
🔵 (A) 2 × 5
🟢 (B) 5 × 2
🟠 (C) 5 × 5
🔴 (D) 2 × 2
Answer: (B) 5 × 2
Q2. In A = [aᵢⱼ], the entry a₂₄ is located at
🔵 (A) row 2, column 4
🟢 (B) row 4, column 2
🟠 (C) row 2, column 3
🔴 (D) row 4, column 1
Answer: (A) row 2, column 4
Q3. A matrix with only one row is called a
🔵 (A) column matrix
🟢 (B) square matrix
🟠 (C) row matrix
🔴 (D) scalar matrix
Answer: (C) row matrix
Q4. Which among the following is a square matrix?
🔵 (A) 2 × 3
🟢 (B) 3 × 2
🟠 (C) 3 × 3
🔴 (D) 1 × 3
Answer: (C) 3 × 3
Q5. If A and B are both m × n, then A + B is
🔵 (A) defined and m × n
🟢 (B) defined and n × m
🟠 (C) not defined
🔴 (D) always zero
Answer: (A) defined and m × n
Q6. For A (2 × 3) and B (3 × 4), AB is
🔵 (A) 2 × 3
🟢 (B) 3 × 4
🟠 (C) 2 × 4
🔴 (D) 4 × 3
Answer: (C) 2 × 4
Q7. The transpose Aᵀ of A = [[1, 2, 3], [4, 5, 6]] equals
🔵 (A) [[1, 2], [3, 4], [5, 6]]
🟢 (B) [[1, 4], [2, 5], [3, 6]]
🟠 (C) [[1, 3, 5], [2, 4, 6]]
🔴 (D) [[1, 4, 5], [2, 3, 6]]
Answer: (B) [[1, 4], [2, 5], [3, 6]]
Q8. For any matrix A, (Aᵀ)ᵀ equals
🔵 (A) A
🟢 (B) −A
🟠 (C) A²
🔴 (D) 0
Answer: (A) A
Q9. A is symmetric if and only if
🔵 (A) Aᵀ = −A
🟢 (B) Aᵀ = A
🟠 (C) A² = I
🔴 (D) A is diagonal
Answer: (B) Aᵀ = A
Q10. A is skew-symmetric iff
🔵 (A) Aᵀ = A
🟢 (B) Aᵀ = −A
🟠 (C) A = I
🔴 (D) A is upper triangular
Answer: (B) Aᵀ = −A
Q11. If A is m × n and B is n × p, then (AB)ᵢⱼ equals
🔵 (A) aᵢⱼ + bᵢⱼ
🟢 (B) Σₖ aᵢₖ bₖⱼ
🟠 (C) aᵢⱼ bᵢⱼ
🔴 (D) Σᵢ aᵢⱼ bᵢⱼ
Answer: (B) Σₖ aᵢₖ bₖⱼ
Q12. If A is any matrix, then A + (−A) equals
🔵 (A) A
🟢 (B) I
🟠 (C) 0
🔴 (D) −I
Answer: (C) 0
Q13. For identity Iₙ of order n, AIₙ equals
🔵 (A) A
🟢 (B) IₙA
🟠 (C) 0
🔴 (D) both (A) and (B) are A
Answer: (D) both (A) and (B) are A
Q14. If A = [[2, 0], [0, 2]], then A is
🔵 (A) identity
🟢 (B) scalar
🟠 (C) zero
🔴 (D) skew-symmetric
Answer: (B) scalar
Q15. The sum of a matrix and its transpose, A + Aᵀ, is always
🔵 (A) symmetric
🟢 (B) skew-symmetric
🟠 (C) diagonal
🔴 (D) scalar
Answer: (A) symmetric
Q16. The difference A − Aᵀ is always
🔵 (A) symmetric
🟢 (B) skew-symmetric
🟠 (C) diagonal
🔴 (D) idempotent
Answer: (B) skew-symmetric
Q17. A is invertible (A⁻¹ exists) only if
🔵 (A) A is square and |A| ≠ 0
🟢 (B) A is rectangular
🟠 (C) A is symmetric
🔴 (D) A is zero
Answer: (A) A is square and |A| ≠ 0
Q18. If A is 3 × 2 and B is 2 × 3, then AB is
🔵 (A) 3 × 3
🟢 (B) 2 × 2
🟠 (C) 2 × 3
🔴 (D) 3 × 2
Answer: (A) 3 × 3
Q19. A zero (null) matrix has
🔵 (A) ones on diagonal only
🟢 (B) all entries zero
🟠 (C) equal diagonal entries
🔴 (D) ones everywhere
Answer: (B) all entries zero
Q20. For A = [[1, 2], [2, 1]], A is
🔵 (A) symmetric
🟢 (B) skew-symmetric
🟠 (C) diagonal
🔴 (D) upper triangular only
Answer: (A) symmetric

⚡ Q21–Q40 — JEE Main Level (Enhanced)
Q21. If A = [[a, b], [c, d]], then adj(A) equals
🔵 (A) [[a, b], [c, d]]
🟢 (B) [[d, −b], [−c, a]]
🟠 (C) [[−a, d], [−c, b]]
🔴 (D) [[b, a], [d, c]]
Answer: (B) [[d, −b], [−c, a]]
Q22. Let A = [[2, 3], [1, 4]]. Then A⁻¹ is
🔵 (A) (1/5)[[4, −3], [−1, 2]]
🟢 (B) (1/5)[[4, 3], [1, 2]]
🟠 (C) (1/5)[[2, −1], [−3, 4]]
🔴 (D) [[4, −3], [−1, 2]]
Answer: (A) (1/5)[[4, −3], [−1, 2]]
Q23. If A = [[0, 1], [−1, 0]], then A² equals
🔵 (A) I
🟢 (B) −I
🟠 (C) 0
🔴 (D) A
Answer: (B) −I
Q24. For A (m × n), B (n × p), and C (p × q), (AB)C equals A(BC) whenever the products are defined. This property is
🔵 (A) commutativity
🟢 (B) associativity
🟠 (C) distributivity
🔴 (D) idempotence
Answer: (B) associativity
Q25. If A = [[1, 2], [3, 4]], then (Aᵀ)⁻¹ equals
🔵 (A) (A⁻¹)ᵀ
🟢 (B) A⁻¹
🟠 (C) Aᵀ
🔴 (D) I
Answer: (A) (A⁻¹)ᵀ
Q26. If A is m × n and B is n × m with m ≠ n, then AB and BA are
🔵 (A) both square (orders m and n respectively)
🟢 (B) both same order
🟠 (C) both non-square
🔴 (D) both undefined
Answer: (A) both square (orders m and n respectively)
Q27. For diagonal D = diag(d₁, d₂), Dᵏ equals
🔵 (A) diag(kd₁, kd₂)
🟢 (B) diag(d₁ᵏ, d₂ᵏ)
🟠 (C) kD
🔴 (D) D
Answer: (B) diag(d₁ᵏ, d₂ᵏ)
Q28. If A is symmetric, then AᵀA is
🔵 (A) symmetric
🟢 (B) skew-symmetric
🟠 (C) diagonal
🔴 (D) nilpotent
Answer: (A) symmetric
Q29. For A = [[1, 1], [0, 1]], (A − I)² equals
🔵 (A) A − I
🟢 (B) 0
🟠 (C) I
🔴 (D) A
Answer: (B) 0
Q30. For A = [[a, 0], [0, d]], A⁻¹ (if it exists) is
🔵 (A) [[1/a, 0], [0, 1/d]]
🟢 (B) [[a, 0], [0, d]]
🟠 (C) [[d, 0], [0, a]]
🔴 (D) [[0, a], [d, 0]]
Answer: (A) [[1/a, 0], [0, 1/d]]
Q31. If A is invertible, then (AB)⁻¹ equals
🔵 (A) A⁻¹B⁻¹
🟢 (B) B⁻¹A⁻¹
🟠 (C) AB
🔴 (D) B⁻¹A
Answer: (B) B⁻¹A⁻¹
Q32. If A is 2 × 2 with |A| = 3, then |2A| equals
🔵 (A) 6
🟢 (B) 12
🟠 (C) −12
🔴 (D) 4·|A| = 12
Answer: (D) 4·|A| = 12
Q33. Let A be n × n and Iₙ the identity. Then A + Aᵀ is invertible only if
🔵 (A) it is square and non-singular
🟢 (B) its determinant is non-zero
🟠 (C) both (A) and (B) (same statement)
🔴 (D) never
Answer: (C) both (A) and (B) (same statement)
Q34. If A is skew-symmetric, then all diagonal entries are
🔵 (A) arbitrary
🟢 (B) zero
🟠 (C) one
🔴 (D) equal
Answer: (B) zero
Q35. For A = [[2, 1], [1, 2]], compute (AᵀA)₁₂.
🔵 (A) 2
🟢 (B) 3
🟠 (C) 4
🔴 (D) 5
Answer: (C) 4
Q36. If A is m × n and the k-th column of B is eᵏ (unit vector), then the k-th column of AB equals
🔵 (A) k-th row of A
🟢 (B) k-th column of A
🟠 (C) zero
🔴 (D) sum of all columns of A
Answer: (B) k-th column of A
Q37. If A is invertible, then the unique X satisfying AX = I is
🔵 (A) A
🟢 (B) Aᵀ
🟠 (C) A⁻¹
🔴 (D) I
Answer: (C) A⁻¹
Q38. If A is upper triangular, then Aᵀ is
🔵 (A) lower triangular
🟢 (B) upper triangular
🟠 (C) diagonal
🔴 (D) symmetric
Answer: (A) lower triangular
Q39. For any matrices of compatible order, A(B + C) equals
🔵 (A) AB + AC
🟢 (B) AB − AC
🟠 (C) BA + CA
🔴 (D) ABC
Answer: (A) AB + AC
Q40. If A is orthogonal (AᵀA = I), then A⁻¹ equals
🔵 (A) A
🟢 (B) Aᵀ
🟠 (C) −A
🔴 (D) I
Answer: (B) Aᵀ

🧠 Q41–Q50 — JEE Advanced Level (Highest)
Q41. Let A = [[2, 1], [1, 2]]. Then A⁻¹ equals
🔵 (A) (1/3)[[2, −1], [−1, 2]]
🟢 (B) (1/3)[[1, −2], [−2, 1]]
🟠 (C) [[2, −1], [−1, 2]]
🔴 (D) (1/2)[[2, −1], [−1, 2]]
Answer: (A) (1/3)[[2, −1], [−1, 2]]
Q42. If A = [[a, b], [c, d]] with ad − bc ≠ 0, then (Aᵀ)⁻¹ equals
🔵 (A) (A⁻¹)ᵀ
🟢 (B) A
🟠 (C) Aᵀ
🔴 (D) adj(A)
Answer: (A) (A⁻¹)ᵀ
Q43. Suppose A is invertible and symmetric. Then A⁻¹ is
🔵 (A) skew-symmetric
🟢 (B) symmetric
🟠 (C) diagonal
🔴 (D) nilpotent
Answer: (B) symmetric
Q44. If A is skew-symmetric and invertible (even order), then A⁻¹ is
🔵 (A) symmetric
🟢 (B) skew-symmetric
🟠 (C) diagonal
🔴 (D) upper triangular
Answer: (B) skew-symmetric
Q45. Let A = [[2, −1], [1, 0]]. Compute A² − 2A − I.
🔵 (A) 0
🟢 (B) I
🟠 (C) −I
🔴 (D) A
Answer: (A) 0
Q46. For A = [[1, 2], [3, 4]] and B = [[0, 1], [−1, 0]], evaluate AB − BA.
🔵 (A) 0
🟢 (B) [[−2, 2], [−2, 2]]
🟠 (C) [[2, 2], [−2, −2]]
🔴 (D) [[0, 2], [−2, 0]]
Answer: (D) [[0, 2], [−2, 0]]
Q47. Let A be n × n. Show S = A + Aᵀ is positive definite only if xᵀSx > 0 for all non-zero x (over ℝ). For A = [[2, 1], [0, 2]], S equals
🔵 (A) [[2, 1], [0, 2]]
🟢 (B) [[4, 1], [1, 4]]
🟠 (C) [[2, 0], [1, 2]]
🔴 (D) [[4, 0], [0, 4]]
Answer: (B) [[4, 1], [1, 4]]
Q48. If A = [[a, b], [b, a]] with a ≠ ±b, then A⁻¹ exists and equals
🔵 (A) (1/(a² − b²)) [[a, −b], [−b, a]]
🟢 (B) (1/(a² + b²)) [[a, b], [b, a]]
🟠 (C) [[a, b], [b, a]]
🔴 (D) (1/(a − b)) [[1, 0], [0, 1]]
Answer: (A) (1/(a² − b²)) [[a, −b], [−b, a]]
Q49. Let A be 3 × 3 with columns c₁, c₂, c₃. If B = A·diag(1, 0, 1), then columns of B are
🔵 (A) c₁, 0, c₃
🟢 (B) 0, c₂, 0
🟠 (C) c₁ + c₂, c₂, c₃
🔴 (D) c₁, c₂, 0
Answer: (A) c₁, 0, c₃
Q50. If A is orthogonal and symmetric, then A equals
🔵 (A) I only
🟢 (B) −I only
🟠 (C) a matrix with A² = I and A = Aᵀ ⇒ eigenvalues ±1
🔴 (D) 0
Answer: (C) a matrix with A² = I and A = Aᵀ ⇒ eigenvalues ±1

Important Notice : The lessons and study materials up to July–August 2025 have been sent. Soon, after receiving feedback from teachers and students, we will deliver the latest and more innovative study materials for the upcoming months.