RD Sharma Solutions for Class 12 Maths Chapter 5 Algebra of Matrices – PDF Download

RD Sharma Solutions for Class 12 Maths Chapter 5 – Algebra of Matrices are available here in a detailed and easy-to-understand format. To achieve a strong academic performance in Mathematics, students are encouraged to solve every exercise thoroughly. These solutions are carefully prepared by subject matter experts to help students excel in their board exams.

Regular practice with these step-by-step solutions not only improves accuracy but also enhances problem-solving skills, which are crucial for scoring well. By referring to these RD Sharma Solutions while solving textbook questions, students can reinforce their conceptual understanding and minimize errors.

For effective exam preparation, students can download the free PDF of RD Sharma Class 12 Chapter 5 – Algebra of Matrices from the link provided below and use it as a reliable learning resource.

Download RD Sharma Solutions for Class 12 Maths Chapter 5 Algebra of Matrices PDF

RD Sharma Class 12 Maths Solutions Chapter 5 Algebra of Matrices cover all the questions from the textbook, crafted by expert Mathematics teachers at Infinity Learn. Download our free PDF of Chapter 5 Algebra of Matrices RD Sharma Solutions for Class 12 to boost your performance in board exams and competitive exams.

Access answers to RD Sharma Solutions For Class 12 Chapter 5 – Algebra of Matrices

1. If a matrix has 8 elements, what are the possible orders it can have?

Ans. The number of elements in a matrix is the product of rows and columns.

Possible orders are:

1 × 8, 2 × 4, 4 × 2, 8 × 1

2. Given A = [[2, 3, -5], [1, 4, 9], [0, 7, -2]] and B = [[2, -1], [-3, 4], [1, 2]]

(i) a₂₂ + b₂₁ = 4 + (-3) = 1

(ii) a₁₁ × b₁₁ + a₂₂ × b₂₂ = 2×2 + 4×4 = 20

3. Compute: [[3, -2], [1, 4]] + [[-2, 4], [1, 3]]

Answer: [[1, 2], [2, 7]]

4. Add: [[2, 1, 3], [0, 3, 5], [-1, 2, 5]] + [[1, -2, 3], [2, 6, 1], [0, -3, 1]]

Answer: [[3, -1, 6], [2, 9, 6], [-1, -1, 6]]

5. Find 2A − 3B, where A = [[2, 4], [3, 2]], B = [[1, 3], [-2, 5]]

2A = [[4, 8], [6, 4]], 3B = [[3, 9], [-6, 15]]

Result: [[1, -1], [12, -11]]

6. Find B − 4C, where C = [[-2, 5], [3, 4]]

4C = [[-8, 20], [12, 16]]

B − 4C = [[9, -17], [-14, -11]]

7. Find 3A − C

3A = [[6, 12], [9, 6]]

3A − C = [[8, 7], [6, 2]]

8. Find 3A − 2B + 3C

3A = [[6, 12], [9, 6]]

2B = [[2, 6], [-4, 10]], 3C = [[-6, 15], [9, 12]]

Result: [[-2, 21], [22, 8]]

9. Verify (A^T)^T = A for A = [[1, 2], [3, 4]]

A^T = [[1, 3], [2, 4]]

(A^T)^T = [[1, 2], [3, 4]] = A

10. Verify (A + B)^T = A^T + B^T

A = [[1, 2], [3, 4]], B = [[5, 6], [7, 8]]

(A + B)^T = [[6, 10], [8, 12]]

A^T + B^T = [[6, 10], [8, 12]]

11. Verify (AB)^T = B^TA^T

A = [[1, 2], [3, 4]], B = [[5, 6], [7, 8]]

AB = [[19, 22], [43, 50]] ⇒ (AB)^T = [[19, 43], [22, 50]]

B^T = [[5, 7], [6, 8]], A^T = [[1, 3], [2, 4]]

B^TA^T = [[19, 43], [22, 50]] = (AB)^T