Ch 3: Matrices (NCERT Solutions)

Solution:

(i) A + B = 2+1   4+3
3-2   2+5 = 3   7
1   7

(ii) A - B = 2-1   4-3
3-(-2)   2-5 = 1   1
5   -3

(iii) 3A - C = 6   12
9   6 - -2   5
3   4 = 6+2   12-5
9-3   6-4 = 8   7
6   2

(iv) AB = 2(1)+4(-2)   2(3)+4(5)
3(1)+2(-2)   3(3)+2(5) = 2-8   6+20
3-4   9+10 = -6   26
-1   19

(v) BA = 1(2)+3(3)   1(4)+3(2)
-2(2)+5(3)   -2(4)+5(2) = 2+9   4+6
-4+15   -8+10 = 11   10
11   2

Solution:

(i) a   b
-b   a + a   b
b   a = a+a   b+b
-b+b   a+a = 2a   2b
0   2a

(ii) a²+b²   b²+c²
a²+c²   a²+b² + 2ab   2bc
-2ac   -2ab = a²+b²+2ab   b²+c²+2bc
a²+c²-2ac   a²+b²-2ab = (a+b)²   (b+c)²
(a-c)²   (a-b)²

(iii) (Simple addition of 3x3 matrices, adding corresponding elements) = 11   11   0
16   5   21
5   10   9

(iv) cos²x   sin²x
sin²x   cos²x + sin²x   cos²x
cos²x   sin²x = cos²x+sin²x   sin²x+cos²x
sin²x+cos²x   cos²x+sin²x = 1   1
1   1

Solution:

(i) a   b
-b   a a   -b
b   a = a²+b²   -ab+ab
-ab+ab   b²+a² = a²+b²   0
0   a²+b²

(ii) 1
2
3 (3x1 matrix) × 2   3   4 (1x3 matrix) = (3x3 matrix) = 2   3   4
4   6   8
6   9   12

Solution:

A + B = 1+3   2-1   -3+2
5+4   0+2   2+5
1+2   -1+0   1+3 = 4   1   -1
9   2   7
3   -1   4

B - C = 3-4   -1-1   2-2
4-0   2-3   5-2
2-1   0+2   3-3 = -1   -2   0
4   -1   3
1   2   0

LHS: A + (B - C) = 1-1   2-2   -3+0
5+4   0-1   2+3
1+1   -1+2   1+0 = 0   0   -3
9   -1   5
2   1   1

RHS: (A + B) - C = 4-4   1-1   -1-2
9-0   2-3   7-2
3-1   -1+2   4-3 = 0   0   -3
9   -1   5
2   1   1
LHS = RHS. Verified.

Solution:

Multiply matrix A by 3 (denominators cancel out):
3A = 2   3   5
1   2   4
7   6   2

Multiply matrix B by 5:
5B = 2   3   5
1   2   4
7   6   2

3A - 5B = 2-2   3-3   5-5
1-1   2-2   4-4
7-7   6-6   2-2 = 0   0   0
0   0   0
0   0   0 (Null Matrix O)

Solution:

Multiply the scalars into the matrices:
= cos²θ   cosθsinθ
-cosθsinθ   cos²θ + sin²θ   -sinθcosθ
sinθcosθ   sin²θ

Add the matrices:
= cos²θ+sin²θ   cosθsinθ-sinθcosθ
-cosθsinθ+sinθcosθ   cos²θ+sin²θ

Since cos²θ + sin²θ = 1, we get:
= 1   0
0   1 (Identity Matrix I)

Solution:

Add the two equations: (X+Y) + (X-Y) = 2X
2X = 7+3   0+0
2+0   5+3 = 10   0
2   8
X = 5   0
1   4

Subtract the two equations: (X+Y) - (X-Y) = 2Y
2Y = 7-3   0-0
2-0   5-3 = 4   0
2   2
Y = 2   0
1   1

Quick Solutions:

Q8. Find X if Y = 3   2
1   4 and 2X + Y = 1   0
-3   2.
2X = 1   0
-3   2 - 3   2
1   4 = -2   -2
-4   -2 → X = -1   -1
-2   -1

Q9. Find x, y if 21   3
0   x + y   0
1   2 = 5   6
1   8.
2+y   6
1   2x+2 = 5   6
1   8
2+y = 5 → y = 3. 2x+2 = 8 → x = 3.

Q10, Q11, Q12. Use the same method of scalar multiplication and element-wise equating.
Q12: 3x   y
z   w = x+4   6+x+y
-1+z+w   2w+3
3x = x + 4 → x = 2
3w = 2w + 3 → w = 3
3y = 6 + x + y → 2y = 6 + 2 → y = 4
3z = -1 + z + w → 2z = -1 + 3 → z = 1

Solution:

F(x) × F(y) = cos x   -sin x   0
sin x   cos x   0
0   0   1 × cos y   -sin y   0
sin y   cos y   0
0   0   1

Multiplying row by column:
R1C1 = cos(x)cos(y) - sin(x)sin(y) = cos(x+y)
R1C2 = -cos(x)sin(y) - sin(x)cos(y) = -sin(x+y)
R2C1 = sin(x)cos(y) + cos(x)sin(y) = sin(x+y)
R2C2 = -sin(x)sin(y) + cos(x)cos(y) = cos(x+y)

Resulting Matrix = cos(x+y)   -sin(x+y)   0
sin(x+y)   cos(x+y)   0
0   0   1 = F(x+y). Proved.

Solution:

First, find A² = A × A:
A² = 2(2)+0+1(1)   2(0)+0-1(1)   2(1)+0+1(0)
2(2)+1(2)+3(1)   2(0)+1(1)-3(1)   2(1)+1(3)+0
1(2)-1(2)+0   0-1(1)+0   1(1)-3+0 = 5   -1   2
9   -2   5
0   -1   -2

Now calculate A² - 5A + 6I:
= 5   -1   2
9   -2   5
0   -1   -2 - 10   0   5
10   5   15
5   -5   0 + 6   0   0
0   6   0
0   0   6

Adding the elements:
= 1   -1   -3
-1   -1   -10
-5   4   4

Solution:

Find A² = 3   -2
4   -2 3   -2
4   -2 = 9-8   -6+4
12-8   -8+4 = 1   -2
4   -4

Set up the equation: A² = kA - 2I
1   -2
4   -4 = 3k   -2k
4k   -2k - 2   0
0   2

1   -2
4   -4 = 3k-2   -2k
4k   -2k-2

Equate any element (e.g., R2C1): 4k = 4 → k = 1.
(Checking R1C1: 3(1)-2 = 1, it holds true).

Solution:

Let tan(α/2) = t. Use trig identities: cosα = (1-t²)/(1+t²) and sinα = 2t/(1+t²).

LHS = I + A = 1   -t
t   1

RHS = (I - A) × Matrix
I - A = 1   t
-t   1
Multiply by (1-t²)/(1+t²)   -2t/(1+t²)
2t/(1+t²)   (1-t²)/(1+t²)

R1C1 = 1((1-t²)/(1+t²)) + t(2t/(1+t²)) = (1-t²+2t²)/(1+t²) = (1+t²)/(1+t²) = 1
R1C2 = 1(-2t/(1+t²)) + t((1-t²)/(1+t²)) = (-2t+t-t³)/(1+t²) = -t(1+t²)/(1+t²) = -t
R2C1 = -t((1-t²)/(1+t²)) + 1(2t/(1+t²)) = (-t+t³+2t)/(1+t²) = t(1+t²)/(1+t²) = t
R2C2 = -t(-2t/(1+t²)) + 1((1-t²)/(1+t²)) = (2t²+1-t²)/(1+t²) = (1+t²)/(1+t²) = 1

RHS = 1   -t
t   1 = LHS. Proved.

Solution setup:

Q19. Trust Fund: Rs 30,000 is to be invested in two types of bonds paying 5% and 7% interest.
Let investment matrix = x   30000-x (1x2 matrix).
Interest matrix = 0.05
0.07 (2x1 matrix).
For Rs 1800 total interest: x   30000-x 0.05
0.07 = 1800.
0.05x + 2100 - 0.07x = 1800 → 0.02x = 300 → x = 15,000. So Rs 15,000 in both.

Q20. Bookshop: Chemistry (10 dz), Physics (8 dz), Eco (10 dz). Prices: Rs 80, 60, 40.
Quantity matrix Q = 120   96   120 (Converted dozens to units).
Price matrix P = 80
60
40.
Total amount = Q × P = 120(80) + 96(60) + 120(40) = 9600 + 5760 + 4800 = Rs 20,160.

Solution:

Given orders: X(2×n), Y(3×k), Z(2×p), W(n×3), P(p×k).

Q21: Restriction on n, k, p so that PY + WY will be defined.
For P(p×k) × Y(3×k) to be defined, columns of P = rows of Y → k = 3.
Order of PY = p×k = p×3.
For W(n×3) × Y(3×k) to be defined, col of W = row of Y (3=3, true).
Order of WY = n×k = n×3.
For PY + WY, orders must be equal: p×3 = n×3 → p = n.
Answer: (A) k = 3, p = n

Q22: If n = p, then the order of the matrix 7X - 5Z is:
X is 2×n. Z is 2×p.
Since n = p, they have the same order (2×n or 2×p).
The result of subtraction maintains the same order.
Answer: (B) 2 × n

Solution:

(i) Let A = 5
1/2
-1. The transpose changes columns into rows.
A' = 5   1/2   -1

(ii) Let A = 1   -1
2   3. Interchanging rows and columns:
A' = 1   2
-1   3

(iii) Let A = -1   5   6
√3   5   6
2   3   -1. Interchanging rows and columns:
A' = -1   √3   2
5   5   3
6   6   -1

Solution:

(i) Verification of (A + B)' = A' + B'
LHS: First find A + B = -5   3   -2
6   9   9
-1   4   2
(A + B)' = -5   6   -1
3   9   4
-2   9   2

RHS: Find A' and B' separately and add them.
A' = -1   5   -2
2   7   1
3   9   1,    B' = -4   1   1
1   2   3
-5   0   1
A' + B' = -5   6   -1
3   9   4
-2   9   2
LHS = RHS. Verified.

(ii) Verification of (A - B)' = A' - B'
LHS: Find A - B = 3   1   8
4   5   9
-3   -2   0
(A - B)' = 3   4   -3
1   5   -2
8   9   0

RHS: A' - B' = -1-(-4)   5-1   -2-1
2-1   7-2   1-3
3-(-5)   9-0   1-1 = 3   4   -3
1   5   -2
8   9   0
LHS = RHS. Verified.

Solution:

First, we need to find A from A'. (A')' = A.
A = 3   -1   0
4   2   1

LHS: (A + B) = 3-1   -1+2   0+1
4+1   2+2   1+3 = 2   1   1
5   4   4
(A + B)' = 2   5
1   4
1   4

RHS: We have A'. Find B' = -1   1
2   2
1   3
A' + B' = 3-1   4+1
-1+2   2+2
0+1   1+3 = 2   5
1   4
1   4
LHS = RHS. Verified.

Solution:

Find A from A': A = -2   1
3   2

Calculate 2B = -2   0
2   4

Find A + 2B = -2-2   1+0
3+2   2+4 = -4   1
5   6

Now, take the transpose: (A + 2B)' = -4   5
1   6

Solution:

LHS: (AB)'
First find AB (multiplying a 3x1 and 1x3 matrix gives a 3x3 matrix):
AB = 1(-1)   1(2)   1(1)
-4(-1)   -4(2)   -4(1)
3(-1)   3(2)   3(1) = -1   2   1
4   -8   -4
-3   6   3
(AB)' = -1   4   -3
2   -8   6
1   -4   3

RHS: B'A'
B' = -1
2
1, A' = 1   -4   3
B'A' = -1(1)   -1(-4)   -1(3)
2(1)   2(-4)   2(3)
1(1)   1(-4)   1(3) = -1   4   -3
2   -8   6
1   -4   3
LHS = RHS. Verified.

Solution:

A' = cos α   -sin α
sin α   cos α

Multiply A' × A:
cos α   -sin α
sin α   cos α × cos α   sin α
-sin α   cos α

R1C1 = cos²α + sin²α = 1
R1C2 = cosαsinα - sinαcosα = 0
R2C1 = sinαcosα - cosαsinα = 0
R2C2 = sin²α + cos²α = 1

A'A = 1   0
0   1 = I. Verified.

Solution:

A matrix is symmetric if A' = A.
Taking the transpose of A:
A' = 1   -1   5
-1   2   1
5   1   3
Since A' = A, the given matrix is symmetric.

Solution:

A' = 1   6
5   7

(i) Let P = A + A' = 1+1   5+6
6+5   7+7 = 2   11
11   14
Taking transpose: P' = 2   11
11   14. Since P = P', it is a symmetric matrix.

(ii) Let Q = A - A' = 1-1   5-6
6-5   7-7 = 0   -1
1   0
Taking transpose: Q' = 0   1
-1   0 = -(0   -1
1   0) = -Q.
Since Q' = -Q, it is a skew symmetric matrix.

Solution:

A' = 0   -a   -b
a   0   -c
b   c   0

Calculating 1/2(A + A'):
A + A' = 0+0   a-a   b-b
-a+a   0+0   c-c
-b+b   -c+c   0+0 = 0   0   0
0   0   0
0   0   0
Therefore, 1/2(A + A') = O (Null Matrix).

Calculating 1/2(A - A'):
A - A' = 0-0   a-(-a)   b-(-b)
-a-a   0-0   c-(-c)
-b-b   -c-c   0-0 = 0   2a   2b
-2a   0   2c
-2b   -2c   0
Therefore, 1/2(A - A') = 0   a   b
-a   0   c
-b   -c   0 = A.

Solution:

Formula: Any square matrix A can be expressed as P + Q, where P = 1/2(A+A') [Symmetric] and Q = 1/2(A-A') [Skew-Symmetric].

(i) For A = 3   5
1   -1:
A' = 3   1
5   -1
P = 1/2(A+A') = 1/26   6
6   -2 = 3   3
3   -1
Q = 1/2(A-A') = 1/20   4
-4   0 = 0   2
-2   0
Verification: P + Q = 3   3
3   -1 + 0   2
-2   0 = A.

(ii) For A = 6   -2   2
-2   3   -1
2   -1   3:
A' = 6   -2   2
-2   3   -1
2   -1   3
Notice that A' = A. The matrix is already symmetric.
P = 1/2(A+A') = A = 6   -2   2
-2   3   -1
2   -1   3
Q = 1/2(A-A') = 0   0   0
0   0   0
0   0   0
Sum = P + Q = A.

Solution:

Given: A' = A and B' = B. We need to check the nature of (AB - BA).
Take the transpose: (AB - BA)' = (AB)' - (BA)'
Using Reversal Law (XY)' = Y'X':
= B'A' - A'B'
= BA - AB     (Since A'=A and B'=B)
= -(AB - BA)
Since the transpose of the matrix equals its negative, the matrix is skew-symmetric.
Correct Answer: (A) Skew symmetric matrix

Solution:

Find A': A' = cos α   sin α
-sin α   cos α

Add A and A':
A + A' = cos α + cos α   -sin α + sin α
sin α - sin α   cos α + cos α = 2cos α   0
0   2cos α

We are given that A + A' = I. So equate the matrix to 1   0
0   1.
2cos α = 1
cos α = 1/2
We know that cos(π/3) = 1/2. Therefore, α = π/3.

Correct Answer: (B) π/3

Solution:

We write A = IA → 1   -1
2   3 = 1   0
0   1 A

Step 1: R₂ → R₂ - 2R₁
1   -1
0   5 = 1   0
-2   1 A

Step 2: R₂ → R₂ / 5
1   -1
0   1 = 1   0
-2/5   1/5 A

Step 3: R₁ → R₁ + R₂
1   0
0   1 = 3/5   1/5
-2/5   1/5 A

Hence, A⁻¹ = 3/5   1/5
-2/5   1/5

Solution:

A = IA → 2   1
1   1 = 1   0
0   1 A

Step 1: R₁ ↔ R₂ (Interchange rows to get 1 at a₁₁)
1   1
2   1 = 0   1
1   0 A

Step 2: R₂ → R₂ - 2R₁
1   1
0   -1 = 0   1
1   -2 A

Step 3: R₂ → (-1)R₂
1   1
0   1 = 0   1
-1   2 A

Step 4: R₁ → R₁ - R₂
1   0
0   1 = 1   -1
-1   2 A

Hence, A⁻¹ = 1   -1
-1   2

Solution:

A = IA → 1   3
2   7 = 1   0
0   1 A

R₂ → R₂ - 2R₁:   1   3
0   1 = 1   0
-2   1 A

R₁ → R₁ - 3R₂:   1   0
0   1 = 7   -3
-2   1 A

Hence, A⁻¹ = 7   -3
-2   1

Direct Verified Answers:

(Board Tip: For a 2x2 matrix a   b
c   d, the inverse is d   -b
-c   a divided by determinant (ad-bc). You can use this to quickly verify your row operations!)

• Q4. A = 2   3
  5   7 → A⁻¹ = -7   3
  5   -2
• Q5. A = 2   1
  7   4 → A⁻¹ = 4   -1
  -7   2
• Q6. A = 2   5
  1   3 → A⁻¹ = 3   -5
  -1   2
• Q7. A = 3   1
  5   2 → A⁻¹ = 2   -1
  -5   3
• Q8. A = 4   5
  3   4 → A⁻¹ = 4   -5
  -3   4
• Q9. A = 3   10
  2   7 → A⁻¹ = 7   -10
  -2   3
• Q10. A = 3   -1
  -4   2 → A⁻¹ = 1   1/2
  2   3/2
• Q11. A = 2   -6
  1   -2 → A⁻¹ = -1   3
  -1/2   1
• Q13. A = 2   -3
  -1   2 → A⁻¹ = 2   3
  1   2

Solution:

For Q12: Let A = IA → 6   -3
-2   1 = 1   0
0   1 A
Apply R₁ → R₁ + 3R₂ :
0   0
-2   1 = 1   3
0   1 A
Since all elements in the first row of the LHS matrix are zero, A⁻¹ does not exist.

For Q14: Let B = IB → 2   1
4   2 = 1   0
0   1 B
Apply R₂ → R₂ - 2R₁ :
2   1
0   0 = 1   0
-2   1 B
Since all elements in the second row of the LHS matrix are zero, B⁻¹ does not exist.

Solution Summary:

We write A = IA and apply row operations to make LHS an identity matrix.

1. R₁ → R₁ + R₂ - R₃ → Gets a '1' in a₁₁ position.
2. R₂ → R₂ - 2R₁ and R₃ → R₃ - 3R₁ → Makes column 1 zeros below a₁₁.
3. R₂ → R₂ / 5 → Gets a '1' in a₂₂ position.
4. R₁ → R₁ + R₂ and R₃ → R₃ + 5R₂ → Clears column 2.
5. R₃ → R₃ / (-2) → Gets a '1' in a₃₃ position.
6. R₁ → R₁ - 2R₃ and R₂ → R₂ + R₃ → Clears column 3.

Final Inverse Matrix: A⁻¹ = -2/5   0   3/5
-1/5   1/5   0
2/5   1/5   -2/5

Solution Summary:

Using A = IA, the key row operations sequence is:

1. R₂ → R₂ + 3R₁ and R₃ → R₃ - 2R₁ → Clears column 1 below the pivot.
2. R₂ → R₂ / 9 → Normalizes row 2.
3. R₁ → R₁ - 3R₂ and R₃ → R₃ + R₂ → Clears column 2.
4. Multiply R₃ by appropriate reciprocal to make a₃₃ = 1.
5. Clear column 3 using R₃.

Final Inverse Matrix: A⁻¹ = 1   -2/5   -3/5
-2/5   4/25   11/25
-3/5   1/25   9/25

Solution Summary:

Using A = IA, the key row operations sequence is:

1. R₁ → 3R₁ - R₂ → Gets a '1' in a₁₁ position.
2. R₂ → R₂ - 5R₁ → Clears column 1.
3. R₂ → R₂ / 6 → Normalizes row 2.
4. R₃ → R₃ - R₂ → Clears column 2 below diagonal.
5. R₃ → R₃ × 6/13 → Gets '1' in a₃₃.
6. Back-substitute to clear column 3.

Final Inverse Matrix: A⁻¹ = 3   -1   1
-15   6   -5
5   -2   2

Solution:

By the fundamental definition of invertible matrices, a square matrix A has an inverse B if and only if their product (in any order) yields the Identity Matrix (I).

Therefore, AB = I and BA = I.

Correct Answer: (D) AB = BA = I

Solution (By Principle of Mathematical Induction):

Step 1 (Base Case, n=1):
LHS = (aI + bA)¹ = aI + bA.
RHS = a¹I + 1(a⁰)bA = aI + (1)bA = aI + bA.
Since LHS = RHS, it is true for n=1.

Step 2 (Assumption): Let it be true for n = k.
(aI + bA)ᵏ = aᵏI + kaᵏ⁻¹bA    --- (Equation 1)

Step 3 (Prove for n = k+1):
We need to prove: (aI + bA)ᵏ⁺¹ = aᵏ⁺¹I + (k+1)aᵏbA
LHS = (aI + bA)ᵏ⁺¹ = (aI + bA)ᵏ × (aI + bA)
Substituting from Equation 1:
= (aᵏI + kaᵏ⁻¹bA) × (aI + bA)
= aᵏ⁺¹I² + aᵏb(IA) + kaᵏb(AI) + kaᵏ⁻¹b²A²

We know that I² = I, and IA = AI = A.
Now find A²: A² = 0   1
0   0 × 0   1
0   0 = 0   0
0   0 = O (Null Matrix).

Substitute these values back:
LHS = aᵏ⁺¹I + aᵏbA + kaᵏbA + O
= aᵏ⁺¹I + aᵏbA(1 + k)
= aᵏ⁺¹I + (k+1)aᵏbA = RHS.
By PMI, the statement is true for all n ∈ N. Proved.

Solution (By PMI):

For n=1:
RHS element = 3¹⁻¹ = 3⁰ = 1. Therefore A¹ matches the given matrix A. True for n=1.

Assume true for n=k:
Aᵏ = 3ᵏ⁻¹   3ᵏ⁻¹   3ᵏ⁻¹
3ᵏ⁻¹   3ᵏ⁻¹   3ᵏ⁻¹
3ᵏ⁻¹   3ᵏ⁻¹   3ᵏ⁻¹

Prove for n=k+1:
Aᵏ⁺¹ = Aᵏ × A
Multiplying row by column, each element will be:
(3ᵏ⁻¹ × 1) + (3ᵏ⁻¹ × 1) + (3ᵏ⁻¹ × 1) = 3(3ᵏ⁻¹) = 3ᵏ.
So, Aᵏ⁺¹ = 3ᵏ   3ᵏ   3ᵏ
3ᵏ   3ᵏ   3ᵏ
3ᵏ   3ᵏ   3ᵏ, which matches the formula 3⁽ᵏ⁺¹⁾⁻¹. Proved.

Solution (By PMI):

For n=1:
A¹ = 1+2(1)   -4(1)
1   1-2(1) = 3   -4
1   -1 = A. True for n=1.

Assume true for n=k: Aᵏ = 1+2k   -4k
k   1-2k.

Prove for n=k+1:
Aᵏ⁺¹ = Aᵏ × A = 1+2k   -4k
k   1-2k × 3   -4
1   -1
R1C1 = 3(1+2k) - 4k = 3 + 6k - 4k = 3 + 2k = 1 + 2(k+1)
R1C2 = -4(1+2k) + 4k = -4 - 8k + 4k = -4 - 4k = -4(k+1)
R2C1 = 3k + 1 - 2k = k + 1
R2C2 = -4k - (1-2k) = -4k - 1 + 2k = -1 - 2k = 1 - 2(k+1)
This matches the exact structure for n=k+1. Proved.

Solution:

Given: A and B are symmetric, so A' = A and B' = B.
We need to find the transpose of (AB - BA).
(AB - BA)' = (AB)' - (BA)'     [Using property: (X-Y)' = X' - Y']
= B'A' - A'B'     [Using Reversal Law: (XY)' = Y'X']
= BA - AB     [Since A'=A and B'=B]
= -(AB - BA)
Since the transpose of the matrix is equal to its negative, AB - BA is a skew symmetric matrix.

Solution:

Let's take the transpose of B'AB:
(B'AB)' = [B'(AB)]' = (AB)'(B')' = (B'A')B = B'A'B.

Case 1: If A is symmetric (A' = A)
Substitute A' with A in our derived equation:
(B'AB)' = B'(A)B = B'AB. Since its transpose equals itself, B'AB is symmetric.

Case 2: If A is skew symmetric (A' = -A)
Substitute A' with -A in our derived equation:
(B'AB)' = B'(-A)B = -B'AB. Since its transpose equals its negative, B'AB is skew symmetric.

Solution:

A' = Transpose of A = 0   x   x
2y   y   -y
z   -z   z.
Multiply A' × A and equate it to the Identity Matrix I = 1   0   0
0   1   0
0   0   1.

Multiply the diagonal elements to form equations:
R1C1: 0(0) + x(x) + x(x) = 2x² = 1 → x² = 1/2 → x = ±1/√2
R2C2: (2y)(2y) + y(y) + (-y)(-y) = 4y² + y² + y² = 6y² = 1 → y = ±1/√6
R3C3: z(z) + (-z)(-z) + z(z) = z² + z² + z² = 3z² = 1 → z = ±1/√3

Solution:

Multiply the 1st (1x3) and 2nd (3x3) matrices:
[1(1)+2(2)+1(1)    1(2)+2(0)+1(0)    1(0)+2(1)+1(2)]
= 6   2   4

Now multiply this resulting 1x3 matrix with the 3rd (3x1) matrix:
6   2   4 0
2
x = [6(0) + 2(2) + 4(x)] = [4 + 4x]

Equate the final 1x1 matrix to the zero matrix:
4 + 4x = 0 → 4x = -4 → x = -1

Solution:

First, find A² = A × A:
A² = 3   1
-1   2 × 3   1
-1   2 = 9-1   3+2
-3-2   -1+4 = 8   5
-5   3

Calculate 5A = 15   5
-5   10, and 7I = 7   0
0   7

Substitute into LHS: A² - 5A + 7I
= 8   5
-5   3 - 15   5
-5   10 + 7   0
0   7
= 8-15+7   5-5+0
-5-(-5)+0   3-10+7 = 0   0
0   0 = O. Proved.

Solution:

Multiply the 1st and 2nd matrix:
[x(1)+(-5)0+(-1)2    x(0)+(-5)2+(-1)0    x(2)+(-5)1+(-1)3]
= x-2   -10   2x-8

Now multiply with the 3rd matrix:
x-2   -10   2x-8 x
4
1
= [(x-2)x + (-10)(4) + (2x-8)(1)]
= x² - 2x - 40 + 2x - 8 = x² - 48

Equate to zero: x² - 48 = 0 → x² = 48 → x = ±4√3

Solution:

(a) Total Revenue:
Quantity Matrix Q = 10000   2000   18000
6000   20000   8000
Selling Price SP = 2.50
1.50
1.00
Revenue = Q × SP = 10000(2.5)+2000(1.5)+18000(1)
6000(2.5)+20000(1.5)+8000(1) = 46000
53000
Market I Revenue = Rs 46,000. Market II Revenue = Rs 53,000.

(b) Gross Profit:
Cost Price CP = 2.00
1.00
0.50
Total Cost = Q × CP = 10000(2)+2000(1)+18000(0.5)
6000(2)+20000(1)+8000(0.5) = 31000
36000
Profit = Revenue - Cost = 46000 - 31000
53000 - 36000 = 15000
17000
Gross Profit for Market I = Rs 15,000. Market II = Rs 17,000.

Solution:

Matrix (2×3) is given on the RHS. The 2nd matrix on the LHS is (2×3). For the product to be valid and result in a 2x3 matrix, X must be a (2×2) matrix.
Let X = a   b
c   d

a   b
c   d 1   2   3
4   5   6 = a+4b   2a+5b   3a+6b
c+4d   2c+5d   3c+6d

Equating elements with the RHS matrix:
a + 4b = -7
2a + 5b = -8
Multiply first equation by 2: 2a + 8b = -14. Subtracting gives 3b = -6 → b = -2. Putting b in eq 1 gives a = 1.

c + 4d = 2
2c + 5d = 4
Multiply first equation by 2: 2c + 8d = 4. Subtracting gives 3d = 0 → d = 0. Putting d in eq 1 gives c = 2.

Therefore, Matrix X = 1   -2
2   0

Solution (By PMI):

Part 1: Prove ABⁿ = BⁿA
For n=1, AB¹ = B¹A → AB = BA (Given condition). True.
Assume true for n=k: ABᵏ = BᵏA.
Prove for n=k+1:
LHS = ABᵏ⁺¹ = A(Bᵏ × B) = (ABᵏ)B
Substitute using our assumption: = (BᵏA)B
Use associativity and the given AB=BA: = Bᵏ(AB) = Bᵏ(BA) = (BᵏB)A = Bᵏ⁺¹A = RHS. Proved.

Part 2: Prove (AB)ⁿ = AⁿBⁿ
For n=1, (AB)¹ = A¹B¹. True.
Assume true for n=k: (AB)ᵏ = AᵏBᵏ.
Prove for n=k+1:
LHS = (AB)ᵏ⁺¹ = (AB)ᵏ(AB)
Substitute using assumption: = (AᵏBᵏ)(AB) = Aᵏ(BᵏA)B
Using the result from Part 1 (BᵏA = ABᵏ): = Aᵏ(ABᵏ)B = (AᵏA)(BᵏB) = Aᵏ⁺¹Bᵏ⁺¹ = RHS. Proved.

Solution:

Find A² = α   β
γ   -α α   β
γ   -α = α²+βγ   αβ-αβ
αγ-αγ   βγ+α² = α²+βγ   0
0   α²+βγ

Since A² = I, we equate it to 1   0
0   1.
α² + βγ = 1 → Move 1 to RHS: 1 - α² - βγ = 0.

Correct Answer: (C) 1 - α² - βγ = 0

Solution:

Since A is symmetric, A' = A.
Since A is skew symmetric, A' = -A.
Equating both: A = -A → A + A = O → 2A = O → A = O.
Correct Answer: (B) A is a zero matrix.

Solution:

Use the binomial expansion for matrices (this is valid because I and A commute, i.e., IA = AI):
(I + A)³ = I³ + 3I²A + 3IA² + A³
= I + 3A + 3A² + A³

We are given A² = A. Therefore, A³ = A² × A = A × A = A² = A.
Substitute these simplified values back:
(I + A)³ = I + 3A + 3(A) + A = I + 7A.

Now evaluate the full expression: (I + A)³ - 7A = (I + 7A) - 7A = I.

Correct Answer: (C) I