Determinants

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

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

[∵ (a – b)(a + b) = (a² – b²)]

= x³ – x² + x + x² – x + 1 – x² + 1

= x³ – x² + 2

Let

Expanding |A| along C₁, we get

= (a + x) [(a + y)(a + z) – yz] – y [(x)(a + z) – xz] + z [xy – x (a + y)]

= (a + x) [a² + az + ya + yz – yz] – y [ax + xz – xz] + z [xy – xa + xy]

= (a + x) [a² + az + ya] – y [ax] + z [– xa]

= a(a² + az + ya) + x(a² + az + ya) – yax – zxa

= a³ + a²z + ya² + xa² + xaz + xya – yax – zxa

= a³ + a²z + ya² + xa²

= a² (a + z + y + x)

Let

Expanding |A| along C₁, we get

= 0 – xy² [(x²y)(0) – (x²z)(yz²)] + xz² [(x²y)(zy²) – 0]

= – xy² [– (x²yz³)] + xz² [(x²zy³)]

= x³y³z³ + x³z³y³

= 2x³y³z³

Let

By applying C₁→ C₁ + C₂ + C₃, we have

Taking (x + y + z) common from Column C₁, we get

By applying R₂→ R₂ – R₁, we get

By applying R₃→ R₃ – R₁, we get

Applying C₂→ C₂ – C₃, we get

Now, expanding along C₁, we get

= (x + y + z) [1×{(3y)(2z + x) – (-3z)(x – y)}]

= (x + y + z) [6yz + 3yx + (3z)(x – y)]

= (x + y + z) [6yz + 3yx + 3zx – 3zy]

= (x + y + z) [3yz + 3zx + 3yx]

= 3(x + y + z)(yz + zx + yx)

By applying C₁→ C₁ + C₂ + C₃, we get

Taking (3x + 4) common from first column, we get

By applying R₂→ R₂ – R₁, we get

By applying R₃→ R₃ – R₁, we get

Now, expanding along first column, we get

= (3x + 4) [1×{(16) – 0}]

= (3x + 4)(16)

= 16(3x + 4)

By applying R₁→ R₁ + R₂ + R₃, we get

Taking (a + b +c) common from the first row, we get

By applying C₂→ C₂ – C₁, we get

By applying C₃→ C₃ – C₁, we get

Now, expanding along first row, we get

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

= (a + b + c)[(a + b + c)²]

= (a + b + c)³

Taking LHS,

Firstly, multiply and divide R₁, R₂, R₃ by x, y, z respectively, we get

[rearrange the terms]

Taking xyz common from the first and second column, we get

Applying C₃→ C₃ + C₁, we get

Taking common (xy + yz + xz) common from C₃, we get

If any two columns (or rows) of a determinant are identical (all corresponding elements are same), then the value of determinant is zero.

Here, C₂ and C₃ are identical.

Hence,

∴ LHS = RHS

Hence Proved

Taking LHS,

By applying R₁→ R₁ + R₂ + R₃, we get

Taking 2 common from the first row, we get

Applying R₁→ R₁ – R₂, we get

Applying R₃→ R₃ - R₁, we get

Applying R₂→ R₂ – R₁, we get

Taking y, z, x common from R₁, R₂ and R₃ respectively, we get

Expanding along C₁, we get

= 2xyz [(1){(1) – 0} – (1){0 – 1} + 0}]

= 2xyz [1 + 1]

= 4xyz

= RHS

Hence,

∴ LHS = RHS

Hence Proved

Taking LHS,

Applying R₁→ R₁ – R₂, we get

[∵(a² – b²) = (a – b)(a + b)]

Taking (a – 1) common from the first row, we get

Applying R₂→ R₂ – R₃, we get

Taking (a – 1) common from the second row, we get

Now, expanding along C₃, we get

= (a – 1)² [1{(a + 1) – 2}]

= (a – 1)² [a + 1 – 2]

= (a – 1)³

= RHS

Hence, LHS = RHS

Hence Proved

Given: A + B + C = 0

To Prove:

Taking LHS,

Expanding along the first row, we get

= [1{1 – cos²A} – cos C {cos C – cos B cos A} + cos B {cos C cos A – cos B}]

= {1 – cos²A} – {cos²C – cos A cos B cos C} + {cos A cos B cos C – cos²B}

= {sin²A} – cos²C + cos A cos B cos C + cos A cos B cos C – cos²B

[∵ cos²x + sin²x = 1]

= sin²A – cos²C – cos²B + 2cos A cos B cos C

= -(cos²B – sin²A)– cos²C + 2cos A cos B cos C

= -[cos(B + A)cos(B – A)] + cos C [2 cos A cos B – cos C]

[∵ cos²B – sin²A = cos(B + A)cos(B – A)]

= -[cos(B + A)cos(B – A)] + cos C [cos(A + B) + cos(A – B) – cos C]…(i)

[∵ 2cos A cos B = cos(A + B) + cos(A – B)]

It is given that A + B + C = 0

⇒ A + B = - C

Putting the value of A + B in eq (i), we get

= -[cos(- C) cos(B – A)] + cos C [cos(-C) + cos (A – B) – cos C]

= -cos C cos(B – A) + cos C[cos C + cos(A – B) – cos C]

[∵ cos(-C) = cos C]

Now, cos(A – B) = cos A cos B + sin A sin B

= -cos C{cos B cos A + sin B sin A} + cos C [cos A cos B + sin A sin B]

= 0 = RHS

Hence Proved

The coordinates of the vertices of an equilateral triangle are (x₁, y₁), (x₂, y₂), (x₃, y₃).

So, the area of triangle with given vertices is given by

Given that the length of the sides of an equilateral triangle = a

Also, area of equilateral triangle =

Now, squaring both the sides, we get

Hence Proved

We have,

Expanding along R₁, we get

⇒ (1){-6 – {(-7) cos2θ}} – 1{8 – 7cos2θ} + sin3θ {28 – 21} = 0

⇒ – 6 + 7cos2θ – 8 + 7cos2θ + 7sin3θ = 0

⇒ 14cos2θ + 7sin3θ – 14 = 0

⇒ 2cos2θ + sin3θ – 2 = 0

Now, we know that

cos 2θ = 1 – 2sin²θ

sin 3θ = 3sinθ – 4sin³θ

⇒ 2(1 – 2sin²θ) + (3sinθ – 4sin³θ) – 2 = 0

⇒ 2 – 4sin²θ + 3sinθ – 4sin³θ – 2 = 0

⇒ -2 + 4sin²θ - 3sinθ + 4sin³θ + 2 = 0

⇒ sinθ (4sinθ – 3 + 4sin²θ) = 0

⇒ sinθ (4sin²θ – 6sinθ + 2sinθ – 3) = 0

⇒ sinθ [2sinθ(2sinθ – 3) + 1(2sinθ – 3)] = 0

⇒ sinθ (2sinθ + 1)(2sinθ – 3) = 0

⇒ sinθ = 0 or 2sinθ + 1 = 0 or 2sinθ – 3 = 0

⇒ θ = nπ or 2sinθ = -1 or 2sinθ = 3

⇒

⇒ θ = nπ ; m, n ∈ Z

We have,

By applying C₁→ C₁ + C₂ + C₃, we get

Taking (12 + x) common from the first column, we get

By applying C₂→ C₂ + C₃, we get

Applying R₂→ R₂ – R₃, we get

Applying R₃→ R₃ – R₁, we get

Expanding along first column, we get

⇒ (12 + x)[(1){0 – (2x)(-2x)}] = 0

⇒ (12 + x)(4x²) = 0

⇒ 12 + x = 0 or 4x² = 0

⇒ x = -12 or x = 0

Hence, the value of x = -12 and 0

Given: a₁, a₂…, a_r are in G.P

We know that, a_r+1 = AR^(r+1)-1 = AR^r …(i)

[∵a_n = ar^n-1, where a = first term and r = common ratio]

where A = First term of given G.P

and R = common ratio of G.P

…[from(i)]

Taking AR^r, AR^r+6 and AR^r+10 common from R₁, R₂ and R₃ respectively, we get

If any two columns (or rows) of a determinant are identical (all corresponding elements are same), then the value of determinant is zero.

Here, R₁ and R₂ are identical.

Hence Proved

Given points are (a + 5, a – 4), (a – 2, a + 3) and (a, a)

To Prove: these points don’t lie on straight line for any value of a

So, we have to show that these points form a triangle.

Area of triangle:-

Applying R₂→ R₂ – R₁, we get

Applying R₃→ R₃ – R₁, we get

Now, expanding along third column, we get

Hence, given points form a triangle i.e. points do not lie on a straight line.

Hence Proved

We have,

Applying C₂→ C₂ – C₁, we get

Taking common cos B – cos A from second column, we get

Applying C₃→ C₃ – C₁, we get

⇒

Taking common cos C – cos A from column third, we get

Now, expanding along first row, we get

⇒ (cos B – cos A)(cos C – cos A)[(1){cos C + cos A + 1 – (cos B + cos A + 1)}] = 0

⇒ (cos B – cos A)(cos C – cos A)[cos C + cos A + 1 – cos B – cos A – 1] = 0

⇒ (cos B – cos A)(cos C – cos A)(cos C – cos B) = 0

⇒ cos B – cos A = 0 or cos C – cos A = 0 or cos C – cos B = 0

⇒ cos B = cos A or cos C = cos A or cos C = cos B

⇒ B = A or C = A or C = B

Hence, ΔABC is an isosceles triangle.

We have,

We have to find A^-1 and

Firstly, we find |A|

Expanding |A| along C₁, we get

= 0 – 1 {0 – 1} + 1 {1 – 0}

= 1 + 1

= 2

Now, we have to find adj A and for that we have to find co-factors:

Now, we have to show that

= A^-1

Hence Proved

We have,

We have to find A^-1 and

Firstly, we find |A|

Expanding |A| along C₁, we get

= (-1 + 2) + 2 (0) + 0

= 1

Now, we have to find adj A and for that we have to find co-factors:

Now, the system of linear equation is

x – 2y = 10

2x – y – z = 8

-2y + z = 7

We know that, AX = B

Here,

and we can see that this matrix is the transpose of the given matrix. So, transpose of A^-1 is

⇒ X = A^-1B

∴ x = 0, y = -5 and z = -3

Given system of equations is:

3x + 2y – 2z = 3,

x + 2y + 3z = 6,

2x – y + z = 2

We know that,

AX = B

i.e.

∴ X = A^-1 B

So, firstly, we have to find the A^-1 and

Firstly, we find |A|

Expanding |A| along C₁, we get

= 3(2 + 3) – 1(2 – 2) + 2(6 + 4)

= 3(5) + 2(10)

= 15 + 20

= 35

Now, we have to find adj A and for that we have to find co-factors:

Now, X = A^-1B

∴ x = 1, y = 1 and z = 1

We have,

BA = 6I …(i)

Now, given system of equations is:

y + 2z = 7,

x – y = 3,

2x + 3y + 4z = 17

So,

Applying R₁→ R₂, we get

Applying R₂→ R₃, we get

Now, …(ii)

So, BA = 6I [from eq(i)]

and

Now, putting the value of B^-1, in eq. (ii), we get

⇒

∴ x = 4 , y = 2 and z = -1

Let

Applying C₁→ C₁ + C₂ + C₃, we get

Taking (a + b + c) common from the first column, we get

Now, Expanding along C₁, we get

= (a + b + c)[(1)(bc – a²) – (1)(b² – ac) + (1)(ba – c²)]

= (a + b + c)[bc – a² – b² + ac + ab – c²]

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

[∵ (a – b)² = a² + b² – 2ab]

Given that Δ = 0

⇒ (a + b + c)[(a – b)² + (b – c)² + (c – a)²] = 0

Either (a + b + c) = 0 or (a – b)² + (b – c)² + (c – a)² = 0

but it is given that (a + b + c) ≠ 0

∴(a – b)² + (b – c)² + (c – a)² = 0

⇒ a – b = b – c = c – a = 0

⇒ a = b = c

Hence Proved

We have,

Applying R₁→ R₁ – R₂, we get

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

Applying R₂→ R₂ – R₃, we get

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

Applying C₁→ C₁ + C₂ + C₃, we get

Now, expanding along C₁, we get

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

= (a + b + c)²[ab + bc + ca – (a² + b² + c²){(cb – ac – b² + ab – (a + c² – 2ac)}]

= (a + b + c)²[ab + bc + ca – (a² + b² + c²){(cb – ac – b² + ab – a - c² + 2ac)}]

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

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

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

Hence, given determinant is divisible by (a + b + c) and Quotient is (a + b + c)[ab + bc + ca – (a² + b² + c²)

Given: x + y + z = 0

To Prove:

Taking LHS,

Expanding along the first row, we get

= xa{(za)(ya) – (xc)(xb)} – (yb){(yc)(ya) – (zb)(xb)} + (zc){(yc)(xc) – (zb)(za)}

= xa{a²yz – x²bc} – yb{y²ac – b²xz} + zc{c²xy – z²ab}

= a³xyz – x³abc – y³abc + b³xyz + c³xyz – z³abc

= xyz(a³ + b³ + c³) – abc(x³ + y³ + z³)

It is given that x + y + z = 0

⇒ x³ + y³ + z³ = 3xyz

= xyz(a³ + b³ + c³) – abc (3xyz)

= xyz(a³ + b³ + c³ – 3abc)

Hence Proved

We have,

⇒ (2x)(x) – (5)(8) = (6)(3) – (7)(-2)

⇒ 2x² – 40 = 18 – (-14)

⇒ 2x² – 40 = 18 + 14

⇒ x² – 20 = 9 + 7

⇒ x² – 20 = 16

⇒ x² = 16 + 20

⇒ x² = 36

⇒ x = √36

⇒ x = ±6

Hence, the correct option is (c)

We have,

Applying C₂→ C₂ + C₃, we get

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

Applying C₁ → C₁ – C₃, we get

Expanding along first row, we get

= (a + b + c)[(-b){c – b} – (1){-c² – (-ab)} + a{-c – (-a)}]

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

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

= -abc + ab² + ac² – a²b – a²c + a³ – b²c + b³ + bc² – ab² – abc + a²b – bc² + b²c + c³ – abc – ac² + a²c

= a³ + b³ + c³ – 3abc

Hence, the correct option is (c)

We know that, the area of a triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) is given by

[given]

Now, expanding along second column, we get

⇒ -(k) {-3 – 3} = 18

⇒ -k (-6) = 18

⇒ 6k = 18

⇒ k = 3

Hence, the correct option is (b)

We have,

Taking (b – a) common from C₁ and C₃, we get

Applying C₁→ C₁ – C₃, we get

If any two columns (or rows) of a determinant are identical (all corresponding elements are same), then the value of determinant is zero.

Here, C₁ and C₂ are identical.

Hence, the correct option is (d).

We have,

Applying C₁→ C₁ + C₂ + C₃, we get

Taking (2cos X + sin X) common from the first column, we get

Applying R₂→ R₂ – R₁, we get

Applying R₃→ R₃ – R₁, we get

Expanding |A| along C₁, we get

⇒ (2cos X + sin X) [(1){(sin X – cos X)(sin X – cos X)}]

⇒ (2cos X + sin X)(sin X – cos X)² = 0

⇒ 2cos X = -sin X or (sin X – cos X)² = 0

⇒ tan X = -2 or tan X = 1

but tan X = -2 is not possible as for

So, tan X = 1

Hence, only one real distinct root exist.

Hence, the correct option is (c)

We have,

Expanding along C₁, we get

= [(-1){1 – cos²A} – cos C{-cos C – cos Acos B} + cos B{cos A cos C + cos B}]

= -1 + cos²A + cos²C + cos A cos B cos C + cos A cos B cos C + cos²B

= -1 + cos²A + cos²B + cos²C + 2cos A cos B cos C

Here, we use formula

1 + cos2A = 2cos²A

Taking L.C.M, we get

Now, we use the formula:

cos(A + B) cos(A – B) = 2cos Acos B

so, cos2A + cos2B = 2 cos(A + B) cos(A – B)

…(i)

Since, A, B and C are the angles of a triangle and we know that the sum of the angles of a triangle = 180° or π

⇒ A + B + C = π

⇒ A + B = π – C

Putting the value of (A+B) in eq. (i), we get

[∵ cos(π – x) = -cos X]

= -cos C{cos(A – B) – cos C} + 2cos Acos Bcos C

= -cos C[cos(A – B) – cos{π – (A + B)}] + 2cos Acos Bcos C

= -cos C[cos(A – B) + cos(A + B)] + + 2cos Acos Bcos C

= -cos C[2cos Acos B] + 2cos Acos Bcos C

= 0

We have,

Dividing R₂ and R₃ by ‘t’

Expanding along the first row, we get

= (1)(1 – 2) + (1)(2 – 1)

= - 1 + 1

= 0

Hence, the correct option is (A)

Given

Performing operations C₁→ C₂ – C₃ and C₂→ C₂ – C₃

= 0 – 0 + 1 (sin θ. cos θ)

Multiply and divide by 2,

= 1/2 (2sin θ cos θ)

We know that 2 sin θ cos θ = sin 2θ

= 1/2 (sin 2θ)

Since the maximum value of sin 2θ is 1, θ = 45°.

∴ Δ = 1/2 (sin 2(45°))

= 1/2 sin 90°

= 1/2 (1)

∴ Δ = 1/2

Given

Option (A):

= 0 – 0 + (a – b) [2a (a + c) – 0 (a + b)]

= (a – b) [2a² + 2ac – 0]

= (a – b) (2a² + 2ac) ≠ 0

Option (B):

= 0 – (b - a) [(b + a) (0) – (b – c) (2b)] + 0

= - (b – a) [0 - 2b² + 2bc]

= (a – b) (2b² – 2bc) ≠ 0

Option (C):

= 0 + a [a (0) – (-bc)] – b [ac – b (0)]

= a [bc] – b [ac]

= abc – abc = 0

Option (D):

= 0 – (1 - a) [(1 + a) (0) – (1 – c) (1 + b)] + (1 – b) [ (1 + a) (1 + c) – 0 (1 + b)]

= - (1 – a) [- (1 – c) (1 + b)] + (1 – b) [(1 + a) (1 + c)]

= (1 – a) (1 – c) (1 + b) + (1 – b) (1 + a) (1 + c) ≠ 0

Hence, option (C) satisfies.

Given

⇒ |A| = 2 (6 – 5) - λ (0 – 5) + (-3) (0 – 2)

= 2 + 5λ + 6

= 5λ + 8

We know that A^-1 exists if A is non – singular matrix i.e. |A| ≠ 0.

∴ 5λ + 8 ≠ 0

⇒ 5λ ≠ -8

So, A^-1 exists if and only if .

Given A and B are invertible matrices.

Consider (AB) B^-1 A^-1

⇒ (AB) B^-1 A^-1 = A(BB^-1) A^-1

= AIA^-1 = (AI) A^-1

= AA^-1 = I

⇒ (AB)^-1 = B^-1 A^-1 … option (C)

Also AA^-1 = I

⇒ |AA^-1| = |I|

⇒ |A| |A^-1| = 1

∴ det (A)^-1 = [det (A)]^-1 … (B)

We know that

⇒ adj A = |A|. A^-1 … option (A)

But

∴ (A + B)^-1 ≠ B^-1 + A^-1

Hence, option (D) is not correct.

Given

Applying C₁→ C �₁ – C₃ and C₂→ C₂ – C₃,

⇒

Expanding along R₁,

⇒ x [y (1 + z) + z] – 0 + 1 (yz) = 0

⇒ xy + xyz + xz + yz = 0

Dividing by xyz on both sides,

Hence, option (D) satisfies.

Given matrix

= x [x² – (x + y) (x + 2y)] – (x + y) [(x + 2y) (x) – (x + y)²] + (x + 2y) [(x + 2y)² – x (x + y)]

= x [x² – x² – 3xy – 2y²] – (x + y) [x² + 2xy – x² – 2xy – y²] + (x + 2y) [x² + 4xy + 4y² – x² – xy]

= x [-3xy – 2y²] – (x + y) [-y²] + (x + 2y) [3xy + 4y²]

= -3x²y – 2xy² + xy² + y³ +3x²y + 4xy² + 6xy² + 8y³

= 9y³ + 9xy²

= 9y² (x + y) … option (B)

Hence, option B satisfies.

Given

= 1 [2a² – (-4)] + 2 [4a – 0] + 5 [8 – 0] = 86

= 1 [2a² + 4] + 2 [4a] + 5 [8] = 86

= 2a² + 4 + 8a + 40 = 86

= 2a² + 8a + 44 = 86

= 2a² + 8a = 42

= 2 (a² + 4a) = 42

= (a² + 4a) = 21

⇒ a² + 4a – 21 = 0

⇒ (a + 7) (a – 3) = 0

∴ a = -7 or 3

Sum of these numbers is -7 + 3 = -4. [Option (C)]

If A is a matrix of order 3 × 3, then |3A| = 27 |A|.

Explanation:

We know that if A = [a_ij]_3×3, then |k.A| = k³|A|

∴ |3A| = 3³ |A| = 27 |A|

If A is invertible matrix of order 3 × 3, then |A^-1| = |A|^-1

Explanation:

Given A is invertible matrix of order 3 × 3.

We know that AA^-1 = I.

⇒|A||A^-1| = 1

|A|^-1

If x, y, z ∈ R, then the value of determinant is equal to 0.

Explanation:

Given

Performing the operation C₁→ C₁ – C₂,

We know that (a + b)² – (a – b)² = 4ab.

C₁ and C₃ are proportional to each other.

If cos 2θ = 0, then

Explanation:

Given cos 2θ = 0

⇒ cos 2θ = cos π/2

⇒ 2θ = π/2

∴ θ = π/4

Then

If A is a matrix of order 3 × 3, then (A²)^-1 = (A^-1)².

Explanation:

We know that if A is a matrix of order 3 × 3, then (A²)^-1 = (A^-1)².

If A is a matrix of order 3 × 3, then number of minors in determinant of A are 9.

Explanation:

In a 3 × 3 matrix, there are 9 elements.

So, there are 9 minors of these elements.

The sum of the products of elements of any row with the co-factors of corresponding elements is equal to Δ (Determinant).

Explanation:

If ,

then |A| = a₁₁C₁₁ + a₁₂C₁₂ + a₁₃C₁₃

= Sum of products of elements of R₁ with their corresponding cofactors

= Δ (Determinant)

If x = -9 is a root of , then other two roots are 2 and 7.

Explanation:

Given x = -9 is a root of .

⇒ x [x² – 12] – 3 [2x – 14] + 7 [12 – 7x] = 0

⇒ x³ – 12x – 6x + 42 + 84 – 49x = 0

⇒ x³ – 67x + 126 =0

Here, 126 × 1 = 9 × 2 × 7

For x = 2,

⇒ 2³ – 67 (2) + 126 = 134 – 134 = 0

∴ x = 2 is one root.

For x = 7,

⇒ 7³ – 67 (7) + 126 = 469 – 469 = 0

∴ x = 7 is also one root.

Explanation:

Given

Performing the operation C₁→ C₁ – C₃

Taking (z – x) common from C₁,

= (z – x) [1 [0 – (y – z) (z – y)] - (xyz) [0 - (y - z)] + (x – z) [(z – y) – 0]]

= (z – x) (z – y) (-y + z – xyz + x – z)

= (z – x) (z – y) (x – y – xyz)

= (z – x) (y – z) (y – x + xyz)

If , then A = 0.

Explanation:

Given

Here R₁ and R₃ are identical.

∴ A = 0

True

We know that (Aⁿ)^-1 = (A^-1)ⁿ, where n∈N.

∴ (A³)^-1 = (A^-1)³

False

We know that if A is a non-singular square matrix, then for any scalar a, aA is invertible such that

i.e. , where a is any non-zero scalar.

In the above statement a is any real number So, we can conclude that above statement is false.

False

Given A is non-singular matrix.

We know that AA^-1 = I.

⇒|A||A^-1| = 1

∴ |A^-1| = 1/|A| = |A|^-1

True

We know that |AB| = |A|. |B| and if A = [a_ij]_3×3, then |k.A| = k³|A|.

∴ |3A| = 27 |AB|

= 27 |A| |B|

= 27 (5) (3)

= 405

Hence proved.

The above statement is true.

True

Let A be the determinant.

∴ |A| = 12

We know that if A is a square matrix of order n, then |adj A| = |A|^n-1

For n = 3, |adj A| = |A|^3-1

= |A|²

= 12² = 144

Hence the above statement is true.

True

Since a, b, c are in AP, 2b = a + c.

Performing the operation R₁→ R₁ + R₃

Since 2b = a + c,

Here R₁ and R₂ are proportional to each other,

⇒ 0 = 0

Hence the above statement is true.

False

We know that if A is a square matrix of order n, then |adj A| = |A|^n-1

Here n =2,

⇒ |adj A| = |A|^n-1 = |A|

Hence the statement is false.

True

Here in first determinant C₁ and C₃ are identical.

Taking cos A and cos B common from C₂ and C₃.

Here C₂ and C₃ are identical.

= 0

Hence the statement is true.

True

Given

Splitting first row,

Splitting second row,

Similarly, we can split these 4 determinants in 8 determinants by splitting each one in two determinants further.

So given statement is true.

True

Given

We have to prove that

Performing the operation,

Taking 2 common from C₁ and performing the operation C₁ → C₁ – C₂ and C₂→ C₂ – C₃

Here in the second determinant C₁ and C₂ ae identical.

Here in the second determinant C₁ and C₃ are identical.

= 2 (16) = 32

Hence the above statement is true.

True

Given

Performing operations R₂→ R₂ – R₁ and R₃→ R₃ – R₁

= cos θ. sin θ

Multiply and divide by 2,

= 1/2 (2sin θ cos θ)

We know that 2 sin θ cos θ = sin 2θ

= 1/2 (sin 2θ)

Since the maximum value of sin 2θ is 1, θ = 45°.

∴ Δ = 1/2 (sin 2(45°))

= 1/2 sin 90°

= 1/2 (1)

∴ Δ = 1/2

Hence the above statement is true.