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

The RD Sharma Solutions for Class 12 Maths Chapter 5 – Algebra of Matrices are available here in PDF format. Students who have difficulty solving the exercise problems in this chapter can easily download the solutions PDF for RD Sharma Solutions. These answers are prepared by expert maths teachers, following the latest CBSE guidelines and updated marking schemes. The solutions are given step-by-step so students can understand the concepts more easily.

The RD Sharma Solutions for Class 12 are designed in a clear and simple way, closely following the CBSE Class 12 Maths syllabus. Students getting ready for their exams can use these PDFs as a helpful study tool to boost their preparation. To understand the topics in this chapter better, students can download the RD Sharma Solutions for Class 12 Chapter 5 Algebra of Matrices PDF from the links below.

RD Sharma Solutions for Class 12 – Algebra of Matrices

Algebra of matrices helps us work with matrices, which are special arrangements of numbers in rows and columns. These concepts are useful in higher-level maths like calculus and computer science. The RD Sharma Solutions for Class 12 explain important ideas like matrix addition, subtraction, multiplication, and the properties of matrices. This chapter teaches students how to solve problems about matrices with easy examples and clear explanations.

Download RD Sharma Class 12 Chapter Algebra of Matrices PDF Solutions

Students can quickly download the RD Sharma Solutions PDF. It covers all exercises, solved examples, and extra practice questions. These PDF solutions are perfect for quick revision and help students get a strong understanding of the algebra of matrices, preparing them well for board exams and entrance tests.

RD Sharma Solutions for Algebra of Matrices: Solved Questions

These RD Sharma Solutions for Class 12 will help students practice and become better at solving problems on algebra of matrices. Use these solutions to improve your skills and get ready for exams with confidence.

Question 1: Matrix Orders with Given Elements

Problem:

If a matrix has 8 elements, what are the possible orders it can have? What about a matrix with 5 elements?

Solution:

For 8 elements:

Step 1: If a matrix has order m×n, then total elements = m × n

Step 2: We need m × n = 8

Step 3: Find all factor pairs of 8:

• 1 × 8 = 8 → Order can be 1×8
• 2 × 4 = 8 → Order can be 2×4
• 4 × 2 = 8 → Order can be 4×2
• 8 × 1 = 8 → Order can be 8×1

Answer: Possible orders are 1×8, 2×4, 4×2, 8×1

For 5 elements:

Step 1: We need m × n = 5

Step 2: Since 5 is prime, its only factors are 1 and 5

Step 3: Factor pairs of 5:

• 1 × 5 = 5 → Order can be 1×5
• 5 × 1 = 5 → Order can be 5×1

Answer: Possible orders are 1×5, 5×1

Question 2: Constructing Matrices with Given Elements

Problem:

Construct a 2×3 matrix whose elements aij are given by: (i) aij = i × j; (ii) aij = 2i − j

Solution:

Part (i): aij = i × j

Step 1: For a 2×3 matrix, i ranges from 1 to 2, j ranges from 1 to 3

Step 2: Calculate each element:

• a11 = 1 × 1 = 1
• a12 = 1 × 2 = 2
• a13 = 1 × 3 = 3
• a21 = 2 × 1 = 2
• a22 = 2 × 2 = 4
• a23 = 2 × 3 = 6

Step 3: Form the matrix:

A = [1 2 3]
[2 4 6]

Part (ii): aij = 2i − j

Step 1: Calculate each element:

• a11 = 2(1) − 1 = 1
• a12 = 2(1) − 2 = 0
• a13 = 2(1) − 3 = −1
• a21 = 2(2) − 1 = 3
• a22 = 2(2) − 2 = 2
• a23 = 2(2) − 3 = 1

Step 2: Form the matrix:

A = [1 0 -1]
[3 2 1]

Question 3: Matrix Element Operations

Problem:

If A = [[2, 3, -5], [1, 3, 4], [0, 7, -2]] and B = [[2, -1], [-3, 4], [2, 1]], find (i) a22 + b21, (ii) a11 × b11 + a22 × b22

Solution:

Step 1: Identify the elements from matrices A and B:

From matrix A:

• a11 = 2 (element in row 1, column 1)
• a22 = 3 (element in row 2, column 2)

From matrix B:

• b11 = 2 (element in row 1, column 1)
• b21 = -3 (element in row 2, column 1)
• b22 = 4 (element in row 2, column 2)

Part (i): a22 + b21

Step 2: a22 + b21 = 3 + (-3) = 0

Part (ii): a11 × b11 + a22 × b22

Step 3: a11 × b11 + a22 × b22 = 2 × 2 + 3 × 4 = 4 + 12 = 16

Answers: (i) 0, (ii) 16

Question 4: Orders of Row and Column Matrices

Problem:

Let A be a matrix of order 3×4. If R1 denotes the first row of A and C2 denotes its second column, determine the orders of matrices R1 and C2.

Solution:

Step 1: Matrix A has order 3×4, meaning it has 3 rows and 4 columns

Step 2: Find order of R1 (first row):

• R1 contains all elements from the first row
• Since A has 4 columns, R1 has 4 elements
• R1 is a row matrix with 1 row and 4 columns
• Order of R1 = 1×4

Step 3: Find order of C2 (second column):

• C2 contains all elements from the second column
• Since A has 3 rows, C2 has 3 elements
• C2 is a column matrix with 3 rows and 1 column
• Order of C2 = 3×1

Answer: Order of R1 is 1×4, Order of C2 is 3×1

Question 5: Proving Matrix Identity

Problem:

If A and B are square matrices of the same order and AB = BA, show that (A + B)² = A² + 2AB + B²

Solution:

Step 1: Start with the left side: (A + B)²

(A + B)² = (A + B)(A + B)

Step 2: Apply distributive property of matrix multiplication:

(A + B)(A + B) = A(A + B) + B(A + B)

= AA + AB + BA + BB

= A² + AB + BA + B²

Step 3: Use the given condition AB = BA:

Since AB = BA, we can substitute:

A² + AB + BA + B² = A² + AB + AB + B²

= A² + 2AB + B²

Therefore, (A + B)² = A² + 2AB + B²

Note: This identity holds only when AB = BA (commutative matrices)

Question 6: Matrix Addition Associativity

Verify that (A + B) + C = A + (B + C) for given matrices A, B, and C

Solution:

Step 1: Let's use example matrices to verify:

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

Step 2: Calculate (A + B) + C:

A + B = [1+3 2+4] = [4 6]
[3+1 4+2] [4 6]

(A + B) + C = [4 6] + [5 6] = [4+5 6+6] = [9 12]
[4 6] [7 8] [4+7 6+8] [11 14]

Step 3: Calculate A + (B + C):

B + C = [3+5 4+6] = [8 10]
[1+7 2+8] [8 10]

A + (B + C) = [1 2] + [8 10] = [1+8 2+10] = [9 12]
[3 4] [8 10] [3+8 4+10] [11 14]

Step 4: Compare results:

(A + B) + C = [9 12] = A + (B + C)
[11 14]

General Proof: For any matrices A, B, C of the same order, each element satisfies:

[(A + B) + C]ij = (aij + bij) + cij = aij + (bij + cij) = [A + (B + C)]ij

Problem: Find the matrices X and Y such that: X + Y = A and X − Y = B, where A and B are given matrices

Solution:

Step 1: Set up the system of equations:

X + Y = A ... (1)
X − Y = B ... (2)

Step 2: Solve for X by adding equations (1) and (2):

(X + Y) + (X − Y) = A + B
X + Y + X − Y = A + B
2X = A + B
X = (A + B)/2

Step 3: Solve for Y by subtracting equation (2) from (1):

(X + Y) − (X − Y) = A − B
X + Y − X + Y = A − B
2Y = A − B
Y = (A − B)/2

Step 4: Example with specific matrices:

Let A = [4 6] and B = [2 2]
[8 10] [4 2]

X = (A + B)/2 = ([4 6] + [2 2])/2 = [6 8]/2 = [3 4]
([8 10] [4 2]) [12 12] [6 6]

Y = (A − B)/2 = ([4 6] − [2 2])/2 = [2 4]/2 = [1 2]
([8 10] [4 2]) [4 8] [2 4]

Verification:
X + Y = [3 4] + [1 2] = [4 6] = A ✓
[6 6] [2 4] [8 10]

X − Y = [3 4] − [1 2] = [2 2] = B ✓
[6 6] [2 4] [4 2]

Problem:

For what values of x, y, z does the equation A = B hold, given:
A = [[x−2, 3, 2z], [18, y+2, 6z]]
B = [[y, z, 6], [6y, x, 2y]]

Step 1: For matrices to be equal, corresponding elements must be equal:

Element (1,1): x − 2 = y ... (1)
Element (1,2): 3 = z ... (2)
Element (1,3): 2z = 6 ... (3)
Element (2,1): 18 = 6y ... (4)
Element (2,2): y + 2 = x ... (5)
Element (2,3): 6z = 2y ... (6)

Step 2: Solve equation (2):

z = 3

Step 3: Verify with equation (3):

2z = 2(3) = 6 ✓

Step 4: Solve equation (4):

18 = 6y
y = 3

Step 5: Verify with equation (6):

6z = 6(3) = 18
2y = 2(3) = 6
But 18 ≠ 6, so we need to recheck...

Step 6: Solve equation (4) again:

18 = 6y → y = 3

Step 7: From equation (6): 6z = 2y

6(3) = 2y → 18 = 2y → y = 9

Step 8: We have a contradiction. Let's solve systematically:

From (4): y = 3
From (2): z = 3
From (1): x − 2 = y = 3 → x = 5
From (5): y + 2 = x → 3 + 2 = 5 ✓

Step 9: Check equation (6):

6z = 6(3) = 18
2y = 2(3) = 6
Since 18 ≠ 6, the matrices cannot be equal with any real values of x, y, z.

Answer: No solution exists. The matrices A and B cannot be equal for any real values of x, y, z.

Question 9: Matrix Sum and Product

Compute the sum and product of matrices:
A = [[3, -2], [1, 3]] and B = [[-2, 4], [2, 6]]

Sum: A + B

Step 1: Add corresponding elements:

A + B = [3+(-2) -2+4] = [1 2]
[1+2 3+6 ] [3 9]

Product: A × B

Step 2: Multiply matrices using row-by-column method:

A × B = [[3, -2], [1, 3]] × [[-2, 4], [2, 6]]

Step 3: Calculate each element:

Element (1,1): 3×(-2) + (-2)×2 = -6 + (-4) = -10
Element (1,2): 3×4 + (-2)×6 = 12 + (-12) = 0
Element (2,1): 1×(-2) + 3×2 = -2 + 6 = 4
Element (2,2): 1×4 + 3×6 = 4 + 18 = 22

Step 4: Form the product matrix:

A × B = [-10 0]
[4 22]

Answers:
Sum: A + B = [1 2]
[3 9]
Product: A × B = [-10 0]
[4 22]

Question 10: Matrix Application - Shopkeeper Bills

Three shopkeepers A, B, and C buy different numbers of notebooks, pens, and pencils. Given the price of each, use matrix multiplication to find the bill for each shopkeeper.

Step 1: Set up the problem with example data:

Quantities purchased (Shopkeeper × Items):

Q = [10 5 20] ← A: 10 notebooks, 5 pens, 20 pencils
[15 8 25] ← B: 15 notebooks, 8 pens, 25 pencils
[12 6 30] ← C: 12 notebooks, 6 pens, 30 pencils

Prices per item:

P = [50] ← Notebook: ₹50
[20] ← Pen: ₹20
[5] ← Pencil: ₹5

Step 2: Calculate bills using matrix multiplication B = Q × P:

For Shopkeeper A:
Bill_A = 10×50 + 5×20 + 20×5 = 500 + 100 + 100 = ₹700

For Shopkeeper B:
Bill_B = 15×50 + 8×20 + 25×5 = 750 + 160 + 125 = ₹1035

For Shopkeeper C:
Bill_C = 12×50 + 6×20 + 30×5 = 600 + 120 + 150 = ₹870

Step 3: Matrix form of the solution:

Bills = [10 5 20] × [50] = [700]
[15 8 25] [20] [1035]
[12 6 30] [5] [870]

Answer:
Shopkeeper A's bill: ₹700
Shopkeeper B's bill: ₹1035
Shopkeeper C's bill: ₹870

General Method: If Q is the quantity matrix (m×n) where m = number of shopkeepers, n = number of items, and P is the price vector (n×1), then the bill vector B (m×1) is given by B = Q × P.