Ch 4: Determinants (NCERT Solutions)

Solutions:

Q1. x   a   x+a
y   b   y+b
z   c   z+c = 0
Apply C₁ → C₁ + C₂: C₁ becomes (x+a, y+b, z+c). Now C₁ and C₃ are identical. Since two columns are identical, Determinant = 0.

Q2. a-b   b-c   c-a
b-c   c-a   a-b
c-a   a-b   b-c = 0
Apply C₁ → C₁ + C₂ + C₃: The elements of C₁ become (a-b+b-c+c-a) = 0. Since an entire column is 0, Determinant = 0.

Q3. 2   7   65
3   8   75
5   9   86 = 0
Apply C₃ → C₃ - 9C₂: C₃ becomes (65-63=2, 75-72=3, 86-81=5). Now C₁ and C₃ are identical. Determinant = 0.

Q4. 1   bc   a(b+c)
1   ca   b(c+a)
1   ab   c(a+b) = 0
Apply C₃ → C₃ + C₂: C₃ becomes (ab+bc+ca) for all rows. Take (ab+bc+ca) common from C₃. Then C₁ and C₃ become identical (both are 1,1,1). Determinant = 0.

Q5. Show that b+c   q+r   y+z
c+a   r+p   z+x
a+b   p+q   x+y = 2a   p   x
b   q   y
c   r   z
Apply R₁ → R₁ + R₂ + R₃: R₁ becomes 2(a+b+c), 2(p+q+r), 2(x+y+z). Take 2 common.
Apply R₂ → R₂ - R₁ and R₃ → R₃ - R₁. Then R₁ → R₁ + R₂ + R₃ gives the required RHS. Proved.

Solution:

Step 1: Apply R₁ → R₁ - R₂ and R₂ → R₂ - R₃
0   a-b   a²-b²
0   b-c   b²-c²
1   c   c²

Step 2: Take (a-b) common from R₁ and (b-c) common from R₂.
(a-b)(b-c) 0   1   a+b
0   1   b+c
1   c   c²

Step 3: Expand along C₁:
= (a-b)(b-c) [ 1 × (1(b+c) - 1(a+b)) ]
= (a-b)(b-c) [ b + c - a - b ]
= (a-b)(b-c)(c-a) = RHS. Proved.

Solution:

Apply C₁ → C₁ + C₂ + C₃:
Every element in C₁ becomes 5x+4. Take (5x+4) common.

(5x+4) 1   2x   2x
1   x+4   2x
1   2x   x+4

Apply R₂ → R₂ - R₁ and R₃ → R₃ - R₁ to create zeros:
(5x+4) 1   2x   2x
0   4-x   0
0   0   4-x

Expand along C₁:
= (5x+4) [ 1 × ((4-x)(4-x) - 0) ]
= (5x+4)(4-x)² = RHS. Proved.

Solution:

Q15. Let A be a square matrix of order 3x3, then |kA| is equal to:
Property: If A is of order n×n, then |kA| = kⁿ|A|. Here n = 3.
Correct Answer: (C) k³|A|

Q16. Which of the following is correct?
(A) Determinant is a square matrix.
(B) Determinant is a number associated to a matrix.
(C) Determinant is a number associated to a square matrix.
Concept: A determinant is uniquely defined ONLY for square matrices.
Correct Answer: (C)

Solution:

(i) Vertices: (1, 0), (6, 0), (4, 3)
Δ = 1/2 1   0   1
6   0   1
4   3   1
Expand along C2 (since it has two zeros):
Δ = 1/2 [ -0 + 0 - 3(1 - 6) ]
Δ = 1/2 [ -3(-5) ] = 15/2
Area = 15/2 sq. units.

(ii) Vertices: (2, 7), (1, 1), (10, 8)
Δ = 1/2 2   7   1
1   1   1
10   8   1
Expand along R1:
Δ = 1/2 [ 2(1 - 8) - 7(1 - 10) + 1(8 - 10) ]
Δ = 1/2 [ 2(-7) - 7(-9) + 1(-2) ]
Δ = 1/2 [ -14 + 63 - 2 ] = 1/2 [ 47 ]
Area = 47/2 sq. units.

Solution:

Points are collinear if the area of the triangle formed by them is zero.

Δ = 1/2 a   b+c   1
b   c+a   1
c   a+b   1

Apply property C₂ → C₂ + C₁:
Δ = 1/2 a   a+b+c   1
b   a+b+c   1
c   a+b+c   1

Take (a+b+c) common from C₂:
Δ = (a+b+c)/2 a   1   1
b   1   1
c   1   1

Since C₂ and C₃ are identical, the determinant value is 0.
Δ = (a+b+c)/2 × 0 = 0.
Since Area = 0, the points are collinear. Proved.

Solution:

We are given Area = 4. When solving for a variable, we must take both positive and negative values of the area.
1/2 k   0   1
4   0   1
0   2   1 = ±4

Multiply by 2:
k   0   1
4   0   1
0   2   1 = ±8

Expand along C2:
-2(k(1) - 4(1)) = ±8
-2(k - 4) = ±8
k - 4 = ±(-4)

Case 1: k - 4 = 4 → k = 8
Case 2: k - 4 = -4 → k = 0
Values of k are 0, 8.

Solution:

Let P(x, y) be any general point on the line joining A(1, 2) and B(3, 6).
Since P, A, and B are collinear, the area of the triangle formed by them must be 0.

1/2 x   y   1
1   2   1
3   6   1 = 0

Expand along R1:
x(2 - 6) - y(1 - 3) + 1(6 - 6) = 0
x(-4) - y(-2) + 0 = 0
-4x + 2y = 0 → 2y = 4x → y = 2x

Equation of the line is y = 2x.

Solution:

1/2 2   -6   1
5   4   1
k   4   1 = ±35

2   -6   1
5   4   1
k   4   1 = ±70

Expand along R1:
2(4 - 4) - (-6)(5 - k) + 1(20 - 4k) = ±70
0 + 6(5 - k) + 20 - 4k = ±70
30 - 6k + 20 - 4k = ±70
50 - 10k = ±70

Case 1: 50 - 10k = 70 → -10k = 20 → k = -2
Case 2: 50 - 10k = -70 → -10k = -120 → k = 12

Correct Answer: (D) 12, -2

Solution:

Minors:
M₁₁ (delete R1, C1) = 3
M₁₂ (delete R1, C2) = 0
M₂₁ (delete R2, C1) = -4
M₂₂ (delete R2, C2) = 2

Cofactors A_ij = (-1)^(i+j) × M_ij:
A₁₁ = (-1)²(3) = 3
A₁₂ = (-1)³(0) = 0
A₂₁ = (-1)³(-4) = -(-4) = 4
A₂₂ = (-1)⁴(2) = 2

Solution:

Minors:
M₁₁ = (1×1 - 0) = 1  |  M₁₂ = (0 - 0) = 0  |  M₁₃ = (0 - 0) = 0
M₂₁ = 0  |  M₂₂ = 1  |  M₂₃ = 0
M₃₁ = 0  |  M₃₂ = 0  |  M₃₃ = 1

Cofactors:
A₁₁ = (-1)²(1) = 1  |  A₁₂ = (-1)³(0) = 0  |  A₁₃ = 0
A₂₁ = 0  |  A₂₂ = 1  |  A₂₃ = 0
A₃₁ = 0  |  A₃₂ = 0  |  A₃₃ = 1

Solution:

Elements of second row are: a₂₁ = 2, a₂₂ = 0, a₂₃ = 1.

Find their Cofactors:
A₂₁ = (-1)³ 3   8
2   3 = - (9 - 16) = -(-7) = 7
A₂₂ = (-1)⁴ 5   8
1   3 = (15 - 8) = 7
A₂₃ = (-1)⁵ 5   3
1   2 = - (10 - 3) = -7

Δ = a₂₁A₂₁ + a₂₂A₂₂ + a₂₃A₂₃
Δ = (2 × 7) + (0 × 7) + (1 × -7)
Δ = 14 + 0 - 7 = 7

Solution:

Elements of C3 are: a₁₃ = yz, a₂₃ = zx, a₃₃ = xy.

Cofactors:
A₁₃ = (-1)⁴ 1   y
1   z = (z - y)
A₂₃ = (-1)⁵ 1   x
1   z = -(z - x) = (x - z)
A₃₃ = (-1)⁶ 1   x
1   y = (y - x)

Δ = yz(z - y) + zx(x - z) + xy(y - x)
Δ = yz² - y²z + zx² - z²x + xy² - x²y
Grouping terms logically to factorize gives:
Δ = (x - y)(y - z)(z - x)

Solution:

The determinant value is the sum of the product of elements of any row (or column) with their corresponding cofactors.
Let's check the options. Only option D matches the elements of a column with their exact corresponding cofactors: a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁.

Correct Answer: (D) a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁

Solution:

Using the 2x2 shortcut: Swap 1 and 4, change signs of 2 and 3.
adj(A) = 4   -2
-3   1

(Long method: Find all 4 cofactors A₁₁=4, A₁₂=-3, A₂₁=-2, A₂₂=1, then write them as columns to get the transpose matrix).

Solution:

First, find the Cofactors A_ij:
A₁₁ = (3 - 0) = 3
A₁₂ = -(2 - (-10)) = -12
A₁₃ = (0 - (-6)) = 6
A₂₁ = -(-1 - 0) = 1
A₂₂ = (1 - (-4)) = 5
A₂₃ = -(0 - 2) = 2
A₃₁ = (-5 - 6) = -11
A₃₂ = -(5 - 4) = -1
A₃₃ = (3 - (-2)) = 5

Adjoint is the transpose of the Cofactor matrix:
adj(A) = 3   1   -11
-12   5   -1
6   2   5

Solution:

Step 1: Find |A|
|A| = (2 × -6) - (3 × -4) = -12 + 12 = 0.
So, |A|I = 0 1   0
0   1 = 0   0
0   0 = O.

Step 2: Find adj(A)
Using shortcut: Swap 2 and -6, change signs of 3 and -4.
adj(A) = -6   -3
4   2

Step 3: Verify Products
A(adj A) = 2   3
-4   -6 -6   -3
4   2
= -12+12   -6+6
24-24   12-12 = 0   0
0   0 = O.
Since both equal the Null matrix O, Verified.

Solution:

Formula: A⁻¹ = 1/|A| × adj(A)

|A| = (2 × 3) - (-2 × 4) = 6 + 8 = 14.
Since |A| ≠ 0, inverse exists.

adj(A) = 3   2
-4   2

A⁻¹ = 1/14 3   2
-4   2

Solution Summary:

Step 1: Find Determinant |A|
Expand along C1 (has two zeros): |A| = 1(10 - 0) = 10. Since |A| ≠ 0, A⁻¹ exists.

Step 2: Find Cofactors
A₁₁=10, A₁₂=0, A₁₃=0
A₂₁=-10, A₂₂=5, A₂₃=0
A₃₁=2, A₃₂=-4, A₃₃=2

Step 3: Write Adjoint & Inverse
adj(A) = 10   -10   2
0   5   -4
0   0   2

A⁻¹ = 1/|A| × adj(A) = 1/10 10   -10   2
0   5   -4
0   0   2

Solution:

Part 1: Show A² - 5A + 7I = O
A² = 3   1
-1   2 3   1
-1   2 = 8   5
-5   3
8   5
-5   3 - 15   5
-5   10 + 7   0
0   7 = 0   0
0   0 = O. Proved.

Part 2: Find A⁻¹ using equation (No Determinant Method!)
A² - 5A + 7I = O
Multiply entire equation by A⁻¹:
A(A⁻¹A) - 5(A⁻¹A) + 7(A⁻¹I) = O
A - 5I + 7A⁻¹ = O
7A⁻¹ = 5I - A
7A⁻¹ = 5   0
0   5 - 3   1
-1   2 = 2   -1
1   3

A⁻¹ = 1/7 2   -1
1   3

Solution:

Property: If A is a square matrix of order 'n', then |adj A| = |A|ⁿ⁻¹.
Here n = 3. So |adj A| = |A|³⁻¹ = |A|².
Correct Answer: (B) |A|²

Solution:

Property: We know that AA⁻¹ = I.
Taking determinant on both sides: |AA⁻¹| = |I|
|A| × |A⁻¹| = 1
Therefore, |A⁻¹| = 1 / |A| = 1 / det(A).
Correct Answer: (B) 1/det(A)

Solution Summary & Concepts:

Rule: If |A| ≠ 0, the system is consistent (unique solution). If |A| = 0, we check (adj A)B. If (adj A)B ≠ O, it is inconsistent (no solution).

• Q1: x + 2y = 2, 2x + 3y = 3
  |A| = 1   2
  2   3 = 3 - 4 = -1. Since |A| ≠ 0, system is Consistent.
• Q2: 2x - y = 5, x + y = 4
  |A| = 2   -1
  1   1 = 2 - (-1) = 3. Since |A| ≠ 0, system is Consistent.
• Q3: x + 3y = 5, 2x + 6y = 8
  |A| = 1   3
  2   6 = 6 - 6 = 0.
  Calculate (adj A)B = 6   -3
  -2   1 5
  8 = 6
  -2 ≠ O. System is Inconsistent.
• Q4: x+y+z=1, 2x+3y+2z=2, ax+ay+2az=4
  |A| = 1   1   1
  2   3   2
  a   a   2a = a(6-2) - a(2-2) + 2a(3-2) = a ≠ 0. Consistent.
• Q5: 3x-y-2z=2, 2y-z=-1, 3x-5y=3
  |A| = 0. Further check shows (adj A)B ≠ O. Inconsistent.
• Q6: 5x-y+4z=5, 2x+3y+5z=2, 5x-2y+6z=-1
  |A| = 51 ≠ 0. Consistent.

Solutions:

Q7: 5x + 2y = 4, 7x + 3y = 5
A = 5   2
7   3, X = x
y, B = 4
5
|A| = 15 - 14 = 1 ≠ 0.
A⁻¹ = 1/1 3   -2
-7   5
X = A⁻¹B = 3   -2
-7   5 4
5 = 12-10
-28+25 = 2
-3.
x = 2, y = -3

Q8: 2x - y = -2, 3x + 4y = 3
|A| = 2   -1
3   4 = 8 - (-3) = 11.
X = 1/11 4   1
-3   2 -2
3 = 1/11 -5
12
x = -5/11, y = 12/11

Q9: 4x - 3y = 3, 3x - 5y = 7
|A| = -20 - (-9) = -11.
X = 1/-11 -5   3
-3   4 3
7 = 1/-11 6
19
x = -6/11, y = -19/11

Q10: 5x + 2y = 3, 3x + 2y = 5
|A| = 10 - 6 = 4.
X = 1/4 2   -2
-3   5 3
5 = 1/4 -4
16
x = -1, y = 4

Solution:

Write as AX = B:
A = 2   1   1
1   -2   -1
0   3   -5, X = x
y
z, B = 1
3/2
9

Step 1: Find |A|
|A| = 2(10 + 3) - 1(-5 - 0) + 1(3 - 0) = 26 + 5 + 3 = 34.
Since |A| ≠ 0, the system has a unique solution.

Step 2: Find Cofactors and Adjoint
A₁₁ = 13, A₁₂ = 5, A₁₃ = 3
A₂₁ = 8, A₂₂ = -10, A₂₃ = -6
A₃₁ = 1, A₃₂ = 3, A₃₃ = -5

adj(A) = 13   8   1
5   -10   3
3   -6   -5

Step 3: Find X = A⁻¹B
X = 1/34 13   8   1
5   -10   3
3   -6   -5 1
3/2
9

Multiply row by column:
Row 1: 13(1) + 8(3/2) + 1(9) = 13 + 12 + 9 = 34
Row 2: 5(1) - 10(3/2) + 3(9) = 5 - 15 + 27 = 17
Row 3: 3(1) - 6(3/2) - 5(9) = 3 - 9 - 45 = -51

X = 1/34 34
17
-51 = 1
1/2
-3/2

Final Answer: x = 1, y = 1/2, z = -3/2

Solution Summaries:

Q12: x-y+z=4, 2x+y-3z=0, x+y+z=2
|A| = 10.
adj(A) = 4   2   2
-5   0   5
1   -2   3
X = 1/10 20
-10
10 → x = 2, y = -1, z = 1

Q13: 2x+3y+3z=5, x-2y+z=-4, 3x-y-2z=3
|A| = 40.
adj(A) = 5   3   9
5   -13   1
5   11   -7
X = 1/40 40
80
-40 → x = 1, y = 2, z = -1

Q14: x-y+2z=7, 3x+4y-5z=-5, 2x-y+3z=12
|A| = 4.
adj(A) = 7   1   -3
-19   -1   11
-11   -1   7
X = 1/4 8
4
12 → x = 2, y = 1, z = 3

Solution:

Step 1: Find A⁻¹
|A| = 2(-4 + 4) - (-3)(-6 + 4) + 5(3 - 2) = 0 - 6 + 5 = -1.
Cofactors: A₁₁=0, A₁₂=2, A₁₃=1; A₂₁=-1, A₂₂=-9, A₂₃=-5; A₃₁=2, A₃₂=23, A₃₃=13.
adj(A) = 0   -1   2
2   -9   23
1   -5   13
A⁻¹ = 1/(-1) × adj(A) = 0   1   -2
-2   9   -23
-1   5   -13

Step 2: Solve the equations
Notice that the coefficient matrix of the given equations is exactly the matrix A.
So, AX = B → X = A⁻¹B, where B = 11
-5
-3.

X = 0   1   -2
-2   9   -23
-1   5   -13 11
-5
-3

Multiply row by column:
x = 0(11) + 1(-5) - 2(-3) = -5 + 6 = 1
y = -2(11) + 9(-5) - 23(-3) = -22 - 45 + 69 = 2
z = -1(11) + 5(-5) - 13(-3) = -11 - 25 + 39 = 3

Final Answer: x = 1, y = 2, z = 3

Solution:

Let the cost per kg of onion be Rs x, wheat be Rs y, and rice be Rs z.
Equations form:
4x + 3y + 2z = 60
2x + 4y + 6z = 90
6x + 2y + 3z = 70

AX = B where A = 4   3   2
2   4   6
6   2   3, X = x
y
z, B = 60
90
70

|A| = 4(12 - 12) - 3(6 - 36) + 2(4 - 24) = 0 + 90 - 40 = 50.
A⁻¹ = 1/50 0   -5   10
30   0   -20
-20   10   10

X = 1/50 0   -5   10
30   0   -20
-20   10   10 60
90
70

x = 1/50 [0 - 450 + 700] = 250 / 50 = 5
y = 1/50 [1800 + 0 - 1400] = 400 / 50 = 8
z = 1/50 [-1200 + 900 + 700] = 400 / 50 = 8

Cost of Onion = Rs 5/kg, Wheat = Rs 8/kg, Rice = Rs 8/kg.

Solution:

Let Δ = x   sinθ   cosθ
-sinθ   -x   1
cosθ   1   x

Expand the determinant along Row 1 (R₁):
Δ = x-x   1
1   x - sinθ-sinθ   1
cosθ   x + cosθ-sinθ   -x
cosθ   1

Δ = x(-x² - 1) - sinθ(-x sinθ - cosθ) + cosθ(-sinθ + x cosθ)
Δ = -x³ - x + x sin²θ + sinθcosθ - sinθcosθ + x cos²θ

Cancel out +sinθcosθ and -sinθcosθ:
Δ = -x³ - x + x(sin²θ + cos²θ)
Since sin²θ + cos²θ = 1:
Δ = -x³ - x + x(1) = -x³

Since the final result (-x³) contains no θ terms, the determinant is independent of θ. Proved.

Solution:

Let Δ = a   a²   bc
b   b²   ca
c   c²   ab

Multiply R₁ by a, R₂ by b, and R₃ by c. To balance this, divide the determinant by abc:
Δ = 1/abc a²   a³   abc
b²   b³   abc
c²   c³   abc

Now, take 'abc' common from the third column (C₃):
Δ = (abc)/(abc) a²   a³   1
b²   b³   1
c²   c³   1
Δ = a²   a³   1
b²   b³   1
c²   c³   1

Interchange C₂ and C₃ (Determinant sign changes to -ve):
Δ = - a²   1   a³
b²   1   b³
c²   1   c³

Now, interchange C₁ and C₂ (Determinant sign changes back to +ve):
Δ = 1   a²   a³
1   b²   b³
1   c²   c³ = RHS. Proved.

Solution:

We know the Reversal Law for inverses: (AB)⁻¹ = B⁻¹A⁻¹.
Since we already have A⁻¹, we only need to find B⁻¹ and multiply them.

Step 1: Find B⁻¹
|B| = 1(3 - 0) - 2(-1 - 0) - 2(2 - 0) = 3 + 2 - 4 = 1. (Since |B| ≠ 0, B⁻¹ exists).
Find the cofactors of B to construct adj(B):
adj(B) = 3   2   6
1   1   2
2   2   5
B⁻¹ = 1/|B| × adj(B) = 3   2   6
1   1   2
2   2   5

Step 2: Multiply B⁻¹ and A⁻¹
(AB)⁻¹ = 3   2   6
1   1   2
2   2   5 × 3   -1   1
-15   6   -5
5   -2   2

Multiplying row by column:
R1 = [9-30+30,   -3+12-12,   3-10+12] = [9, -3, 5]
R2 = [3-15+10,   -1+6-4,   1-5+4] = [-2, 1, 0]
R3 = [6-30+25,   -2+12-10,   2-10+10] = [1, 0, 2]

(AB)⁻¹ = 9   -3   5
-2   1   0
1   0   2

Solution:

Let 1/x = p, 1/y = q, and 1/z = r. The equations become:
2p + 3q + 10r = 4
4p - 6q + 5r = 1
6p + 9q - 20r = 2

This is of the form AX = B, where A = 2   3   10
4   -6   5
6   9   -20, X = p
q
r, B = 4
1
2.

Find |A| = 1200. Find adj(A) and A⁻¹.
A⁻¹ = 1/1200 75   150   75
110   -100   30
72   0   -24

X = A⁻¹B → p = 1/2, q = 1/3, r = 1/5.

Since p = 1/x, q = 1/y, r = 1/z, we get:
x = 2, y = 3, z = 5.

Solution:

Since a, b, c are in A.P., we know that 2b = a + c.

Let Δ be the determinant. Apply the operation R₂ → 2R₂ - (R₁ + R₃) to the middle row:
New R₂ elements:
C₁: 2(x+3) - (x+2 + x+4) = 2x + 6 - 2x - 6 = 0
C₂: 2(x+4) - (x+3 + x+5) = 2x + 8 - 2x - 8 = 0
C₃: 2(x+2b) - (x+2a + x+2c) = 2x + 4b - 2x - 2(a+c)
Substitute a+c = 2b: 4b - 2(2b) = 4b - 4b = 0.

Since the entire second row (R₂) becomes zero, the value of the determinant is 0.
Correct Answer: (A) 0

Solution:

Evaluate the determinant |A|:
|A| = 1(1 + sin²θ) - sinθ(-sinθ + sinθ) + 1(sin²θ + 1)
|A| = 1 + sin²θ - 0 + sin²θ + 1
|A| = 2 + 2sin²θ = 2(1 + sin²θ)

We know that for 0 ≤ θ ≤ 2π, the value of sin²θ ranges from 0 to 1. (i.e., 0 ≤ sin²θ ≤ 1)

Multiply the inequality by 2: 0 ≤ 2sin²θ ≤ 2
Add 2 to all sides: 2 ≤ 2 + 2sin²θ ≤ 4

Therefore, the value of |A| lies in the closed interval [2, 4].
Correct Answer: (D) Det(A) ∈ [2, 4]