Determinants

We know that determinant of A is calculated as

Now,

= 2(-1) – 4(-5)

= -2 – (-20)

= -2 + 20

= 18

The determinant of the above matrix is 18.

We know that determinant of A is calculated as

Now,

= cosθ(cosθ) - (-sinθ)(sinθ)

= cos²θ + sin²θ

= 1 [∵ cos²θ + sin²θ = 1]

The determinant of the above matrix is 1.

We know that determinant of A is calculated as

Now,

= (x² – x + 1)(x + 1) - (x - 1)(x + 1)

= (x³ - x² + x + x² – x + 1) - (x² - 1)

= x³ + 1 - x² + 1

= x³ - x² + 2

Ans. The determinant of the above matrix is x³ - x² + 2.

|A| =

We know that determinant of A is calculated as

= 1(2) - 2(4)

= 2 - 8

|A| = -6

LHS: |2A|

= 2(4) - 4(8)

= 8 - 32 = -24

|2A| = -24 …LHS

RHS: 4|A|

4|A|= 4(-6)

= -24

4|A| = -24 …RHS

LHS = RHS

Hence proved.

We know that a determinant of a 3 x 3 matrix is calculated as

= 1[1(4) - 2(0)] – 0 + 1[0-0]

= 1[4 - 0] – 0 + 0

= 4

|A|= 4

LHS: |3A|

= 3[3(12) - 0(6)] – 0 + 3[0 – 0]

= 3(36) – 0 + 0

= 108

|3A| = 108 ----LHS

RHS: 27|A|

27|A| = 27(4)

= 108

27|A| = 108 ----RHS

LHS=RHS

Hence proved.

Now,

We know that a determinant of a 3x3 matrix is calculated as

= 3[0 – (-1)(-5)] +1[0 – (-1)(3)] – 2[0 – 0]

= 3(-5) + 1(3) – 0

= -15 + 3

= -12

The determinant of the above matrix is -12

Now,

We know that a determinant of a 3 x 3 matrix is calculated as

= 3[1 – (-2)(3)] + 4[1 – (-2)(2)] + 5[3-2]

= 3[1 + 6] + 4[1 + 4] + 5[1]

= 3[7] + 4[5] + 5

= 21 + 20 + 5

= 46

The determinant of the above matrix is 46.

Now,

We know that a determinant of a 3 x 3 matrix is calculated as

= 0 – 1[0 – (-3)(-2)] + 2[(-1)(3) – 0]

= 0 – 1[0 - 6] + 2[-3 – 0]

= 0 – 1[-6] + 2[-3]

= 0 + 6 – 6

= 0

The determinant of the above matrix is 0.

Now,

We know that a determinant of a 3 x 3 matrix is calculated as

= 2[0 – (-1)(-5)] + 1[0 – (-1)(3)] – 2[0 – 3(2)]

= 2[0 – 5] + 1[0 + 3] – 2[-6]

= 2[-5] + 1[3] -2[-6]

= -10 + 3 + 12

= 5

The determinant of the above matrix is 5.

GIVEN:

Now,

We know that a determinant of a 3 x 3 matrix is calculated as

.

= 1[-9 – (-3)(4)] – 1[2(-9) – (-3)(5)] – 2[2(4) – 1(5)]

= 1[-9 + 12] – 1[-18 + 15] – 2[8 – 5]

= 1[3] – 1[-3] – 2[3]

= 3 + 3 – 6

= 0

Ans. |A| = 0

We have

We know that determinant of A is calculated as

⇒ 2(1) – 4(5) = 2x(x) – 4(6)

⇒ 2 – 20 = 2x² - 24

⇒ -18 = 2x² – 24

⇒ 2x² = -24 + 18

⇒ 2x² = 6

⇒ x² = 6/2

⇒ x² = 3

x = �√3

Ans. The value of x is �√3

We know that determinant of A is calculated as

We have

⇒ 2(5) – 3(4) = x(5) – 3(2x)

⇒ 10 – 12 = 5x – 6x

⇒ -2 = -x

⇒ x = 2

Ans. The value of x is 2.

We have

We know that determinant of A is calculated as

⇒ x(x) – 2(18) = 6(6) – 2(18)

⇒ x² - 36 = 36 – 36

⇒ x² =36 – 36 + 36

⇒ x² = 36

⇒ x = ±6

Applying Operations C₁→ C₁ + C₂ (i.e. Replacing 1^st column by addition of 1^st and 2^nd column)

C₁→ C₁ - C₃ (i.e. Replacing 1^st column by subtraction of 1^st and 3^rd column)

If any one of the rows or columns of a determinant is 0 then the value of that determinant is 0.

∴ LHS = 0 = RHS

Applying Operation C₁→ C₁ + C₂ (i.e. Replacing 1^st column by addition of 1^st and 2^nd column)

C₁→ C₁ + C₃ (i.e. Replacing 1^st column by addition of 1^st and 3^rd column)

If any one of the rows or columns of a determinant is 0 then the value of that determinant is 0.

∴ LHS = 0 = RHS

Applying Operation C₁→ C₁ + 9C₂ (i.e. Replacing 1^st column by addition of 1^st column and 9 times second column)

C₁→ C₁ - C₃ (i.e. Replacing 1^st column by subtraction of 1^st and 3rd column)

If any one of the rows or columns of a determinant is 0 then the value of that determinant is 0.

∴ LHS = 0 = RHS

C₃→ C₂ + C₃ (i.e. replace 3^rd column by addition of 2^nd and 3^rd column)

Taking ab + bc + ac outside determinant

C₁→ C₁ - C₃ (i.e. replace 1^st column by subtraction of 1^st and 3^rd column)

If any one of the rows or columns of a determinant is 0 then the value of that determinant is 0.

∴ LHS = 0 = RHS

∴

When two determinants are added each of the corresponding elements gets added.

Here we can split the LHS determinant as

For the determinant, u perform the following transformation R₁↔ R₃ (i.e. interchange 1^st row with 3^rd row)

When two particular rows/columns of a determinant are interchanged the value becomes negative 1 times the original value.

R₁↔ R₂ (i.e. interchange 1^st row with 2^nd row)

R1 ↔ R3 (i.e. interchange 1^st row with 3^rd row)

R1 ↔ R2 (i.e. interchange 1^st row with 2^nd row)

∴ LHS = RHS

∴

To Prove:



R₁→ cR₁ (i.e. replace 1^st row by multiplying it with c)

As we are multiplying we should also divide c so that the original given determinant is not changed

R₁→ R₁ - bR₂ (i.e. replace 1^st row by subtraction of 1^st row and b times 2^nd row)

Taking a outside the determinant from 1^st row

If any two rows or columns of a determinant are identical then the value of that determinant is 0 because we get a row or column with all elements 0 when we when we subtract those particular rows/columns here the transformation is R1 → R1 - R3

∴LHS = 0 = RHS

∴

Taking ‘a’, ‘b’ and ‘c’ outside the determinant from 1^st,2^nd and 3^rd column respectively

Taking ‘a’, ‘b’ and ‘c’ outside the determinant from 1^st,2^nd and 3^rd row respectively

.

R1 → R1 + R2 (i.e. Replacing 1^st row by addition of 1^st and 2^nd row)

panding the determinant along 1^st row

∴ LHS = a² b² c² × 2(1 - (-1)) = 4a²b²c² = RHS

∴

R1 → R1 - R2 (i.e. Replacing 1^st row by subtraction of 1^st and 2^nd row)

R2 → R2 - R3 (i.e. Replacing 2^nd row by subtraction of 2^nd and 3^rd row)

Since we know a² - b² = (a + b)(a - b)

Therefore taking (a - b) and (b - c) outside the determinant from 1^st and 2^nd row respectively

R1 → R1 - R2 (i.e. Replacing 1^st row by subtraction of 1^st and 2^nd row)

Expanding the determinant along 1^st column

∴

∴LHS = (a - b)(b - c)(0 - (a - c))

∴LHS = (a - b)(b - c)(c - a) = RHS

∴

C1 → C1 - C2 (i.e. Replacing 1^st column by subtraction of 1^st and 2^nd column)

C2 → C2 - C3 (i.e. Replacing 2^nd column by subtraction of 2^nd and 3^rd column)

We have a³ - b³ = (a - b)(a² + ab + b²) and b³ - c³ = (b - c)(b² + bc + c²)

Therefore taking (a - b) and (b - c) outside the determinant from 1^st and 2^nd column respectively

C1 → C1 - C2 (i.e. Replacing 1^st column by subtraction of 1^st and 2^nd column)

As a² - c²) = (a + c)(a - c) therefore taking (a - c) outside the determinant from 1^st column we get

.

Expanding the determinant along 1^st row

∴ LHS = (a - b)(b - c)(a - c)( - (a + b + c))

Adjusting the minus sign with (a - c)

∴LHS = (a - b)(b - c)(c - a)(a + b + c) = RHS

∴

R1 → R1 - R2 (i.e. Replacing 1^st row by subtraction of 1^st and 2^nd row)

R2 → R2 - R3 (i.e. Replacing 2^nd row by subtraction of 2^nd and 3^rd row)

We know x² - y² = (x + y)(x - y) and y² - z² = (y + z)(y - z)

Therefore taking (x - y) and (y - z) outside the determinant from 1^st and 2^nd row respectively

R1 → R1 - R2 (i.e. Replacing 1^st row by subtraction of 1^st and 2^nd row)

Taking (x - z) outside determinant from 1^st row

C2 → C2 - C3 (i.e. Replacing 2^nd column by subtraction of 2^nd and 3^rd column)

Expanding the determinant along 1^st row

∴ LHS = (x - y)(y - z)(x - z)(z² - xy - xz - yz - z²)

Cancelling z² and adjusting the negative sign with (x - z)

∴LHS = (x - y)(y - z)(z - x)(xy + yz + zx) = RHS

∴

R1 → R1 + R2 + R3 (i.e. replace 1^st row by addition of 1^st, 2^nd and 3^rd row)

Taking 5x + 4 outside the determinant from 1^st row

C2 → C2 - C1 (i.e. replace 2^nd column by subtraction of 2^nd and 1^st column)

C3 → C3 - C1 (i.e. replace 3^rd column by subtraction of 3^rd and 1^st column)

Expanding the determinant along 1^st row

∴ LHS = (5x - 4) (4 - x)² = RHS

∴

R1 → R1 + R2 + R3 (i.e. replace 1^st row by addition of 1^st, 2^nd and 3^rd row)

Taking 3y + k outside the determinant from 1^st row

C2 → C2 - C1 (i.e. replace 2^nd column by subtraction of 2^nd and 1^st column)

C3 → C3 - C1 (i.e. replace 3^rd column by subtraction of 3^rd and 1^st column)

Expanding the determinant along 1^st row

∴ LHS = (3y + k) k² = RHS

∴

R1 → R1 + R2 + R3 (i.e. replace 1^st row by addition of 1^st, 2^nd and 3^rd row)

Taking a + b + c outside the determinant from 1^st row

C2 → C2 - C1 (i.e. replace 2^nd column by subtraction of 2^nd and 1^st column)

C3 → C3 - C1 (i.e. replace 3^rd column by subtraction of 3^rd and 1^st column)

Expanding the determinant along 1^st row

∴ LHS = (a + b + c)³ = RHS

∴

C1 → C1 + C2 + C3 (i.e. replace 1^st column by addition of 1^st, 2^nd and 3^rd column)

Taking 2(x + y + z) outside the determinant from 1^st column

R2 → R2 - R1 (i.e. replace 2^nd row by subtraction of 2^nd and 1^st row)

R3 → R3 - R1 (i.e. replace 3^rd row by subtraction of 3^rd and 1^st row)

Expanding the determinant along 1^st column

∴ LHS = 2(x + y + z)³ = RHS

∴

R1 → R1 - xR2 (i.e. replace 1^st row by subtraction of 1^st row and ‘x’ times 2^nd row)

Taking (1 - x³) outside the determinant from 1^st row

.

Expanding the determinant along 1^st row

HS = (1 - x³)² = RHS

∴

R1 → R1 + bR3 (i.e. replace 1^st row by addition of 1^st row and b times 3^rd row)

R2 → R2 - aR3 (i.e. replace 2^nd row by subtraction of 2^nd row and a times 3^rd row)

.

Taking both (1 + a² + b²) outside the determinant from 1^st and 2^nd row

Expanding the determinant along 1^st row

∴ LHS = (1 + a² + b²)² [1 + a² - b² - (-2b²)]

∴ LHS = (1 + a² + b² )² (1 + a² + b²)

∴ LHS = (1 + a² + b² )³ = RHS

Taking out a, b and c from the determinant from 1^st, 2^nd and 3^rd row respectively.

R2 → R2 - R1 (i.e. replace 2^nd row by subtraction of 2^nd and 1^st row)

R3 → R3 - R1 (i.e. replace 3^rd row by subtraction of 3^rd and 1^st row)

.

C1 → aC1 (i.e. replace 1^st column by ‘a’ times 1^st column)

C2 → bC2 (i.e. replace 2^nd column by ‘b’ times 2^nd column)

→ cC3 (i.e. replace 3^rd column by ‘c’ times 3^rd column)

As we are multiplying by a, b and c we should also divide by a, b and c to keep the original determinant value unchanged.

.

C1 → C1 + C2 + C3 (i.e. replace 1^st column by addition of 1^st, 2^nd and 3^rd column)

Expanding determinant along 1^st column

∴ LHS = (1 + a² + b² + c²) = RHS

∴

Let A be any 3×3 matrix

∴

Taking out k from the determinant from 1^st, 2^nd and 3^rd row

∴

∴ |kA| = k³|A|

Therefore, answer is option (C) k³|A|

Determinant is an operation which we perform on arranged numbers. A square matrix is set of arranged numbers. We perform some operations on a matrix and we get a value that value is called as determinant of that matrix hence determinant is a number associated to square matrix.

Given vertices of the triangle are (1, 0), (6, 0), (4, 3)

Let the vertices of the triangle be given by (x₁, y₁), (x₂, y₂), (x₃, y₃)

Area of triangle is given by Δ =

Area of triangle = Δ =

Expanding the determinant along Row 1

Δ = 1/2 × [1 × (0 × 1 – 3 × 1) – 0 × (6 × 1 – 4 × 1) + 1 × (6 × 3 – 4 × 0)]

Δ = 1/2 × [1 × (0 – 3) – 0 + 1 × (18 – 0)]

Δ = 1/2 × (-3 + 18) = 15/2 sq. units

∴ Δ = 15/2 sq. units = 7.5 sq. units

Given vertices of the triangle are (2, 7), (1, 1), (10, 8)

Let the vertices of the triangle be given by (x₁, y₁), (x₂, y₂), (x₃, y₃)

Area of triangle is given by Δ =

Area of triangle = Δ =

Expanding the determinant along Row 1

Δ = 1/2 × [2 × (1 × 1 – 8 × 1) – 7 × (1 × 1 – 10 × 1) + 1 × (8 × 1 – 1 × 10)]

Δ = 1/2 × [2 × (1 – 8) – 7 × (1 – 10) + 1 × (8 – 10)]

Δ = 1/2 × [2 × (-7) – 7 × (-9) + 1 × (-2)] = 1/2 × (-14 + 63 – 2) sq. units

Δ = 1/2 × 47 sq. units = 47/2 sq. units

∴ Δ = 47/2 sq. units = 23.5 sq. units

Given vertices of the triangle are (–2, –3), (3, 2), (–1, –8)

Let the vertices of the triangle be given by (x₁, y₁), (x₂, y₂), (x₃, y₃)

Area of triangle is given by Δ =

Area of triangle = Δ =

Expanding the determinant along Row 1

Δ = 1/2 × |[(-2) × (2 × 1 – (-8) × 1) – (-3) × (3 × 1 – (-1) × 1) + 1 × (3 × (-8) – 2 × (-1))]|

Δ = 1/2 × |[(-2) × (2 + 8) + 3 × (3 + 1) + 1 × (-24 + 2)]|

Δ = 1/2 × |[(-2) × 10 + 3 × 4 + 1 × (-22)]| = 1/2 ×| (-20 + 12 – 22)| sq. units

Δ = 1/2 × |-30| sq. units = 30/2 sq. units

∴ Δ = 30/2 sq. units = 15 sq. units

Given vertices of the triangle are A (a, b + c), B (b, c + a), C (c, a + b)

Let the vertices of the triangle be given by (x₁, y₁), (x₂, y₂), (x₃, y₃)

Area of triangle is given by Δ =

For points to be collinear area of triangle = Δ = 0

So, we have to show that area of triangle formed by ABC is 0

Area of triangle = Δ =

Expanding the determinant along Row 1

Δ = 1/2 × [a × {(c + a) × 1 – (a + b) × 1) – (b + c) × {b × 1 – c × 1} + 1 × {b × (a + b) – c × (c +a)}]

Δ = 1/2 × [a × (c + a – a – b) – (b + c) × (b – c) + 1 × (ab + b² – c² - ca)]

Δ = 1/2 × [a × (c – b) – (b² – c²) + 1 × (ab + b² – c² – ca)]

Δ = 1/2 × (ac – ab – b² + c² + ab + b² – c² – ca) sq units

Δ = 1/2 × 0 sq units

∴ Δ = 0

∴ Given vertices of the triangle are A (a, b + c), B (b, c + a), C (c, a + b) are collinear

Given vertices of the triangle are (k, 0), (4, 0), (0, 2)

Let the vertices of the triangle be given by (x₁, y₁), (x₂, y₂), (x₃, y₃)

Area of triangle is given by Δ =

Given, Area of triangle = Δ = 4 sq. units

= 4

⇒ 4 = 1/2 |[k × (0 × 1 – 2 × 1) – 0 × (4 × 1 – 0 × 1) + 1 × (4 × 2 – 0 × 0)]|

⇒ 4 = 1/2 × |[k × (0 – 2) – 0 + 1 × (8 – 0)]|

⇒ � 4 × 2 = -2k + 8

⇒ 8 = -2k + 8 and -8 = -2k + 8

⇒ 8 – 8 = -2k and 8 + 8 = 2k

⇒ 2k = 0 and 16 = 2k

⇒ k = 0 and k = 8

Given vertices of the triangle are (–2, 0), (0, 4), (0, k)

Let the vertices of the triangle be given by (x₁, y₁), (x₂, y₂), (x₃, y₃)

Area of triangle is given by Δ =

Given, Area of triangle = Δ = 4 sq. units

= 4

⇒ 4 = 1/2 |[(-2) × (4 × 1 – k × 1) – 0 × (0 × 1 – 0 × 1) + 1 × (0 × k – 0 × 4)]|

⇒ 4 = 1/2 × |[(-2) × (4 – k) – 0 + 1 × (0 – 0)]|

⇒ 4 × 2 = |(-8 + 2k)

⇒ � 8= 2k - 8

⇒ 8 = 2k - 8 and ⇒ -8 = 2k - 8

⇒ 8 + 8 = 2k and ⇒ 8 - 8 = 2k

⇒ k = 16/2 and ⇒ k = 0/2

⇒ k = 8 and ⇒ k = 0

Equation of line joining points (x₁, y₁) & (x₂, y₂) is given by = 0

Given points are (1, 2) and (3, 6)

Equation of line is given by = 0

⇒ 1/2 × [1 × (6 × 1 – y × 1) – 2 × (3 × 1 – x × 1) + 1 × (3 × y – x × 6)] = 0

⇒ [(6 – y) – 2 × (3 – x) + (3y – 6x)] = 0 × 2

⇒ (6 – y – 6 + 2x + 3y – 6x) = 0

⇒ 2y – 4x = 0

⇒ y – 2x = 0 ⇒ y = 2x

∴ Required Equation of line is y = 2x

Equation of line joining points (x₁, y₁) & (x₂, y₂) is given by = 0

Given points are (3, 1) and (9, 3)

Equation of line is given by = 0

⇒ 1/2 × [3 × (3 × 1 – y × 1) – 1 × (9 × 1 – x × 1) + 1 × (9 × y – x × 3)] = 0

⇒ [3 × (3 – y) – 1 × (9 – x) + (9y – 3x)] = 0 × 2

⇒ (9 – 3y – 9 + x + 9y – 3x) = 0

⇒ 6y – 2x = 0

⇒ 2x – 6y = 0 ⇒ x – 3y = 0

∴ Required Equation of line is x – 3y = 0

Given vertices of the triangle are (2, – 6), (5, 4) and (k, 4).

Let the vertices of the triangle be given by (x₁, y₁), (x₂, y₂), (x₃, y₃)

Area of triangle is given by Δ =

Given, Area of triangle = Δ = 35 sq. units

= 35

⇒ � 35 = 1/2 × [2 × (4 × 1 – 4 × 1) – (-6) × (5 × 1 – k × 1) + 1 × (5 × 4 – k × 4)]

⇒ � 35 = 1/2 × [2 × (4 – 4) + 6 × (5 – k) + 1 × (20 – 4k)]

⇒ � 35 × 2 = (2 × 0 + 30 – 6k + 20 – 4k)

⇒ � 70 = 30 – 6k + 20 – 4k

⇒ � 70 = 50 – 10k

⇒ 70 – 50 = -10k and ⇒ -70 – 50 = -10k

⇒ 20 = -10k and ⇒ -120 = -10k

⇒ k = -20/10 and ⇒ k = 120/10

⇒ k = -2 and ⇒ k = 12

Minor: Minor of an element a_ij of a determinant is the determinant obtained by removing i^th row and j^th column in which element a_ij lies. It is denoted by M_ij.

Cofactor: Cofactor of an element a_ij, A_ij = (-1)^i+j M_ij.

Minor of element a_ij = M_ij

a₁₁ = 2, Minor of element a₁₁ = M₁₁ = 3

Here removing 1^st row and 1^st column from the determinant we are left out with 3 so M₁₁ = 3.

Similarly, finding other Minors of the determinant

a₁₂ = -4, Minor of element a₁₂ = M₁₂ = 0

a₂₁ = 0, Minor of element a₂₁ = M₂₁ = -4

a₂₂ = 3, Minor of element a₂₂ = M₂₂ = 2

Cofactor of an element a_ij, A_ij = (-1)^i+j × M_ij

A₁₁ = (-1)¹⁺¹ × M₁₁ = 1 × 3 = 3

A₁₂ = (-1)¹⁺² × M₁₂ = (-1) × 0 = 0

A₂₁ = (-1)²⁺¹ × M₁₁ = (-1) × (-4) = 4

A₂₂ = (-1)²⁺² × M₂₂ = 1 × 2 = 2

Minor of an element a_ij = M_ij

a₁₁ = a, Minor of element a₁₁ = M₁₁ = d

Here removing 1^st row and 1^st column from the determinant we are left out with d so M₁₁ = d.

Similarly, finding other Minors of the determinant

a₁₂ = c, Minor of element a₁₂ = M₁₂ = b

a₂₁ = b, Minor of element a₂₁ = M₂₁ = c

a₂₂ = d, Minor of element a₂₂ = M₂₂ = a

Cofactor of an element a_ij, A_ij = (-1)^i+j × M_ij

A₁₁ = (-1)¹⁺¹ × M₁₁ = 1 × d = d

A₁₂ = (-1)¹⁺² × M₁₂ = (-1) × b = -b

A₂₁ = (-1)²⁺¹ × M₁₁ = (-1) × c = -c

A₂₂ = (-1)²⁺² × M₂₂ = 1 × a = a

Minor of an element a_ij = M_ij

a₁₁ = 1, Minor of element a₁₁ = M₁₁ = = (1 × 1) – (0 × 0) = 1

Here removing 1^st row and 1^st column from the determinant we are left out with the determinant. Solving this we get M₁₁ = 1

Similarly, finding other Minors of the determinant

a₁₂ = 0, Minor of element a₁₂ = M₁₂ = = (0 × 1) – (0 × 0) = 0

a₁₃ = 0, Minor of element a₁₃ = M₁₃ = = (0 × 0) - (1 × 0) = 0

a₂₁ = 0, Minor of element a₂₁ = M₂₁ = = (0 × 1) – (0 × 0) = 0

a₂₂ = 1, Minor of element a₂₂ = M₂₂ = = (1 × 1) – (0 × 0) = 1

a₂₃ = 0, Minor of element a₂₃ = M₂₃ = = (1 × 0) – (0 × 0) = 0

a₃₁ = 0, Minor of element a₃₁ = M₃₁ = = (0 × 0) – (0 × 1) = 0

a₃₂ = 0, Minor of element a₃₂ = M₃₂ = = (1 × 0) – (0 × 0) = 0

a₃₃ = 1, Minor of element a₃₃ = M₃₃ = = (1 × 1) – (0 × 0) = 1

Cofactor of an element a_ij, A_ij = (-1)^i+j × M_ij

A₁₁ = (-1)¹⁺¹ × M₁₁ = 1 × 1 = 1

A₁₂ = (-1)¹⁺² × M₁₂ = (-1) × 0 = 0

A₁₃ = (-1)¹⁺³ × M₁₃ = 1 × 0 = 0

A₂₁ = (-1)²⁺¹ × M₂₁ = (-1) × 0 = 0

A₂₂ = (-1)²⁺² × M₂₂ = 1 × 1 = 1

A₂₃ = (-1)²⁺³ × M₂₃ = (-1) × 0 = 0

A₃₁ = (-1)³⁺¹ × M₃₁ = 1 × 0 = 0

A₃₂ = (-1)³⁺² × M₃₂ = (-1) × 0 = 0

A₃₃ = (-1)³⁺³ × M₃₃ = 1 × 1 = 1

Minor of an element a_ij = M_ij

a₁₁ = 1, Minor of element a₁₁ = M₁₁ = = (5 × 2) – ((-1) × 1) = 10 + 1 = 11

Here removing 1^st row and 1^st column from the determinant we are left out with the determinant. Solving this we get M₁₁ = 11

Similarly, finding other Minors of the determinant

a₁₂ = 0, Minor of element a₁₂ = M₁₂ = = (3 × 2) – ((-1) × 0) = (6 - 0) = 6

a₁₃ = 4, Minor of element a₁₃ = M₁₃ = = (3 × 1) – (5 × 0) = 3 - 0 = 3

a₂₁ = 3, Minor of element a₂₁ = M₂₁ = = (0 × 2) – (4 × 1) = 0 – 4 = -4

a₂₂ = 5, Minor of element a₂₂ = M₂₂ = = (1 × 2) – (4 × 0) = 2 – 0 = 2

a₂₃ = -1, Minor of element a₂₃ = M₂₃ = = (1 × 1) – (0 × 0) = 1

a₃₁ = 0, Minor of element a₃₁ = M₃₁ = = (0 × (-1)) – (4 × 5) = 0 – 20 = -20

a₃₂ = 1, Minor of element a₃₂ = M₃₂ = = (1 × (-1)) – (4 × 3) = -1 – 12 = -13

a₃₃ = 2, Minor of element a₃₃ = M₃₃ = = (1 × 5) – (0 × 3) = (5 – 0) = 5

Cofactor of an element a_ij, A_ij = (-1)^i+j × M_ij

A₁₁ = (-1)¹⁺¹ × M₁₁ = 1 × 11 = 11

A₁₂ = (-1)¹⁺² × M₁₂ = (-1) × 6 = -6

A₁₃ = (-1)¹⁺³ × M₁₃ = 1 × 3 = 3

A₂₁ = (-1)²⁺¹ × M₂₁ = (-1) × (-4) = 4

A₂₂ = (-1)²⁺² × M₂₂ = 1 × 2 = 2

A₂₃ = (-1)²⁺³ × M₂₃ = (-1) × 1 = -1

A₃₁ = (-1)³⁺¹ × M₃₁ = 1 × (-20) = -20

A₃₂ = (-1)³⁺² × M₃₂ = (-1) × (-13) = 13

A₃₃ = (-1)³⁺³ × M₃₃ = 1 × 5 = 5

To evaluate a determinant using cofactors, Let

B =

Expanding along Row 1

B =

B = a₁₁ A₁₁ + a₁₂ A₁₂ + a₁₃ A₁₃

[Where A_ij represents cofactors of a_ij of determinant B.]

B = Sum of product of elements of R₁ with their corresponding cofactors

Similarly, the determinant can be solved by expanding along column

So, B = sum of product of elements of any row or column with their corresponding cofactors

Cofactors of second row

A₂₁ = (-1)²⁺¹ × M₂₁ = (-1) × = (-1) × (3 × 3 – 8 × 2) = (-1) × (-7) = 7

A₂₂ = (-1)²⁺² × M₂₂ = 1 × = (5 × 3 – 8 × 1) = 7

A₂₃ = (-1)²⁺³ × M₂₃ = (-1) × = (-1) × (5 × 2 – 3 × 1) = (-1) × 7 = -7

[Where A_ij = (-1)^i+j × M_ij, M_ij = Minor of i^th row & j^th column]

Therefore,

Δ = a₂₁A₂₁ + a₂₂A₂₂ + a₂₃A₂₃

Δ = 2 × 7 + 1 × (-7) = 14 - 7 = 7

Ans: Δ = 7

To evaluate a determinant using cofactors, Let

B =

Expanding along Row 1

B =

B = a₁₁ A₁₁ + a₁₂ A₁₂ + a₁₃ A₁₃

[Where A_ij represents cofactors of a_ij of determinant B.]

B = Sum of product of elements of R₁ with their corresponding cofactors

Similarly, the determinant can be solved by expanding along column

So, B = sum of product of elements of any row or column with their corresponding cofactors

Cofactors of third column

A₁₃ = (-1)¹⁺³ × M₁₃ = 1 × = 1 × (1 × z – 1 × y) = (z – y)

A₂₃ = (-1)²⁺³ × M₂₃ = (-1) × = (-1) × (1 × z – 1 × x) = - (z - x) = (x - z)

A₃₃ = (-1)³⁺³ × M₃₃ = 1 × = 1 × (1 × y – 1 × x) = (y – x)

[Where A_ij = (-1)^i+j × M_ij, M_ij = Minor of i^th row & j^th column]

Therefore,

Δ = a₁₃A₁₃ + a₂₃A₂₃ + a₃₃A₃₃

Δ = yz (z - y) + zx (x - z) + xy (y - x) = z [y (z - y) + x (x - z)] + xy (y - x)

Δ = z (yz - y² + x² - xz) + xy (y - x) = z [(yz - xz) + (x² - y²)] + xy (y - x)

Δ = z [z × (y - x) + (x + y) × (x - y)] + xy (y - x)

Δ = z × (y - x) × (z – x - y) + xy (y - x)

Δ = (y - x) × (z² – xz – yz + xy)

Δ = (y - x) × [z (z - x) – y (z - x)] = (y - x) × (z - y) × (z - x)

Δ = (x - y) (y - z) (z - x)

Ans: Δ = (x - y) (y - z) (z - x)

Δ =

Expanding along Column 1

Δ =

Δ = a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁

Adjoint of the matrix A = [a_ij]_n×n is defined as the transpose of the matrix [A_ij]_n×n where A_ij is the co-factor of the element a_ij.

Let’s find the cofactors for all the positions first-

Here, A₁₁ = 4, A₁₂ = -3, A₂₁ = -2, A₂₂ = 1.

∴ Adj A =

=

Adjoint of the matrix A = [a_ij]_n×n is defined as the transpose of the matrix [A_ij]_n×n where A_ij is the co-factor of the element a_ij.

Let’s find the cofactors for all the positions first-

Here, A₁₁ = 1{(3×1-0×5)} = 3

Similarly,

A₁₂ = -12, A₁₃ = 6, A₂₁ = 1, A₂₂ = 5, A₂₃ = 2, A₃₁ = -11, A₃₂ = -1, A₃₃ = 5.

Adjoint of the matrix A = [a_ij]_n×n is defined as the transpose of the matrix [A_ij]_n×n where A_ij is the co-factor of the element a_ij.

Let’s find the cofactors for all the positions first-

Here, A₁₁ = -6, A₁₂ = 4, A₂₁ = -3, A₂₂ = 2.

∴ Adj A =

=

So LHS = A(AdjA) =

Also AdjA(A) =

Determinant of A = |A| = 2(-6)-(3)(-4) = 0

So RHS = |A|I = 0

Hence A(AdjA) = AdjA(A) = |A|I = 0 {hence proved}

Adjoint of the matrix A = [a_ij]_n×n is defined as the transpose of the matrix [A_ij]_n×n where A_ij is the co-factor of the element a_ij.

Let’s find the cofactors for all the positions first-

Here, A₁₁ = 0, A₁₂ = -11, A₁₃ = 0, A₂₁ = 3, A₂₂ = 1, A₂₃ = -1, A₃₁ = 2, A₃₂ = 8, A₃₃ = 3.

∴ Adj A =

=

So, LHS = A(AdjA) =

Also AdjA(A) =

Determinant of A = |A| = 11

So RHS = |A|I = .

Hence A(AdjA) = AdjA(A) = |A|I = {hence proved}

We know that

Adjoint of the matrix A = [a_ij]_n×n is defined as the transpose of the matrix [A_ij]_n×n where A_ij is the co-factor of the element a_ij.

Let’s find the cofactors for all the positions first-

Here, A₁₁ = 3, A₁₂ = -4, A₂₁ = 2, A₂₂ = 2.

∴ Adj A =

=

And |A| = 2(3)-(-2)(4) = 14

So .

We know that

Adjoint of the matrix A = [a_ij]_n×n is defined as the transpose of the matrix [A_ij]_n×n where A_ij is the co-factor of the element a_ij.

Let’s find the cofactors for all the positions first-

Here, A₁₁ = 2, A₁₂ = 3, A₂₁ = -5, A₂₂ = -1.

∴ Adj A =

=

And |A| = -1(2)-(-3)(5) = 13

So

Adjoint of the matrix A = [a_ij]_n×n is defined as the transpose of the matrix [A_ij]_n×n where A_ij is the co-factor of the element a_ij.

Let’s find the cofactors for all the positions first-

Here, A₁₁ = 10, A₁₂ = 0, A₁₃ = 0, A₂₁ = -10, A₂₂ = 5, A₂₃ = 0, A₃₁ = 2, A₃₂ = -4, A₃₃ = 2.

∴ Adj A =

And |A| = 10.

Adjoint of the matrix A = [a_ij]_n×n is defined as the transpose of the matrix [A_ij]_n×n where A_ij is the co-factor of the element a_ij.

Let’s find the cofactors for all the positions first-

Here, A₁₁ = -3, A₁₂ = 3, A₁₃ = -9, A₂₁ = 0, A₂₂ = -1, A₂₃ = -2, A₃₁ = 0, A₃₂ = 0, A₃₃ = 3.

∴ Adj A =

And |A| = -3.

.

Adjoint of the matrix A = [a_ij]_n×n is defined as the transpose of the matrix [A_ij]_n×n where A_ij is the co-factor of the element a_ij.

Let’s find the cofactors for all the positions first-

Here, A₁₁ = -1, A₁₂ = -4, A₁₃ = 1, A₂₁ = 5, A₂₂ = 23, A₂₃ = -11, A₃₁ = 3, A₃₂ = 12, A₃₃ = -6.

∴ Adj A =

= .

And |A| = -3.

.

Adjoint of the matrix A = [a_ij]_n×n is defined as the transpose of the matrix [A_ij]_n×n where A_ij is the co-factor of the element a_ij.

Let’s find the cofactors for all the positions first-

Here, A₁₁ = 2, A₁₂ = -9, A₁₃ = -6, A₂₁ = 0, A₂₂ = -2, A₂₃ = -1, A₃₁ = -1, A₃₂ = 3, A₃₃ = 2.

∴ Adj A =

=

And |A| = -1.

Adjoint of the matrix A = [a_ij]_n×n is defined as the transpose of the matrix [A_ij]_n×n where A_ij is the co-factor of the element a_ij.

Let’s find the cofactors for all the positions first-

Here, A₁₁ = -1, A₁₂ = 0, A₁₃ = 0, A₂₁ = 0, A₂₂ = -cosα, A₂₃ = -sinα, A₃₁ = 0, A₃₂ = -sinα, A₃₃ = cosα.

∴ Adj A =

And |A| = 1.

.

We have AB = = (61)(67)-(47)(87) = -2

Here determinant of matrix = |AB|≠ 0 hence (AB)^-1 exists.

Also |A| = 1 ≠ 0 and |B| = -2 ≠ 0.

∴ A^-1 and B^-1 will also exist and are given by-

And hence,

{Hence proved}

We have A² = A.A = .

So A² – 5A + 7I =

Hence A² – 5A + 7I = 0

∴ A.A – 5A = -7I

Now post multiply with A^-1

So A.A.A^-1-5A.A^-1 = -7I.A^-1

→ A.I – 5I = -7I.A^-1 {since A.A^-1 = I}

A – 5I = -7A^-1 {since X.I = X}

We have A² = A.A =

Since A² + aA + bI =

So A² + aA + bI =

Hence 10+3a+b = 0 …(i)

5+a = 0 …(ii)

5+2a+b = 0 …(iii)

From (ii) a = -5

Putting a in (iii) we get b = 5

So a = -5 and b = 5 satisfy the equation.

Here A² = A.A =

And hence A³ = A. A² =

∴ A³– 6A² + 5A + 11 I =

Thus, A³– 6A² + 5A + 11 I = 0

Now, A³– 6A² + 5A + 11 I = 0,

→ (A.A.A)- 6 (A.A) +5A = -11I

Post-multiply with A^-1 on both sides-

→ (A.A.A.A^-1)- 6 (A.A.A^-1) +5A.A^-1 = -11I. A^-1

→ (A.A.I) – 6(A.I) + 5I = -11I. A^-1 {since A.A^-1 = I}

→ (A.A) – 6A +5I = -11A^-1 {since X.I = X}

Hence

Here A² = A.A =

And hence A³ = A. A² =

∴ A³– 6A² + 9A -4I

Thus, A³– 6A² + 9A -4I = 0

Now, A³– 6A² + 9A -4I = 0,

→ (A.A.A)- 6 (A.A) +9A = 4I

Post-multiply with A^-1 on both sides-

→ (A.A.A.A^-1)- 6 (A.A.A^-1) +9A.A^-1 = 4I. A^-1

→ (A.A.I) – 6(A.I) + 9I = 4I. A^-1 {since A.A^-1 = I}

→ (A.A) – 6A +9I = 4A^-1 {since X.I = X}

Hence

For a square matrix of order n×n,

We know that |AdjA| = |A|^n-1

So, |AdjA| = |A|^(3-1) = |A|²

{since adj(A) is of order n and |Adj(A)| = |A|^n-1}

Alternative-

We know that AA^-1 = I

So |A||A^-1| = |I| = 1

The given system of equations is:

x + 2y = 2

2x + 3y = 3

The given system of equations can be written in the form of AX = B, where

Now |A| = 3(1)-2(2) = -1 ≠ 0.

∴ A is a non-singular matrix and hence A-1 exists.

So, The system of equations will be consistent.

The given system of equations is:

2x - y = 5

x + y = 4

The given system of equations can be written in the form of AX = B, where

Now |A| = 2(1)-1(-1) = 3 ≠ 0.

∴ A is a non-singular matrix and hence A^-1 exists.

So, The system of equations will be consistent.

The given system of equations is:

x + 3y = 5

2x + 6y = 8

The given system of equations can be written in the form of AX = B, where

Now |A| = 1(6)-3(2) = 0

∴ A is a singular matrix and hence A^-1 doesn’t exist.

So, The system of equations will be inconsistent.

The given system of equations is:

x + y + z = 1

2x + 3y + 2z = 2

ax + ay + 2az = 4

The given system of equations can be written in the form of AX = B, where

Now |A| = 1(6a-2a)-1(4a-2a) +1(2a-3a) = a ≠ 0

∴ A is a non-singular matrix and hence A^-1 exists.

So the system of equations will be consistent.

The given system of equations is:

3x - y - 2z = 2

0x + 2y - z = -1

3x - 5y +0z = 3

The given system of equations can be written in the form of AX = B, where

Now |A| = 3(-5) + 3(5) = 0

∴ A is a singular matrix and hence A^-1 doesn’t exist.

So the system of equations will be inconsistent.

The given system of equations is:

5x - y + 4z = 5

2x + 3y + 5z = 2

5x - 2y + 6z = -1

The given system of equations can be written in the form of AX = B, where

Now |A| = 5(18+10) + 1(12-25) + 4(-4-15) = 51 ≠ 0

∴ A is a non-singular matrix and hence A^-1 exists.

So, The system of equations will be consistent.

The given system of equations is:

5x + 2y = 4

7x + 3y = 5

The given system of equations can be written in the form of AX = B, where

Now |A| = 3(5)-2(7) = 1 ≠ 0

∴ A is a non-singular matrix and hence A^-1 exists.

Now

So

And

So

Hence x = 2 and y = -3.

The given system of equations is:

2x - y = -2

3x + 4y = 3

The given system of equations can be written in the form of AX = B, where

Now |A| = 2(4)-3(-1) = 11 ≠ 0.

∴ A is a non-singular matrix and hence A^-1 exists.

Now

So

And

So

Hence

The given system of equations is:

4x - 3y = 3

3x - 5y = 7

The given system of equations can be written in the form of AX = B, where

Now |A| = 4(-5)-3(-3) = -11 ≠ 0

∴ A is a non-singular matrix and hence A^-1 exists.

Now

AdjA =

So

And

So

Hence

The given system of equations is:

5x + 2y = 3

3x + 2y = 5

The given system of equations can be written in the form of AX = B, where

Now |A| = 5(2)-2(3) = 4 ≠ 0

∴ A is a non-singular matrix and hence A^-1 exists.

Now

AdjA =

So

And

So

Hence x = -1 and y = 4

The given system of equations is:

2x + y + z = 1

0x + 3y - 5z = 9

The given system of equations can be written in the form of AX = B, where

Now |A| = 2(10+3)-1(-5) +1(3) = 34 ≠ 0

∴ A is a non-singular matrix and hence A^-1 exists.

Now A₁₁ = 13, A₁₂ = 5, A₁₃ = 3, A₂₁ = 8, A₂₂ = -10, A₂₃ = -6, A₃₁ = 1, A₃₂ = 3, A₃₃ = -5

So AdjA =

∴

And hence X = A^-1B

So

Hence

The given system of equations is:

x - y + z = 4

2x + y -3z = 0

x + y + z = 2

The given system of equations can be written in the form of AX = B, where

Now |A| = 1(1+3) + 1(5) + 1(1) = 10 ≠ 0

∴ A is a non-singular matrix and hence A^-1 exists.

Now A₁₁ = 4, A₁₂ = -5, A₁₃ = 1, A₂₁ = 2, A₂₂ = 0, A₂₃ = -2, A₃₁ = 1, A₃₂ = -2, A₃₃ = 3

So AdjA =

And hence X = A^-1B

So

Hence x = 2, y = -1 and z = 1.

The given system of equations is:

2x + 3y + 3z = 5

x – 2y + z = -4

3x - y - 2z = 3

The given system of equations can be written in the form of AX = B, where

Now |A| = 2(4+1)-3(-5) +3(5) = 40 ≠ 0

∴ A is a non-singular matrix and hence A^-1 exists.

Now A₁₁ = 5, A₁₂ = 5, A₁₃ = 5, A₂₁ = 3, A₂₂ = -13, A₂₃ = 11, A₃₁ = 9, A₃₂ = 1, A₃₃ = -7

So AdjA =

∴

And hence X = A^-1B

So

Hence x = 1, y = 2 and z = -1

The given system of equations is:

x - y + 2z = 7

3x + 4y - 5z = -5

2x – y + 3z = 12

The given system of equations can be written in the form of AX = B, where

Now |A| = 1(12-5) + 1(9+10) + 2(-3-8) = 4 ≠ 0

∴ A is a non-singular matrix and hence A^-1 exists.

Now A₁₁ = 7, A₁₂ = -19, A₁₃ = -11, A₂₁ = 1, A₂₂ = -1, A₂₃ = -1, A₃₁ = -3, A₃₂ = 11, A₃₃ = 7

So AdjA =

∴

So

Hence

∴ |A| = 2(-4+4) + 3(-6+4) + 5(3-2)


= -1 ≠ 0

So inverse of A exists,

Now A₁₁ = 0, A₁₂ = 2, A₁₃ = 1, A₂₁ = -1, A₂₂ = -9, A₂₃ = -5, A₃₁ = 2, A₃₂ = 23, A₃₃ = 13

So AdjA =

∴

So .

Hence x = 1, y = 2 and z = 3

Let the cost of onions, wheat and rice per kg be Rs. x, Rs. y and Rs. z respectively.

Then according to the giving situation, it can be represented by a system of equations as-

4x + 3y + 2z = 60

2x + 4y + 6z = 90

6x + 2y + 3z = 70

The system of equations can be written in the form of AX = B, where

Here |A| = 4(12-12)-3(6-36) + 2(4-24) = 50 ≠ 0

Hence A^-1 will exist.

Now, A₁₁ = 0, A₁₂ = 30, A₁₃ = -20, A₂₁ = -5, A₂₂ = 0, A₂₃ = 10, A₃₁ = 10, A₃₂ = -20, A₃₃ = 10

Hence x = 5, y = 8 & z = 8.

Hence, the cost of onions is Rs. 5 per kg, the cost of wheat is Rs. 8 per kg, and the cost of rice is Rs. 8 per kg

Let Δ =

Expanding the above determinant along R₁ i.e. Row 1

Δ = x (-x² – 1) – sinθ (-x × sinθ – cosθ × 1) + cosθ (-sinθ × 1 + x cosθ)

Δ = -x³ – x + xsin²θ + sinθ × cos θ – sinθ × cosθ + x cos²θ

Δ = -x³ – x + x (sin²θ + cos²θ)

Since, sin²θ + cos²θ = 1

∴ Δ = -x³ – x + x

Hence Δ is independent of θ.

Multiplying row 1 with a, row 2 with b and row 3 with c

[R₁→ aR₁, R₂→ bR₂, R₃→ cR₃]

[Taking common abc from C₃]

[Applying C₁↔ C₃]

= [Applying C₂ ↔ C₃]

= RHS

Since, LHS = RHS

∴ the given result is proved.

Let Δ =

Expanding along C₃ we get

Δ = -sinα (-sinβ × sinα sinβ – cosβ × sinα cosβ) – 0 (sinα cosβ × cosα sinβ – cosα cosβ × sinα sinβ) + cosα (cosα cosβ × cosβ – cosα sinβ × (-sinβ))

Δ = -sinα (-sinα sin²β – sinα cos²β) – 0 + cosα (cosα cos²β + cosα sin²β)

Δ = sinα × sinα (sin²β + cos²β) + cosα × cosα (cos²β + sin²β)

[Taking –sinα and cosα common]

Since, sin²β + cos²β = 1

∴ Δ = sin²α + cos²α

Δ = 1 [sin²α + cos²α = 1]

Given, Δ =

Applying Elementary transformations, we get

R₁→ R₁ + R₂ + R₃ we have,

Δ = 2 (a + b + c)

Now applying C₂→ C₂ – C₁ and C₃→ C₃ – C₁

Δ = 2 (a + b + c)

Expanding along R₁

Δ = 2 (a + b + c) [1 {(b – c)(c – b) – (b – a)(c – a)} – 0 + 0]

Δ = 2 (a + b + c) [- b² – c² + 2bc – bc + ba + ac – a²]

Δ = 2 (a + b + c) [ab + bc + ca – a² – b² – c²]

Given that Δ = 0

∴ 2 (a + b + c) [ab + bc + ca – a² – b² – c²] = 0

⇒ Either a + b + c = 0, or ab + bc + ca – a² – b² – c² = 0

Now

ab + bc + ca – a² – b² – c² = 0

Multiplying both sides by -2

⇒ - 2ab – 2bc – 2ca + 2a² + 2b² + 2c² = 0

⇒ a² – 2ab + b² + b² – 2bc + c² + c² – 2ca + a² = 0

⇒ (a – b)² + (b – c)² + (c – a)² = 0

Since, (a – b)², (b – c)², (c – a)² are non-negative

∴ (a – b)² = (b – c)² = (c – a)²

⇒ (a – b) = (b – c) = (c – a)

⇒ a = b = c

Hence, if ∆ = 0, then either a + b + c = 0 or a = b = c

Given,

Applying elementary transformations we get,

R₁→ R₁ + R₂ + R₃

⇒ (3x + a) = 0

C₂→ C₂ – C₁ and C₃→ C₃ – C₁, we get

(3x + a) = 0

Expanding along R₁ we have

(3x + a) [1 (a × a – 0 × 0) – 0 (x × a – 0 × a) + 0 (x × 0 – a × x)] = 0

⇒ (3x + a) [1 × a²] = 0

⇒ a² (3x + a) = 0

Since, a ≠ 0

∴ 3x + a = 0

⇒ x = -a/3

Given,

LHS =

RHS = 4a²b²c²

LHS = Δ =

Taking out common factors a, b and c from C₁, C₂ and C₃, we have

Δ = abc

Applying Elementary Transformations

R₂→ R₂ – R₁ and R₃→ R₃ – R₁

Δ = abc

R₂→ R₂ + R₁

Δ = abc

R₃→ R₃ + R₂

Δ = abc

Δ = 2ab²c

C₂→ C₂ – C₁

Δ = 2ab²c

Expanding along R₃ we get,

Δ = 2ab²c [a (c – a) + a (a + c)]

= 2ab²c [ac – a² + a² + ac]

= 2ab²c (2ac)

= 4a²b²c²

Δ = RHS

∴ LHS = RHS

Hence, Proved

B =

We need to find B^-1

To find the inverse of a matrix we need to find the Adjoint of that matrix

For finding the adjoint of the matrix we need to find its cofactors

Let B_ij denote the cofactors of Matrix B

Minor of an element b_ij = M_{ij �}

b₁₁ = 1, Minor of element b₁₁ = M₁₁ = = (3 × 1) – ((-2) × 0) = 3

b₁₂ = 2, Minor of element b₁₂ = M₁₂ = = ((-1) × 1) – (0 × 0) = -1

b₁₃ = -2, Minor of element b₁₃ = M₁₃ = = ((-1) × (-2)) – (3 × 0) = 2

b₂₁ = -1, Minor of element b₂₁ = M₂₁ = = (2 × 1) – ((-2) × (-2)) = -2

b₂₂ = 3, Minor of element b₂₂ = M₂₂ = = (1 × 1) – ((-2) × 0) = 1

b₂₃ = 0, Minor of element b₂₃ = M₂₃ = = (1 × (-2)) – (2 × 0) = -2

b₃₁ = 0, Minor of element b₃₁ = M₃₁ = = (2 × 0) – ((-2) × 3) = 6

b₃₂ = -2, Minor of element b₃₂ = M₃₂ = = (1 × 0) – ((-2) × (-1)) = -2

b₃₃ = 1, Minor of element b₃₃ = M₃₃ = = (1 × 3) – (2 × (-1)) = 5

Cofactor of an element b_ij, B_ij = (-1)^i+j × M_ij

B₁₁ = (-1)¹⁺¹× M₁₁ = 1 × 3 = 3

B₁₂ = (-1)¹⁺²× M₁₂ = (-1) × (-1) = 1

B₁₃ = (-1)¹⁺³× M₁₃ = 1 × 2 = 2

B₂₁ = (-1)²⁺¹× M₂₁ = (-1) × (-2) = 2

B₂₂ = (-1)²⁺² × M₂₂ = 1 × 1 = 1

B₂₃ = (-1)²⁺³ × M₂₃ = (-1) × (-2) = 2

B₃₁ = (-1)³⁺¹ × M₃₁ = 1 × 6 = 6

B₃₂ = (-1)³⁺² × M₃₂ = (-1) × (-2) = 2

B₃₃ = (-1)³⁺³ × M₃₃ = 1 × 5 = 5

Adj B = =

|B| = 1 (3 × 1 – (-2) × 0) – 2 ((-1) × 1 – 0 × 0) + (-2) ((-1) × (-2) – 3 × 0)

|B| = 3 – 2 ( -1 – 0) – 2 (2 – 0)

|B| = 3 + 2 – 4 = 1

∴ B^-1 = (Adj B)/|B| = /1 =

We know (AB)^-1 = B^-1A^-1

(AB)^-1 = ×

Solving the above matrix we get

(AB)^-1 =

A =

|A| = 1 (3 × 5 – 1 × 1) – (-2) ((-2) × 5 – 1 × 1) + 1 ((-2) × 1 – 3 × 1)

|A| = (15 – 1) + 2 (-10 – 1) + (-2 – 3)

|A| = 14 – 22 – 5 = -13

To find the inverse of a matrix we need to find the Adjoint of that matrix

For finding the adjoint of the matrix we need to find its cofactors

Let A_ij denote the cofactors of Matrix A

Minor of an element a_ij = M_{ij �}

a₁₁ = 1, Minor of element a₁₁ = M₁₁ = = (3 × 5) – (1 × 1) = 14

a₁₂ = -2, Minor of element a₁₂ = M₁₂ = = (-2 × 5) – (1 × 1) = -11

a₁₃ = 1, Minor of element a₁₃ = M₁₃ = = (-2 × 1) – (3 × 1) = -5

a₂₁ = -2, Minor of element a₂₁ = M₂₁ = = ((-2) × 5) – (1 × 1) = -11

a₂₂ = 3, Minor of element a₂₂ = M₂₂ = = (1 × 5) – (1 × 1) = 4

a₂₃ = 1, Minor of element a₂₃ = M₂₃ = = (1 × 1) – ((-2) × 1) = 3

a₃₁ = 1, Minor of element a₃₁ = M₃₁ = = (-2 × 1) – (3 × 1) = -5

a₃₂ = 1, Minor of element a₃₂ = M₃₂ = = (1 × 1) – (1 × (-2)) = 3

a₃₃ = 5, Minor of element a₃₃ = M₃₃ = = (1 × 3) – ((-2) × (-2)) = -1

Cofactor of an element a_ij = A_ij

A₁₁ = (-1)¹⁺¹× 14 = 1 × 14 = 14

A₁₂ = (-1)¹⁺²× (-11) = (-1) × (-11) = 11

A₁₃ = (-1)¹⁺³× (-5) = 1 × (-5) = -5

A₂₁ = (-1)²⁺¹× (-11) = (-1) × (-11) = 11

A₂₂ = (-1)²⁺² × 4 = 1 × 4 = 4

A₂₃ = (-1)²⁺³ × 3 = (-1) × 3 = -3

A₃₁ = (-1)³⁺¹ × (-5) = 1 × (-5) = -5

A₃₂ = (-1)³⁺² × 3 = (-1) × 3 = -3

A₃₃ = (-1)³⁺³ × (-1) = 1 × (-1) = -1

Adj A = =

A^-1 = (Adj A)/|A|

A^-1 = =

(i) |Adj A| = 14(-4 – 9) – 11 (-11 – 15) – 5 (-33 + 20)

= 14 × (-13) – 11 × (-26) – 5 (-13)

= -182 + 286 + 65 = 169

Similarly Finding the Adj (Adj A) as found above

Adj (Adj A) =

[Adj A]^-1 = Adj (Adj A)/|Adj A|

=

=

A^-1 = =

Similarly Finding the Adj (A^-1) as found above

Adj (A^-1) = =

Hence, [Adj A]^-1 = Adj (A^-1)

A =

|A| = 1 (3 × 5 – 1 × 1) – (-2) ((-2) × 5 – 1 × 1) + 1 ((-2) × 1 – 3 × 1)

|A| = (15 – 1) + 2 (-10 – 1) + (-2 – 3)

|A| = 14 – 22 – 5 = -13

To find the inverse of a matrix we need to find the Adjoint of that matrix

For finding the adjoint of the matrix we need to find its cofactors

Let A_ij denote the cofactors of Matrix A

Minor of an element a_ij = M_{ij �}

a₁₁ = 1, Minor of element a₁₁ = M₁₁ = = (3 × 5) – (1 × 1) = 14

a₁₂ = -2, Minor of element a₁₂ = M₁₂ = = (-2 × 5) – (1 × 1) = -11

a₁₃ = 1, Minor of element a₁₃ = M₁₃ = = (-2 × 1) – (3 × 1) = -5

a₂₁ = -2, Minor of element a₂₁ = M₂₁ = = ((-2) × 5) – (1 × 1) = -11

a₂₂ = 3, Minor of element a₂₂ = M₂₂ = = (1 × 5) – (1 × 1) = 4

a₂₃ = 1, Minor of element a₂₃ = M₂₃ = = (1 × 1) – ((-2) × 1) = 3

a₃₁ = 1, Minor of element a₃₁ = M₃₁ = = (-2 × 1) – (3 × 1) = -5

a₃₂ = 1, Minor of element a₃₂ = M₃₂ = = (1 × 1) – (1 × (-2)) = 3

a₃₃ = 5, Minor of element a₃₃ = M₃₃ = = (1 × 3) – ((-2) × (-2)) = -1

Cofactor of an element a_ij = A_ij

A₁₁ = (-1)¹⁺¹× 14 = 1 × 14 = 14

A₁₂ = (-1)¹⁺²× (-11) = (-1) × (-11) = 11

A₁₃ = (-1)¹⁺³× (-5) = 1 × (-5) = -5

A₂₁ = (-1)²⁺¹× (-11) = (-1) × (-11) = 11

A₂₂ = (-1)²⁺² × 4 = 1 × 4 = 4

A₂₃ = (-1)²⁺³ × 3 = (-1) × 3 = -3

A₃₁ = (-1)³⁺¹ × (-5) = 1 × (-5) = -5

A₃₂ = (-1)³⁺² × 3 = (-1) × 3 = -3

A₃₃ = (-1)³⁺³ × (-1) = 1 × (-1) = -1

Adj A = =

A^-1 = (Adj A)/|A|

A^-1 = =

(ii) To find (A^-1)^-1 we have to find out Adj(A^-1)

A^-1 =

|A^-1| = (-1/13)³ [14 (4 × (-1) – (-3) × (-3)) – 11 (11 × (-1) – (-3) × (-5)) + (-5) (11 × (-3) – 4 × (-5))]

|A| = (-1/13)³ [14 (-4 – 9) – 11 (-11 – 15) – 5 (-33 + 20)]

|A| = (-1/13)³ [14 × (-13) – 11 × (-26) – 5 × (-13)]

|A| = (-1/13)³ × 169 = -1/13

Cofactor of an element a_ij = A_ij

A₁₁ = (-1)¹⁺¹× (-1/13) = 1 × (-1/13) = -1/13

A₁₂ = (-1)¹⁺²× (-2/13) = (-1) × (-2/13) = 2/13

A₁₃ = (-1)¹⁺³× (-1/13) = 1 × (-1/13) = -1/13

A₂₁ = (-1)²⁺¹× (-2/13) = (-1) × (-2/13) = 2/13

A₂₂ = (-1)²⁺² × (3/13) = 1 × (-3/13) = -3/13

A₂₃ = (-1)²⁺³ × (1/13) = (-1) × (1/13) = -1/13

A₃₁ = (-1)³⁺¹ × (-1/13) = 1 × (-1/13) = -1/13

A₃₂ = (-1)³⁺² × (1/13) = (-1) × 1/13 = -1/13

A₃₃ = (-1)³⁺³ × (-5/13) = 1 × (-5/13) = -5/13

Adj (A^-1) = =

(A^-1)^-1 = Adj(A^-1)/|A^-1|

(A^-1)^-1 = = = A

∴ (A^-1)^-1 = A

Let Δ =

Applying Elementary Transformations.

Applying R₁→ R₁ + R₂ + R₃, we have

Δ =

Δ = 2 (x + y)

Applying C₂→ C₂ – C₁ and C₃→ C₃ – C₁, we have

Δ = 2 (x + y)

Expanding along R₁, we have

Δ = 2 (x + y) [1 (x × (-x) – (-y) × (x – y)) – 0 + 0]

Δ = 2 (x + y) [-x² + y (x – y)]

Δ = 2 (x + y) [-x² + xy – y²]

Δ = -2 (x + y) [x² – xy + y²]

Δ = -2 (x³ + y³)

Let Δ =

Applying Row transformations

R₂→ R₂ – R₁; R₃→ R₃ – R₁

Δ =

Δ =

Now, Expanding along C₁

Δ = 1 (y × x – 0 × 0) – 0 (x × x – 0 × y) + 0 (x × 0 – y × y)

Δ = 1 (xy) – 0 + 0

Δ = xy

Let Δ =

Applying Row Transformations

R₂→ R₂ – R₁

Δ =

R₃→ R₃ – R₁

Δ =

Taking (β – α)(γ – α) from R₂ and R₃ respectively

Δ = (β – α) (γ – α)

Applying R₃→ R₃ – R₂, we have

Δ = (β – α) (γ – α)

Expanding along R₃, we have

Δ = (β – α) (γ – α) [0 (α² × (-1) – (β + γ) × (β + γ) – (γ – β)((-1) × α – 1 × (β + γ) + 0 (α × (β + γ) – 1 × α²)

Δ = (β – α) (γ – α) [0 – (γ – β)( - α - β – γ) + 0]

Δ = (β – α) (γ – α) (γ – β) (α + β + γ)

Δ = (α – β) (β – γ) (γ – α) (α + β + γ)

Let Δ =

Applying Elementary Row Transformations

R₂→ R₂ – R₁ and R₃→ R₃ – R₁

Δ =

Taking (y – x) and (z – x) common from R₂ and R₃ respectively

Δ = (y – x) (z – x)

Applying R₃→ R₃ – R₂

Δ = (y – x) (z – x)

Taking (z – y) common from R₃

Δ = (y – x) (z – x) (z – y)

Expanding along R₃, we have

Δ = (x – y) (y – z) (z – x) [0 – 1 {x × p(y² + x² + xy) – 1 × (1 + px³)} + p (x + y + z) {x × (y + x) – 1 × x²}

Δ = (x – y) (y – z) (z – x) (-px³ – pxy² – px²y + 1 + px³ + px²y + pxy² + pxyz)

Δ = (x – y) (y – z) (z – x) (1 + pxyz)

Hence, the given result is proved.

Let Δ =

Applying Elementary Transformations

C₁→ C₁ + C₂ + C₃, we have

Δ =

Taking (a + b + c) common from C₁, we get

Δ = (a + b + c)

Applying R₂→ R₂ – R₁ and R₃→ R₃ – R₁

Expanding along C₁

Δ = (a + b + c) [1 × {(2b + a) (2c + a) – (a – b) (a – c)} – 0 + 0]

Δ = (a + b + c) [4bc + 2ab + 2ac + a² – a² + ac + ba – bc]

Δ = (a + b + c) (3ab + 3bc + 3ac)

Δ = 3 (a + b + c) (ab + bc + ca)

Hence, the given result proved.

Let Δ =

Applying Elementary Row Transformations

R₂→ R₂ – 2R₁

Δ =

R₃→ R₃ – 3R₁

Δ =

R₃→ R₃ – 3R₂

Δ =

Expanding Along C₁, we have

Δ = 1 (1 × 1 – 0 × (2 + p)) – 0 + 0

Δ = 1 – 0

Δ = 1

Hence, the given result is proved

Let Δ =

Δ =

Δ =

Applying Elementary Column Transformations

C₁→ C₁ + C₃

Δ =

Since, the two columns are identical

[In a determinant if two columns are identical the the value of determinant is 0]

So, the value of given determinant is 0

∴ Δ = 0

Hence, the given result is proved.

Given System of equations are

Let

∴ Given system of equation becomes

2p + 3q + 10r = 4

4p – 6q + 5r = 1

6p + 9q – 20r = 2

The given System of Equations can be written in the form of AX = B

Here A = and B = , X =

Now we need to find |A|

∴ |A| = = 2 × (120 – 45) – 3 × (- 80 – 30) + 10 × (36 + 36)

= 150 + 330 + 720

= 1200

Since, |A| ≠ 0

∴ A is non-singular. So, it’s inverse exists

Cofactor of an element a_ij = A_ij

A₁₁ = (-1)¹⁺¹× 75 = 1 × 75 = 75

A₁₂ = (-1)¹⁺²× (-110) = (-1) × (-110) = 110

A₁₃ = (-1)¹⁺³× 72 = 1 × (72) = 72

A₂₁ = (-1)²⁺¹× (-150) = (-1) × (-150) = 150

A₂₂ = (-1)²⁺² × (-100) = 1 × (-100) = -100

A₂₃ = (-1)²⁺³ × 0 = (-1) × 0 = 0

A₃₁ = (-1)³⁺¹ × 75 = 1 × 75 = 75

A₃₂ = (-1)³⁺² × (-30) = (-1) × (-30) = 30

A₃₃ = (-1)³⁺³ × (-24) = 1 × (-24) = -24

Adj A = =

A^-1 = (Adj A)/|A|

A^-1 =

Now,

Since, AX = B

∴ X = A^-1B

∴ p = �, q = 1/3, r = 1/5

∴ x = 1/p = 2; y = 1/q = 3; z = 1/r = 5

So, x = 2; y = 3; z = 5

Let Δ =

Since, a, b, c are in A.P.

∴ 2b = a + c

Δ =

Applying Elementary Row Transformations

R₁→ R₁ – R₂ and R₃→ R₃ – R₂

Δ =

R₁→ R₁ + R₃, we have

Δ =

[In a determinant if all elements of a row is 0 then the value of determinant is 0.]

So, here all the elements of first row (R₁) are zero.

∴ Δ = 0

A =

|A| = x × (y × z) = xyz

Since, |A| ≠ 0

A^-1 exists

To find the inverse of a matrix we need to find the Adjoint of that matrix

For finding the adjoint of the matrix we need to find its cofactors

Let A_ij denote the cofactors of Matrix A

Minor of an element a_ij = M_{ij �}

a₁₁ = x, Minor of element a₁₁ = M₁₁ = = (y × z) – (0 × 0) = yz

a₁₂ = 0, Minor of element a₁₂ = M₁₂ = = (0 × z) – (0 × 0) = 0

a₁₃ = 0, Minor of element a₁₃ = M₁₃ = = (0 × 0) – (0 × y) = 0

a₂₁ = 0, Minor of element a₂₁ = M₂₁ = = (0 × z) – (0 × 0) = 0

a₂₂ = y, Minor of element a₂₂ = M₂₂ = = (x × z) – (0 × 0) = xz

a₂₃ = 0, Minor of element a₂₃ = M₂₃ = = (x × 0) – (0 × 0) = 0

a₃₁ = 0, Minor of element a₃₁ = M₃₁ = = (z × 0) – (0 × 0) = 0

a₃₂ = 0, Minor of element a₃₂ = M₃₂ = = (x × 0) – (0 × 0) = 0

a₃₃ = z, Minor of element a₃₃ = M₃₃ = = (x × y) – (0 × 0) = xy

Cofactor of an element a_ij, A_ij = (-1)^i+j × M_ij

A₁₁ = (-1)¹⁺¹× M₁₁ = 1 × yz = yz

A₁₂ = (-1)¹⁺²× M₁₂ = (-1) × 0 = 0

A₁₃ = (-1)¹⁺³× M₁₃ = 1 × 0 = 0

A₂₁ = (-1)²⁺¹× M₂₁ = (-1) × 0 = 0

A₂₂ = (-1)²⁺² × M₂₂ = 1 × xz = xz

A₂₃ = (-1)²⁺³ × M₂₃ = (-1) × 0 = 0

A₃₁ = (-1)³⁺¹ × M₃₁ = 1 × 0 = 0

A₃₂ = (-1)³⁺² × M₃₂ = (-1) × 0 = 0

A₃₃ = (-1)³⁺³ × M₃₃ = 1 × xy = xy

Adj A = =

A^-1 = adj A / |A|

A^-1 = /xyz

A^-1 =

A^-1 = =

The correct answer is A

A =

|A| = 1 (1 × 1 – sin θ × (-sin θ)) – sin θ ((-sin θ) × 1 – (-1) × sin θ) + 1 ((-sin θ) × (-sin θ) – (-1) × 1)

|A| = 1 + sin²θ + sin²θ – sin²θ + sin²θ + 1

|A| = 2 + 2sin²θ

|A| = 2(1 + sin²θ)

Now, 0 ≤ θ ≤ 2π

⇒ sin 0 ≤ sin θ ≤ sin 2π

⇒ 0 ≤ sin²θ ≤ 1

⇒ 1 + 0 ≤ 1 + sin²θ ≤ 1 + 1

⇒ 2 ≤ 2(1 + sin²θ) ≤ 4

∴ Det (A) є [2, 4]