NCERT Solutions for Class 12 Science Math Chapter 4 - Determinants

Answer:

= 2(−1) − 4(−5) = − 2 + 20 = 18

Answer:

(i) = (cos θ)(cos θ) − (−sin θ)(sin θ) = cos² θ+ sin² θ = 1

(ii)

= (x² − x + 1)(x + 1) − (x − 1)(x + 1)

= x³ − x² + x + x² − x + 1 − (x² − 1)

= x³ + 1 − x² + 1

= x³ − x² + 2

Answer:

The given matrix is.

Answer:

The given matrix is.

It can be observed that in the first column, two entries are zero. Thus, we expand along the first column (C₁) for easier calculation.

From equations (i) and (ii), we have:

Hence, the given result is proved.

Answer:

(i) Let.

It can be observed that in the second row, two entries are zero. Thus, we expand along the second row for easier calculation.

(ii) Let.

By expanding along the first row, we have:

(iii) Let

By expanding along the first row, we have:

(iv) Let

By expanding along the first column, we have:

Answer:

Let

By expanding along the first row, we have:

Answer:

(i)

(ii)

Answer:

Answer: B

Hence, the correct answer is B.

Answer:

Answer:

Here, the two rows R₁ and R₃ are identical.

Δ = 0.

Answer:

Answer:

By applying C₃ → C₃ + C_2, we have:

Here, two columns C₁ and C₃ are proportional.

Δ = 0.

Answer:

Applying R₂ → R₂ − R₃, we have:

Applying R₁ ↔R₃ and R₂ ↔R₃, we have:

Applying R₁ → R₁ − R₃, we have:

Applying R₁ ↔R₂ and R₂ ↔R₃, we have:

From (1), (2), and (3), we have:

Hence, the given result is proved.

Answer:

We have,

Here, the two rows R₁ and R₃ are identical.

∴Δ = 0.

Answer:

Applying R₂ → R₂ + R₁ and R₃ → R₃ + R₁, we have:

Answer:

(i)

Applying R₁ → R₁ − R₃ and R₂ → R₂ − R₃, we have:

Applying R₁ → R₁ + R₂, we have:

Expanding along C₁, we have:

Hence, the given result is proved.

(ii) Let.

Applying C₁ → C₁ − C₃ and C₂ → C₂ − C₃, we have:

Applying C₁ → C₁ + C₂, we have:

Expanding along C₁, we have:

Hence, the given result is proved.

Answer:

Applying R₂ → R₂ − R₁ and R₃ → R₃ − R₁, we have:

Applying R₃ → R₃ + R₂, we have:

Expanding along R₃, we have:

Hence, the given result is proved.

Answer:

(i)

Applying R₁ → R₁ + R₂ + R₃, we have:

Applying C₂ → C₂ − C₁, C₃ → C₃ − C₁, we have:

Expanding along C₃, we have:

Hence, the given result is proved.

(ii)

Applying R₁ → R₁ + R₂ + R₃, we have:

Applying C₂ → C₂ − C₁ and C₃ → C₃ − C₁, we have:

Expanding along C₃, we have:

Hence, the given result is proved.

Answer:

(i)

Applying R₁ → R₁ + R₂ + R₃, we have:

Applying C₂ → C₂ − C₁, C₃ → C₃ − C₁, we have:

Expanding along C₃, we have:

Hence, the given result is proved.

(ii)

Applying C₁ → C₁ + C₂ + C₃, we have:

Applying R₂ → R₂ − R₁ and R₃ → R₃ − R₁, we have:

Expanding along R₃, we have:

Hence, the given result is proved.

Answer:

Applying R₁ → R₁ + R₂ + R₃, we have:

Applying C₂ → C₂ − C₁ and C₃ → C₃ − C₁, we have:

Expanding along R₁, we have:

Hence, the given result is proved.

Answer:

Applying R₁ → R₁ + bR₃ and R₂ → R₂ − aR₃, we have:

Expanding along R₁, we have:

Answer:

Taking out common factors a, b, and c from R₁, R₂, and R₃ respectively, we have:

Applying R₂ → R₂ − R₁ and R₃ → R₃ − R₁, we have:

Applying C₁ → aC₁, C₂ → bC_2, and C₃ → cC₃, we have:

Expanding along R₃, we have:

Hence, the given result is proved.

Answer:

Answer: C

A is a square matrix of order 3 × 3.

Hence, the correct answer is C.

Answer:

Answer: C

We know that to every square matrix, of order n. We can associate a number called the determinant of square matrix A, where element of A.

Thus, the determinant is a number associated to a square matrix.

Hence, the correct answer is C.

Answer:

(i) The area of the triangle with vertices (1, 0), (6, 0), (4, 3) is given by the relation,

(ii) The area of the triangle with vertices (2, 7), (1, 1), (10, 8) is given by the relation,

(iii) The area of the triangle with vertices (−2, −3), (3, 2), (−1, −8)

is given by the relation,

Hence, the area of the triangle is.

Answer:

Area of ΔABC is given by the relation,

Thus, the area of the triangle formed by points A, B, and C is zero.

Hence, the points A, B, and C are collinear.

Answer:

We know that the area of a triangle whose vertices are (x₁, y₁), (x₂, y₂), and

(x₃, y₃) is the absolute value of the determinant (Δ), where

It is given that the area of triangle is 4 square units.

∴Δ = ± 4.

(i) The area of the triangle with vertices (k, 0), (4, 0), (0, 2) is given by the relation,

Δ =

∴−k + 4 = ± 4

When −k + 4 = − 4, k = 8.

When −k + 4 = 4, k = 0.

Hence, k = 0, 8.

(ii) The area of the triangle with vertices (−2, 0), (0, 4), (0, k) is given by the relation,

Δ =

∴k − 4 = ± 4

When k − 4 = − 4, k = 0.

When k − 4 = 4, k = 8.

Hence, k = 0, 8.

Answer:

(i) Let P (x, y) be any point on the line joining points A (1, 2) and B (3, 6). Then, the points A, B, and P are collinear. Therefore, the area of triangle ABP will be zero.

Hence, the equation of the line joining the given points is y = 2x.

(ii) Let P (x, y) be any point on the line joining points A (3, 1) and

B (9, 3). Then, the points A, B, and P are collinear. Therefore, the area of triangle ABP will be zero.

Hence, the equation of the line joining the given points is x − 3y = 0.

Answer:

Answer: D

The area of the triangle with vertices (2, −6), (5, 4), and (k, 4) is given by the relation,

It is given that the area of the triangle is ±35.

Therefore, we have:

When 5 − k = −7, k = 5 + 7 = 12.

When 5 − k = 7, k = 5 − 7 = −2.

Hence, k = 12, −2.

The correct answer is D.

Answer:

(i) The given determinant is.

Minor of element a_ij is M_ij.

∴M₁₁ = minor of element a₁₁ = 3

M₁₂ = minor of element a₁₂ = 0

M₂₁ = minor of element a₂₁ = −4

M₂₂ = minor of element a₂₂ = 2

Cofactor of a_ij is A_ij = (−1)^{i + j} M_ij.

∴A₁₁ = (−1)¹⁺¹ M₁₁ = (−1)² (3) = 3

A₁₂ = (−1)¹⁺² M₁₂ = (−1)³ (0) = 0

A₂₁ = (−1)²⁺¹ M₂₁ = (−1)³ (−4) = 4

A₂₂ = (−1)²⁺² M₂₂ = (−1)⁴ (2) = 2

(ii) The given determinant is.

Minor of element a_ij is M_ij.

∴M₁₁ = minor of element a₁₁ = d

M₁₂ = minor of element a₁₂ = b

M₂₁ = minor of element a₂₁ = c

M₂₂ = minor of element a₂₂ = a

Cofactor of a_ij is A_ij = (−1)^{i + j} M_ij.

∴A₁₁ = (−1)¹⁺¹ M₁₁ = (−1)² (d) = d

A₁₂ = (−1)¹⁺² M₁₂ = (−1)³ (b) = −b

A₂₁ = (−1)²⁺¹ M₂₁ = (−1)³ (c) = −c

A₂₂ = (−1)²⁺² M₂₂ = (−1)⁴ (a) = a

Answer:

(i) The given determinant is.

By the definition of minors and cofactors, we have:

M₁₁ = minor of a₁₁=

M₁₂ = minor of a₁₂=

M₁₃ = minor of a₁₃ =

M₂₁ = minor of a₂₁ =

M₂₂ = minor of a₂₂ =

M₂₃ = minor of a₂₃ =

M₃₁ = minor of a₃₁=

M₃₂ = minor of a₃₂ =

M₃₃ = minor of a₃₃ =

A₁₁ = cofactor of a₁₁= (−1)¹⁺¹ M₁₁ = 1

A₁₂ = cofactor of a₁₂ = (−1)¹⁺² M₁₂ = 0

A₁₃ = cofactor of a₁₃ = (−1)¹⁺³ M₁₃ = 0

A₂₁ = cofactor of a₂₁ = (−1)²⁺¹ M₂₁ = 0

A₂₂ = cofactor of a₂₂ = (−1)²⁺² M₂₂ = 1

A₂₃ = cofactor of a₂₃ = (−1)²⁺³ M₂₃ = 0

A₃₁ = cofactor of a₃₁ = (−1)³⁺¹ M₃₁ = 0

A₃₂ = cofactor of a₃₂ = (−1)³⁺² M₃₂ = 0

A₃₃ = cofactor of a₃₃ = (−1)³⁺³ M₃₃ = 1

(ii) The given determinant is.

By definition of minors and cofactors, we have:

M₁₁ = minor of a₁₁=

M₁₂ = minor of a₁₂=

M₁₃ = minor of a₁₃ =

M₂₁ = minor of a₂₁ =

M₂₂ = minor of a₂₂ =

M₂₃ = minor of a₂₃ =

M₃₁ = minor of a₃₁=

M₃₂ = minor of a₃₂ =

M₃₃ = minor of a₃₃ =

A₁₁ = cofactor of a₁₁= (−1)¹⁺¹ M₁₁ = 11

A₁₂ = cofactor of a₁₂ = (−1)¹⁺² M₁₂ = −6

A₁₃ = cofactor of a₁₃ = (−1)¹⁺³ M₁₃ = 3

A₂₁ = cofactor of a₂₁ = (−1)²⁺¹ M₂₁ = 4

A₂₂ = cofactor of a₂₂ = (−1)²⁺² M₂₂ = 2

A₂₃ = cofactor of a₂₃ = (−1)²⁺³ M₂₃ = −1

A₃₁ = cofactor of a₃₁ = (−1)³⁺¹ M₃₁ = −20

A₃₂ = cofactor of a₃₂ = (−1)³⁺² M₃₂ = 13

A₃₃ = cofactor of a₃₃ = (−1)³⁺³ M₃₃ = 5

Answer:

The given determinant is.

We have:

M₂₁ =

∴A₂₁ = cofactor of a₂₁ = (−1)²⁺¹ M₂₁ = 7

M₂₂ =

∴A₂₂ = cofactor of a₂₂ = (−1)²⁺² M₂₂ = 7

M₂₃ =

∴A₂₃ = cofactor of a₂₃ = (−1)²⁺³ M₂₃ = −7

We know that Δ is equal to the sum of the product of the elements of the second row with their corresponding cofactors.

∴Δ = a₂₁A₂₁ + a₂₂A₂₂ + a₂₃A₂₃ = 2(7) + 0(7) + 1(−7) = 14 − 7 = 7

Answer:

The given determinant is.

We have:

M₁₃ =

M₂₃ =

M₃₃ =

∴A₁₃ = cofactor of a₁₃ = (−1)¹⁺³ M₁₃ = (z − y)

A₂₃ = cofactor of a₂₃ = (−1)²⁺³ M₂₃ = − (z − x) = (x − z)

A₃₃ = cofactor of a₃₃ = (−1)³⁺³ M₃₃ = (y − x)

We know that Δ is equal to the sum of the product of the elements of the second row with their corresponding cofactors.

Hence,

Answer:

Answer: D

We know that:

Δ = Sum of the product of the elements of a column (or a row) with their corresponding cofactors

∴Δ = a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁

Hence, the value of Δ is given by the expression given in alternative D.

The correct answer is D.

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

From (1) and (2), we have:

(AB)⁻¹ = B⁻¹A⁻¹

Hence, the given result is proved.

Answer:

Answer:

We have:

Comparing the corresponding elements of the two matrices, we have:

Hence, −4 and 1 are the required values of a and b respectively.

Answer:

From equation (1), we have:

Answer:

From equation (1), we have:

Answer:

Answer: B

We know that,

Hence, the correct answer is B.

Answer:

Since A is an invertible matrix,

Hence, the correct answer is B.

Answer:

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

∴ A is non-singular.

Therefore, A⁻¹ exists.

Hence, the given system of equations is consistent.

Answer:

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

∴ A is non-singular.

Therefore, A⁻¹ exists.

Hence, the given system of equations is consistent.

Answer:

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

∴ A is a singular matrix.

Thus, the solution of the given system of equations does not exist. Hence, the system of equations is inconsistent.

Answer:

The given system of equations is:

x + y + z = 1

2x + 3y + 2z = 2

ax + ay + 2az = 4

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

∴ A is non-singular.

Therefore, A⁻¹ exists.

Hence, the given system of equations is consistent.

Answer:

The given system of equations is:

3x − y − 2z = 2

2y − z = −1

3x − 5y = 3

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

∴ A is a singular matrix.

Thus, the solution of the given system of equations does not exist. Hence, the system of equations is inconsistent.

Answer:

The given system of equations is:

5x − y + 4z = 5

2x + 3y + 5z = 2

5x − 2y + 6z = −1

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

∴ A is non-singular.

Therefore, A⁻¹ exists.

Hence, the given system of equations is consistent.

Answer:

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

Thus, A is non-singular. Therefore, its inverse exists.

Answer:

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

Thus, A is non-singular. Therefore, its inverse exists.

Answer:

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

Thus, A is non-singular. Therefore, its inverse exists.

Answer:

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

Thus, A is non-singular. Therefore, its inverse exists.

Answer:

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

Thus, A is non-singular. Therefore, its inverse exists.

Answer:

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

Thus, A is non-singular. Therefore, its inverse exists.

Answer:

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

Thus, A is non-singular. Therefore, its inverse exists.

Answer:

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

Thus, A is non-singular. Therefore, its inverse exists.

Answer:

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

Answer:

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

Then, the given situation can be represented by a system of equations as:

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

Now,

X = A⁻¹ B

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.

Answer:

Hence, Δ is independent of θ.

Answer:

Hence, the given result is proved.

Answer:

Expanding along C₃, we have:

Answer:

Expanding along R₁, we have:

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

Answer:

Answer:

Expanding along R₃, we have:

Hence, the given result is proved.

Answer:

We know that.

Answer:

(i)

We have,

(ii)

Answer:

Expanding along R₁, we have:

Answer:

Expanding along C₁, we have:

Answer:

Expanding along R₃, we have:

Hence, the given result is proved.

Answer:

Expanding along R₃, we have:

Hence, the given result is proved.

Answer:

Expanding along C₁, we have:

Hence, the given result is proved.

Answer:

Expanding along C₁, we have:

Hence, the given result is proved.

Answer:

Hence, the given result is proved.

Answer:

Let

Then the given system of equations is as follows:

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

A

Thus, A is non-singular. Therefore, its inverse exists.

Now,

A₁₁ = 75, A₁₂ = 110, A₁₃ = 72

A₂₁ = 150, A₂₂ = −100, A₂₃ = 0

A₃₁ = 75, A₃₂ = 30, A₃₃ = − 24

Answer:

Answer: A

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

Hence, we have Δ = 0.

The correct answer is A.

Answer:

Answer: A

The correct answer is A.

Answer:

Answer: D


Now, $0\le \theta \le 2\mathrm{\pi }$
$\Rightarrow -1\le \sin \theta \le 1$

The correct answer is D.