DETERMINANTS

Exercise 4.1

Q1: Evaluate the determinants

| 2 4 |
| -5 -1 |

Solution:
Determinant = (2)(-1) − (4)(-5)
= −2 + 20
= 18.

_____________________________________________________________

Q2: (i)

| cos θ − sin θ |
| sin θ cos θ |

Solution:
Determinant = (cos θ)(cos θ) − (− sin θ)(sin θ)
= cos² θ + sin² θ
= 1.

_____________________________________________________________

Q2: (ii)

| x² − x + 1 x − 1 |
| x + 1 x + 1 |

Solution:
Determinant = (x² − x + 1)(x + 1) − (x − 1)(x + 1)
= (x + 1)[(x² − x + 1) − (x − 1)]
= (x + 1)[x² − x + 1 − x + 1]
= (x + 1)(x² − 2x + 2).

_____________________________________________________________

Q3: If A = | 1 2 |
     | 4 2 | , then show that | 2A | = 4 | A |

Solution:

A = | 1 2 |
  | 4 2 |

First find | A |:

| A | = (1 × 2) − (2 × 4)
| A | = 2 − 8
| A | = −6

Now find 2A:

2A = | 2 4 |
   | 8 4 |

Now find | 2A |:

| 2A | = (2 × 4) − (4 × 8)
| 2A | = 8 − 32
| 2A | = −24

Now compare:

4 | A | = 4 × (−6) = −24

Therefore,

| 2A | = 4 | A |

_____________________________________________________________

Q4: If A = | 1 0 1 |
     | 0 1 2 |
     | 0 0 4 | , then show that | 3A | = 27 | A |

Solution:

Since A is an upper triangular matrix, its determinant is the product of diagonal elements.

| A | = 1 × 1 × 4 = 4

Now find 3A:

3A = | 3 0 3 |
   | 0 3 6 |
   | 0 0 12 |

Now find | 3A |:

| 3A | = 3 × 3 × 12
| 3A | = 108

Now compare:

27 | A | = 27 × 4 = 108

Therefore,

| 3A | = 27 | A |

_____________________________________________________________

Q5: Evaluate the following determinants:

(i) | 3 −1 −2 |
   | 0 0 −1 |
   | 3 −5 0 |

Solution:

Expand along the second row (because it has two zeros):

| A | = 0 × C₂₁ + 0 × C₂₂ + (−1) × C₂₃

Now find the cofactor C₂₃:

C₂₃ = (−1)²⁺³ × | 3 −1 |
         | 3 −5 |

C₂₃ = − (3 × (−5) − (−1 × 3))
C₂₃ = − (−15 + 3)
C₂₃ = − (−12)
C₂₃ = 12

Now,

| A | = (−1) × 12
| A | = −12

_____________________________________________________________

(ii) | 3 −4 5 |
   | 1 1 −2 |
   | 2 3 1 |

Solution:

Expand along the first row:

| A | = 3 × | 1 −2 | − (−4) × | 1 −2 | + 5 × | 1 1 |
   | 3 1 |    | 2 1 |   | 2 3 |

Now evaluate each 2 × 2 determinant:

First minor:

| 1 −2 | = (1 × 1) − (−2 × 3) = 1 + 6 = 7
| 3 1 |

Second minor:

| 1 −2 | = (1 × 1) − (−2 × 2) = 1 + 4 = 5
| 2 1 |

Third minor:

| 1 1 | = (1 × 3) − (1 × 2) = 3 − 2 = 1
| 2 3 |

Now substitute:

| A | = 3(7) + 4(5) + 5(1)
| A | = 21 + 20 + 5
| A | = 46

_____________________________________________________________

(iii)

| 0 1 2 |
| −1 0 −3 |
| −2 3 0 |

Solution (expand along the first row):

Determinant = a11·C11 − a12·C12 + a13·C13
= 0 · (…) − 1 · det([−1 −3; −2 0]) + 2 · det([−1 0; −2 3])

Compute the 2×2 minors:

det([−1 −3; −2 0]) = (−1)(0) − (−3)(−2) = 0 − 6 = −6.
det([−1 0; −2 3]) = (−1)(3) − (0)(−2) = −3.

Now substitute:

Determinant = 0 − 1·(−6) + 2·(−3)
= 0 + 6 − 6
= 0.

Therefore |A| = 0.

_____________________________________________________________

(iv)

| 2 −1 −2 |
| 0 2 −1 |
| 3 −5 0 |

Solution (expand along the first row):

Determinant = a11·M11 − a12·M12 + a13·M13
= 2·det([ 2 −1; −5 0 ]) − (−1)·det([ 0 −1; 3 0 ]) + (−2)·det([ 0 2; 3 −5 ])

Compute the 2×2 minors:

M11 = det([ 2 −1; −5 0 ]) = (2)(0) − (−1)(−5) = 0 − 5 = −5.
M12 = det([ 0 −1; 3 0 ]) = (0)(0) − (−1)(3) = 3.
M13 = det([ 0 2; 3 −5 ]) = (0)(−5) − (2)(3) = −6.

Now substitute:

Determinant = 2·(−5) − (−1)·(3) + (−2)·(−6)
= −10 + 3 + 12
= 5.

Therefore |A| = 5.

_____________________________________________________________

Q6: If A = | 1 1 −2 |
     | 2 1 −3 |
     | 5 4 −9 | , find | A |

Solution:

Expand along the first row:

| A | = 1 × | 1 −3 | − 1 × | 2 −3 | + (−2) × | 2 1 |
     | 4 −9 |   | 5 −9 |    | 5 4 |

Now evaluate each 2 × 2 determinant:

First minor:
| 1 −3 | = (1)(−9) − (−3)(4) = −9 + 12 = 3
| 4 −9 |

Second minor:
| 2 −3 | = (2)(−9) − (−3)(5) = −18 + 15 = −3
| 5 −9 |

Third minor:
| 2 1 | = (2)(4) − (1)(5) = 8 − 5 = 3
| 5 4 |

Now substitute:

| A | = 1(3) − 1(−3) + (−2)(3)
| A | = 3 + 3 − 6
| A | = 0

_____________________________________________________________

Q7: Find values of x, if

(i) | 2 4 | = | 2x 4 |
  | 5 1 | | 6 x |

Solution:

Left side determinant:

| 2 4 | = (2)(1) − (4)(5) = 2 − 20 = −18
| 5 1 |

Right side determinant:

| 2x 4 | = (2x)(x) − (4)(6) = 2x² − 24
| 6 x |

Since both determinants are equal:

2x² − 24 = −18
2x² = 6
x² = 3
x = ±√3

_____________________________________________________________

(ii) | 2 3 | = | x 3 |
  | 4 5 | | 2x 5 |

Solution:

Left side determinant:

| 2 3 | = (2)(5) − (3)(4) = 10 − 12 = −2
| 4 5 |

Right side determinant:

| x 3 | = (x)(5) − (3)(2x) = 5x − 6x = −x
| 2x 5 |

Since both determinants are equal:

−x = −2
x = 2

_____________________________________________________________

Q8: If | x 2 | = | 6 2 | , then x is equal to
   | 1 8x| | 18 6 |

(A) 6
(B) ± 6
(C) − 6
(D) 0

Solution:

Find the determinant of the left side:

| x 2 | = (x)(x) − (2)(18)
|18 x | = x² − 36

Find the determinant of the right side:

| 6 2 | = (6)(6) − (2)(18)
|18 6 | = 36 − 36
= 0

Since both determinants are equal:

x² − 36 = 0
x² = 36
x = ± 6

Correct Answer: (B) ± 6

_____________________________________________________________

Exercise 4.2

Q1: Find area of the triangle with vertices at the point given in each of the following:

(i) (1, 0), (6, 0), (4, 3)

(ii) (2, 7), (1, 1), (10, 8)

(iii) (−2, −3), (3, 2), (−1, −8)

Formula used:

Area of triangle
= 1/2 | x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂) |

(i) Vertices: (1, 0), (6, 0), (4, 3)

Area
= 1/2 | 1(0 − 3) + 6(3 − 0) + 4(0 − 0) |
= 1/2 | −3 + 18 + 0 |
= 1/2 (15)

Area = 15/2 square units

_____________________________________________________________

(ii) Vertices: (2, 7), (1, 1), (10, 8)

Area
= 1/2 | 2(1 − 8) + 1(8 − 7) + 10(7 − 1) |
= 1/2 | 2(−7) + 1 + 10(6) |
= 1/2 | −14 + 1 + 60 |
= 1/2 (47)

Area = 47/2 square units

_____________________________________________________________

(iii) Vertices: (−2, −3), (3, 2), (−1, −8)

Area
= 1/2 | −2(2 − (−8)) + 3((−8) − (−3)) + (−1)((−3) − 2) |
= 1/2 | −2(10) + 3(−5) + (−1)(−5) |
= 1/2 | −20 − 15 + 5 |
= 1/2 (30)

Area = 15 square units

_____________________________________________________________

Q2: Show that points
A (a, b + c), B (b, c + a), C (c, a + b) are collinear.

Solution:

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

Area = 1/2

|a b + c 1
b c + a 1
c a + b 1 |

Evaluate the determinant:

= a[(c + a) − (a + b)] − (b + c)[b − c] + 1[b(a + b) − c(c + a)]

= a(c − b) − (b + c)(b − c) + [ab + b² − c² − ac]

= a(c − b) − (b² − c²) + a(b − c) + (b² − c²)

= a(c − b + b − c)

= 0

Since the area is zero, the given points are collinear.

_____________________________________________________________

Q3: Find values of k if area of triangle is 4 sq. units and vertices are:

(i) (k, 0), (4, 0), (0, 2)

Solution:

Area
= 1/2 | k(0 − 2) + 4(2 − 0) + 0(0 − 0) |
= 1/2 | −2k + 8 |

Given area = 4,

1/2 | −2k + 8 | = 4
| −2k + 8 | = 8

Case 1: −2k + 8 = 8
−2k = 0
k = 0

Case 2: −2k + 8 = −8
−2k = −16
k = 8

Therefore,
k = 0 or k = 8

_____________________________________________________________

(ii) (−2, 0), (0, 4), (0, k)

Solution:

Area
= 1/2 | −2(4 − k) + 0(k − 0) + 0(0 − 4) |
= 1/2 | −8 + 2k |

Given area = 4,

1/2 | −8 + 2k | = 4
| −8 + 2k | = 8

Case 1: −8 + 2k = 8
2k = 16
k = 8

Case 2: −8 + 2k = −8
2k = 0
k = 0

Therefore,
k = 0 or k = 8

_____________________________________________________________

Q4:

(i) Find equation of line joining (1, 2) and (3, 6) using determinants.

Solution:

The equation of a line through points (x₁, y₁) and (x₂, y₂) is:

| x y 1 |
| x₁ y₁ 1 | = 0
| x₂ y₂ 1 |

Substituting (1, 2) and (3, 6):

| x y 1 |
| 1 2 1 | = 0
| 3 6 1 |

Expanding:

x(2 − 6) − y(1 − 3) + 1(6 − 6) = 0
x(−4) − y(−2) + 0 = 0
−4x + 2y = 0

Equation of the line:

2y = 4x
y = 2x

_____________________________________________________________

(ii) Find equation of line joining (3, 1) and (9, 3) using determinants.

Solution:

Using determinant form:

| x y 1 |
| 3 1 1 | = 0
| 9 3 1 |

Expanding:

x(1 − 3) − y(3 − 9) + 1(9 − 9) = 0
x(−2) − y(−6) + 0 = 0
−2x + 6y = 0

Equation of the line:

6y = 2x
y = x / 3

_____________________________________________________________

Q5: If area of triangle is 35 sq. units with vertices (2, −6), (5, 4) and (k, 4). Then k is

Solution:

Notice the points (5, 4) and (k, 4) have the same y-coordinate, so the base of the triangle along the line y = 4 has length |k − 5|.
The vertical distance (height) from (2, −6) up to y = 4 is 4 − (−6) = 10.

Area = 1/2 × base × height
35 = 1/2 × |k − 5| × 10
35 = 5 · |k − 5|
|k − 5| = 7

So k − 5 = 7 or k − 5 = −7
Hence k = 12 or k = −2.

Therefore the required values are k = 12 and k = −2.

Answer: D (12, -2)

_____________________________________________________________

Exercise 4.3

Q1: Write Minors and Cofactors of the elements of following determinants:

1. (i)

| 2 –4 |
| 0 3 |

Solution:

M11 = 3, M12 = 0, M21 = –4, M22 = 2,
A11 = 3, A12 = 0, A21 = 4, A22 = 2

(ii)
| a c |
| b d |

Solution:

M11 = d, M12 = b, M21 = c, M22 = a

A11 = d, A12 = –b, A21 = –c, A22 = a

_____________________________________________________________

2. (i)
| 1 0 0 |
| 0 1 0 |
| 0 0 1 |

Solution:

M11 = 1, M12 = 0, M13 = 0,
M21 = 0, M22 = 1, M23 = 0,
M31 = 0, M32 = 0, M33 = 1

A11 = 1, A12 = 0, A13 = 0,
A21 = 0, A22 = 1, A23 = 0,
A31 = 0, A32 = 0, A33 = 1

_____________________________________________________________

2. (ii)
  
| 1 0 4 |
| 3 5 –1 |
| 0 1 2 |

M11 = 11, M12 = 6, M13 = 3,
M21 = –4, M22 = 2, M23 = 1,
M31 = –20, M32 = –13, M33 = 5

A11 = 11, A12 = –6, A13 = 3,
A21 = 4, A22 = 2, A23 = –1,
A31 = –20, A32 = 13, A33 = 5

_____________________________________________________________

Q3: Using cofactors of elements of second row, evaluate Δ =

| 5 3 8 |
| 2 0 1 |
| 1 2 3 |

Solution:

Second row elements: 2, 0, 1

Step 1: Cofactors of the second row

C21 = (–1)^(2+1) × determinant of
| 3 8 |
| 2 3 |
= –1 × (3×3 – 8×2)
= –1 × (9 – 16)
= 7

C22 = (–1)^(2+2) × determinant of
| 5 8 |
| 1 3 |
= 1 × (5×3 – 8×1)
= 15 – 8
= 7

C23 = (–1)^(2+3) × determinant of
| 5 3 |
| 1 2 |
= –1 × (5×2 – 3×1)
= –1 × (10 – 3)
= –7

Step 2: Expansion using second row

Δ = 2×C21 + 0×C22 + 1×C23
Δ = 2×7 + 0 + (–7)
Δ = 14 – 7
Δ = 7

Final Answer: Δ = 7

_____________________________________________________________

Q4: Using cofactors of elements of third column, evaluate Δ where

| 1 x yz |
| 1 y zx |
| 1 z xy |

Third column elements: yz, zx, xy

Step 1 — Minors and cofactors of third column

Minor for element at (1,3):
M13 = determinant of the 2×2 matrix formed by removing row 1 and column 3, i.e.
| 1 y |
| 1 z |
M13 = 1·z − y·1 = z − y
Sign factor = (−1)^(1+3) = (−1)^4 = 1
C13 = 1 × M13 = z − y

Minor for element at (2,3):
M23 = determinant of
| 1 x |
| 1 z |
M23 = 1·z − x·1 = z − x
Sign factor = (−1)^(2+3) = (−1)^5 = −1
C23 = −1 × M23 = −(z − x) = x − z

Minor for element at (3,3):
M33 = determinant of
| 1 x |
| 1 y |
M33 = 1·y − x·1 = y − x
Sign factor = (−1)^(3+3) = (−1)^6 = 1
C33 = 1 × M33 = y − x

Step 2 — Expand Δ along the third column

Δ = (element 1,3)·C13 + (element 2,3)·C23 + (element 3,3)·C33

Substitute elements and cofactors:

Δ = yz·(z − y) + zx·(x − z) + xy·(y − x)

Step 3 — Simplify and factorise

Expand (or group) and factor:

yz(z − y) + zx(x − z) + xy(y − x)
= y z z − y^2 z + z x x − z^2 x + x y y − x^2 y
= z^2 y − y^2 z + x^2 z − x z^2 + x y^2 − x^2 y

This expression factorises to the product of three simple factors:

Δ = (x − y) (y − z) (z − x)

Final Answer: (x − y) (y − z) (z − x)

_____________________________________________________________

Q5: We have a 3×3 determinant:

Δ =
| a11 a12 a13 |
| a21 a22 a23 |
| a31 a32 a33 |

Aij = cofactor of aij.

Key fact of determinants

A determinant can always be expanded along any column.

For expansion along the first column, the formula is:

Δ = a11·A11 + a21·A21 + a31·A31

This is a basic property:

Element × its cofactor, added down the chosen column.

Now check the options.

Option D is:

a11 A11 + a21 A21 + a31 A31

This matches exactly the expansion of Δ along the first column.

Therefore, the correct answer is: (D)

_____________________________________________________________