Why should students practise Important Questions for CBSE Class 12 Mathematics Chapter 3 for the board exam?

Mathematics is a subject that can be mastered only when the concepts are clear and the questions are practised several times. That is why practising questions from authentic sources becomes important. Even while doing so, students can focus on perfecting their answering skills for important topics. Important Questions Class 12 Mathematics Chapter 3 will give them a fair idea of what kinds of questions can be asked in the exam. It will help them solve their doubts related to the concepts and questions of the chapter.

Where can students practise Important Questions on CBSE Class 12 Mathematics Chapter 3 Matrices?

CBSE Class 12 students can refer to the website of Extramarks to access the important questions on Matrices. The question set has been prepared by subject experts to guide students during their exam preparation. Extramarks has carefully prepared a comprehensive question set which students can practise and feel confident in the exam. These questions will prove beneficial for the exam preparation and for revising the important concepts of the chapter effectively.

Why is it important for students to solve extra questions on matrices?

Solving extra questions is the key to scoring higher grades in the exam. CBSE Class 12 Mathematics Chapter 3 Important Questions demonstrate the type of questions that can be asked in the board exams. So, students will be familiar with all the possible types of questions which will make them more confident. They do not need to spend time checking various sources to collect questions of past years. Instead, they can utilise this time to practise these questions. In addition to that, these important questions will help them revise the chapter’s most important concepts, which is crucial for their preparation.

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CBSE Important Questions Class 12 Maths Chapter 3 – Matrices

Class 12 Maths Chapter 3 Matrices is part of the CBSE Class 12 Maths syllabus. A matrix is a rectangular arrangement of numbers in rows and columns. Students learn the different types of matrices, how to add, subtract and multiply matrices, the transpose of a matrix, symmetric and skew-symmetric matrices, elementary operations, and how to find the inverse of a matrix.

These CBSE important questions for Class 12 Maths Chapter 3 cover the order of a matrix, constructing a matrix from a given rule, matrix addition and multiplication, the transpose and its properties, symmetric and skew-symmetric matrices, and finding unknown values by equating two equal matrices. Every question has a clear step-by-step solution in simple English, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. Define a matrix. If a matrix has 12 elements, what are the possible orders it can have?

Answer: A matrix is a rectangular arrangement of numbers (or functions) in rows and columns, enclosed in brackets. A matrix with m rows and n columns is said to be of order m × n and has mn elements.
For 12 elements: We need all pairs (m, n) with m × n = 12:
1 × 12, 2 × 6, 3 × 4, 4 × 3, 6 × 2, 12 × 1
So there are 6 possible orders.

Q2. Construct a 2 × 2 matrix A = [aᵢⱼ] whose elements are given by aᵢⱼ = (i + j)²/2.

Answer: Compute each element:
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
So the matrix is

2	9/2
9/2	8

Q3. Write the condition for two matrices to be (i) added and (ii) multiplied. What is the order of the product?

Answer: (i) Addition: Two matrices can be added only if they are of the same order. We then add the corresponding elements.
(ii) Multiplication: The product AB exists only if the number of columns of A equals the number of rows of B.
If A is of order m × n and B is of order n × p, then AB exists and its order is m × p.
Note that matrix multiplication is not commutative: AB and BA are usually different, and one may exist while the other does not.

Q4. If A is a 3 × 4 matrix and B is a matrix such that A′B and BA′ are both defined, what is the order of B?

Answer: A is of order 3 × 4, so A′ (its transpose) is of order 4 × 3.
Let B be of order m × n.
For A′B to be defined: columns of A′ = rows of B, so 3 = m.
For BA′ to be defined: columns of B = rows of A′, so n = 4.
Therefore the order of B is 3 × 4.

Q5. Define symmetric and skew-symmetric matrices. What can you say about the diagonal elements of a skew-symmetric matrix?

Answer: A square matrix A is symmetric if A′ = A, that is aᵢⱼ = aⱼᵢ for all i and j.
A square matrix A is skew-symmetric if A′ = −A, that is aᵢⱼ = −aⱼᵢ.
Diagonal elements: For a skew-symmetric matrix, putting i = j gives aᵢᵢ = −aᵢᵢ, so 2aᵢᵢ = 0, which means aᵢᵢ = 0.
So all diagonal elements of a skew-symmetric matrix are zero.

Q6. Express the matrix A = [[2, 4], [1, 3]] as the sum of a symmetric and a skew-symmetric matrix.

Answer: Any square matrix can be written as
A = ½(A + A′) + ½(A − A′)
where the first part is symmetric and the second is skew-symmetric.
Here A′ = [[2, 1], [4, 3]].
Symmetric part ½(A + A′) = ½[[4, 5], [5, 6]] = [[2, 5/2], [5/2, 3]]
Skew-symmetric part ½(A − A′) = ½[[0, 3], [−3, 0]] = [[0, 3/2], [−3/2, 0]]
Adding the two gives back A, as required.

Q7. If A = [[1, 2], [3, 4]] and B = [[0, 1], [1, 0]], find AB and BA. Are they equal?

Answer: AB: Multiply row by column.
Row 1 of A with the columns of B: (1×0 + 2×1) = 2 and (1×1 + 2×0) = 1
Row 2 of A: (3×0 + 4×1) = 4 and (3×1 + 4×0) = 3
So AB = [[2, 1], [4, 3]]
BA: Row 1 of B with columns of A: (0×1 + 1×3) = 3 and (0×2 + 1×4) = 4
Row 2 of B: (1×1 + 0×3) = 1 and (1×2 + 0×4) = 2
So BA = [[3, 4], [1, 2]]
No, AB ≠ BA. This confirms that matrix multiplication is not commutative.

Q8. If [[x + 3, z + 4], [6, a − 1]] = [[0, 6], [3y − 2, 2c + 2]], find the values of x, y, z, a and c.

Answer: Two matrices are equal only when all their corresponding elements are equal.
x + 3 = 0, so x = −3
z + 4 = 6, so z = 2
6 = 3y − 2, so 3y = 8, giving y = 8/3
a − 1 = 2c + 2, which relates a and c as a = 2c + 3

Q9. Write the properties of the transpose of a matrix.

Answer: For matrices A and B of suitable orders and a scalar k:
• (A′)′ = A
• (kA)′ = kA′
• (A + B)′ = A′ + B′
• (AB)′ = B′A′ — note that the order gets reversed
The last property is the one students most often get wrong, so it is worth remembering carefully.

Q10. If A = [[2, 3], [1, −4]] and B = [[1, −2], [−1, 3]], verify that (AB)′ = B′A′.

Answer: First compute AB:
Row 1: (2×1 + 3×−1) = −1 and (2×−2 + 3×3) = 5
Row 2: (1×1 + −4×−1) = 5 and (1×−2 + −4×3) = −14
AB = [[−1, 5], [5, −14]], so (AB)′ = [[−1, 5], [5, −14]]
Now compute B′A′:
B′ = [[1, −1], [−2, 3]] and A′ = [[2, 1], [3, −4]]
B′A′ row 1: (1×2 + −1×3) = −1 and (1×1 + −1×−4) = 5
row 2: (−2×2 + 3×3) = 5 and (−2×1 + 3×−4) = −14
B′A′ = [[−1, 5], [5, −14]]
Both sides are equal, so (AB)′ = B′A′ is verified.

Q11. Name the different types of matrices with a one-line description of each.

Answer: • Row matrix: has only one row
• Column matrix: has only one column
• Square matrix: number of rows equals number of columns
• Diagonal matrix: a square matrix in which all non-diagonal elements are zero
• Scalar matrix: a diagonal matrix in which all diagonal elements are equal
• Identity matrix: a scalar matrix in which every diagonal element is 1
• Zero matrix: every element is zero

Q12. If A = [[0, 1], [0, 0]], show that A² = O but A ≠ O. What does this tell us about matrices?

Answer: Compute A² = A × A:
Row 1: (0×0 + 1×0) = 0 and (0×1 + 1×0) = 0
Row 2: (0×0 + 0×0) = 0 and (0×1 + 0×0) = 0
So A² = [[0, 0], [0, 0]] = O, the zero matrix, even though A itself is not the zero matrix.
What this shows: For ordinary numbers, if a² = 0 then a = 0. For matrices this is not true. Similarly, AB = O does not mean that A = O or B = O. Matrix algebra does not follow all the rules of ordinary numbers.

Related Study Material on Extramarks

Q1.

$\begin{array}{l} If [\begin{matrix} 3 a - b & a + 3 b \\ 2 c ? d & 4 c + d \end{matrix}] = [\begin{matrix} 5 & 5 \\ 10 & 2 \end{matrix}], then the value \\ of a, b, c and d is \end{array}$

Opt

1. a = 1, b = -2, c = 2 and d = -3.

2. a = 1, b = -2, c = 2 and d = -6.

3. a = -1, b = -8, c = 2 and d = -6.

4. a = -1, b = 8, c = 2 and d = -3.

Ans

$\begin{array}{l} As the given matrices are equal, therefore, their corresponding \\ elements must be equal. Comparing the corresponding elements, \\ we get \\ 3 a b = 5, a + 3 b = ? 5, \\ 2 c d = 10 ? and ? 4 c + d = 2 \\ Solving these equations, we get \\ a = 1, b = ? 2 \\ c = 2, d = ? 6 \end{array}$

Q2.

$\begin{array}{l} Let A = [\begin{array}{l} 01 \\ 00 \end{array}], show that {(aI + bA)}^{n} ? = a^{n} I + {na}^{n 1} bA, where I is the identity \\ matrix of order 2 and n N . \end{array}$

Opt.

$\begin{array}{l} We shall prove the result by using the principle of mathematical induction . \\ For n = 1, we have \\ P (1) : (al + bA) = al + ba A \\ = al + bA \end{array}$

Ans.

$\begin{array}{l} We shall prove the result by using the principle of mathematical induction . \\ For n = 1, we have \\ P (1) : (al + bA) = al + ba A \\ = al + bA \\ Therefore, the result is true for n = 1. \\ Let the result be true for n = k \\ so, \\ P {(al + bA)}^{k} = a^{k} l + {ka}^{k 1} bA Now, we will prove that it is true for n = k +1 \\ {(al + bA)}^{k + 1} = {(al + bA)}^{k} (al + bA) \\ = (a^{k} + {ka}^{k 1} bA) (al + bA) \\ = a^{k + 1} l + {ka}^{k} bA + a^{k} bAl + {ka}^{k 1} b^{2} A^{2} \\ {(al + bA)}^{k + 1} = a^{k + 1} + (k + 1) a^{k} bA + {ka}^{k 1} b^{2} A^{2} \dots.. (1) \\ Now, A^{2} = [\begin{array}{l} 0 1 \\ 0 0 \end{array}] [\begin{array}{l} 0 1 \\ 0 0 \end{array}] \\ = [\begin{array}{l} 0 1 \\ 0 0 \end{array}] = 0 \\ From equation (1), we have \\ {(al + bA)}^{k + 1} = a^{k + 1} + (k + 1) a^{k} bA + {ka}^{k 1} b^{2} (0) \\ = a^{k + 1} + (k + 1) a^{k} bA \\ Therefore, the result is true for n = k +1. \\ Thus, by the Principle of Mathematical Induction, we have \\ {(aI + bA)}^{n} = a^{n} I + {na}^{n 1} bA, for all n ˆˆ N . \end{array}$

Q3. A matrix has 24 elements. What are the possible orders it can have?

Opt. The possible orders of the matrix are:

1 24, 2 12, 3 8, 4 6, 6 4, 8 3, 12 2 and 24 1.

Ans. The possible orders of the matrix are:

1 24, 2 12, 3 8, 4 6, 6 4, 8 3, 12 2 and 24 1.

Q4.

$\begin{array}{l} Express A = [\begin{array}{l} 331 \\ 221 \\ 452 \end{array}] as the sum of a symmetric \\ and a skew - symmetric matrix . \end{array}$

Ans.

$\begin{array}{l} Here A ‘ = [\begin{array}{l} 3 2 4 \\ 3 2 5 \\ 1 1 2 \end{array}] \\ Let P = \frac{1}{2} (A + A ‘) \\ \frac{1}{2} (A + A ‘) = \frac{1}{2} [\begin{array}{l} 3 3 1 \\ 2 2 1 \\ 4 5 2 \end{array}] + [\begin{array}{l} 3 2 4 \\ 3 2 5 \\ 1 1 2 \end{array}] \\ = \frac{1}{2} [\begin{array}{l} 6 1 5 \\ 1 4 4 \\ 5 4 4 \end{array}] = [\begin{array}{l} 3 \frac{1}{2} \frac{5}{2} \\ \frac{1}{2} 2 2 \\ \frac{5}{2} 2 2 \end{array}] \\ Now P ‘ = [\begin{array}{l} 3 \frac{1}{2} \frac{5}{2} \\ \frac{1}{2} 2 2 \\ \frac{5}{2} 2 2 \end{array}] = P \\ Thus P = \frac{1}{2} [A A ‘] \\ \frac{1}{2} [A A ‘] = \frac{1}{2} [[\begin{array}{l} 3 3 1 \\ 2 2 1 \\ 4 5 2 \end{array}] [\begin{array}{l} 3 2 4 \\ 3 2 5 \\ 1 1 2 \end{array}]] \\ = \frac{1}{2} [[\begin{array}{l} 0 5 3 \\ 2 0 6 \\ 3 6 0 \end{array}] = [\begin{array}{l} 0 \frac{5}{2} \frac{3}{2} \\ \frac{5}{2} 3 0 \\ \frac{3}{2} 3 0 \end{array}]] \\ Then Q ‘ = [\begin{array}{l} 0 \frac{5}{2} \frac{3}{2} \\ \frac{5}{2} 0 3 \\ \frac{3}{2} 3 0 \end{array}] = Q \\ Thus Q = \frac{1}{2} (A A ‘) is a skew symmetric matrix . \\ Now, P + Q = [\begin{array}{l} 3 \frac{1}{2} \frac{5}{2} \\ \frac{1}{2} 2 2 \\ \frac{5}{2} 2 2 \end{array}] + [\begin{array}{l} 0 \frac{5}{2} \frac{3}{2} \\ \frac{5}{2} 0 3 \\ \frac{3}{2} 3 0 \end{array}] = [\begin{array}{l} 3 3 1 \\ 2 2 1 \\ 4 5 2 \end{array}] = A \end{array}$

CBSE Important Questions Class 12 Maths Chapter 3 – Matrices

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