**NCERT Solutions for Class 12 Math Chapter 4 – Determinants**

**Chapter 4 – Determinants**

#### Question 1:

Evaluate the determinants in Exercises 1 and 2.

#### Answer:

= 2(−1) − 4(−5) = − 2 + 20 = 18

#### Question 2:

Evaluate the determinants in Exercises 1 and 2.

(i) (ii)

#### Answer:

(i) = (cos *θ*)(cos *θ*) − (−sin *θ*)(sin *θ*) = cos^{2} *θ*+ sin^{2} *θ* = 1

(ii)

= (*x*^{2} − *x* + 1)(*x* + 1) − (*x* − 1)(*x* + 1)

= *x*^{3} − *x*^{2} + *x* + *x*^{2} − *x* + 1 − (*x*^{2} − 1)

= *x*^{3} + 1 − *x*^{2} + 1

= *x*^{3} − *x*^{2} + 2

#### Question 3:

If, then show that

#### Answer:

The given matrix is.

#### Question 4:

If, then show that

#### 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 (C_{1}) for easier calculation.

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

Hence, the given result is proved.

#### Question 5:

Evaluate the determinants

(i) (iii)

(ii) (iv)

#### 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:

#### Page No 109:

#### Question 6:

If, find.

#### Answer:

Let

By expanding along the first row, we have:

#### Question 7:

Find values of *x*, if

(i)

2451=2x46x(ii)

2345=x32x5

#### Answer:

(i)

(ii)

#### Question 8:

If, then *x* is equal to

(A) 6 (B) ±6 (C) −6 (D) 0

#### Answer:

**Answer: B**

Hence, the correct answer is B.

#### Page No 119:

#### Question 1:

Using the property of determinants and without expanding, prove that:

#### Answer:

#### Question 2:

Using the property of determinants and without expanding, prove that:

#### Answer:

Here, the two rows R_{1} and R_{3} are identical.

Δ = 0.

#### Question 3:

Using the property of determinants and without expanding, prove that:

#### Answer:

#### Question 4:

Using the property of determinants and without expanding, prove that:

#### Answer:

By applying C_{3 }→ C_{3} + C_{2, }we have:

Here, two columns C_{1} and C_{3 }are proportional.

Δ = 0.

#### Question 5:

Using the property of determinants and without expanding, prove that:

#### Answer:

Applying R_{2} → R_{2} − R_{3}, we have:

Applying R_{1} ↔R_{3} and R_{2} ↔R_{3}, we have:

Applying R_{1 }→ R_{1} − R_{3}, we have:

Applying R_{1} ↔R_{2} and R_{2} ↔R_{3}, we have:

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

Hence, the given result is proved.

#### Page No 120:

#### Question 6:

By using properties of determinants, show that:

#### Answer:

We have,

Here, the two rows R_{1} and R_{3 }are identical.

∴Δ = 0.

#### Question 7:

By using properties of determinants, show that:

#### Answer:

Applying R_{2 }→ R_{2} + R_{1} and R_{3 }→ R_{3} + R_{1}, we have:

#### Question 8:

By using properties of determinants, show that:

(i)

(ii)

#### Answer:

(i)

Applying R_{1} → R_{1} − R_{3 }and R_{2} → R_{2} − R_{3}, we have:

Applying R_{1} → R_{1} + R_{2}, we have:

Expanding along C_{1}, we have:

Hence, the given result is proved.

(ii) Let.

Applying C_{1} → C_{1} − C_{3 }and C_{2} → C_{2} − C_{3}, we have:

Applying C_{1} → C_{1} + C_{2}, we have:

Expanding along C_{1}, we have:

Hence, the given result is proved.

#### Question 9:

By using properties of determinants, show that:

#### Answer:

Applying R_{2} → R_{2} − R_{1 }and R_{3} → R_{3} − R_{1}, we have:

Applying R_{3} → R_{3} + R_{2}, we have:

Expanding along R_{3}, we have:

Hence, the given result is proved.

#### Question 10:

By using properties of determinants, show that:

(i)

(ii)

#### Answer:

(i)

Applying R_{1} → R_{1} + R_{2 }+ R_{3}, we have:

Applying C_{2} → C_{2} − C_{1}, C_{3} → C_{3} − C_{1}, we have:

Expanding along C_{3}, we have:

Hence, the given result is proved.

(ii)

Applying R_{1} → R_{1} + R_{2 }+ R_{3}, we have:

Applying C_{2} → C_{2} − C_{1 }and C_{3} → C_{3} − C_{1}, we have:

Expanding along C_{3}, we have:

Hence, the given result is proved.

#### Question 11:

By using properties of determinants, show that:

(i)

(ii)

#### Answer:

(i)

Applying R_{1} → R_{1} + R_{2 }+ R_{3}, we have:

Applying C_{2} → C_{2} − C_{1}, C_{3} → C_{3} − C_{1}, we have:

Expanding along C_{3}, we have:

Hence, the given result is proved.

(ii)

Applying C_{1} → C_{1} + C_{2 }+ C_{3}, we have:

Applying R_{2} → R_{2} − R_{1 }and R_{3} → R_{3} − R_{1}, we have:

Expanding along R_{3}, we have:

Hence, the given result is proved.

#### Page No 121:

#### Question 12:

By using properties of determinants, show that:

#### Answer:

Applying R_{1} → R_{1} + R_{2 }+ R_{3}, we have:

Applying C_{2} → C_{2} − C_{1 }and C_{3} → C_{3} − C_{1}, we have:

Expanding along R_{1}, we have:

Hence, the given result is proved.

#### Question 13:

By using properties of determinants, show that:

#### Answer:

Applying R_{1} → R_{1} + *b*R_{3 }and R_{2} → R_{2} − *a*R_{3}, we have:

Expanding along R_{1}, we have:

#### Question 14:

By using properties of determinants, show that:

#### Answer:

Taking out common factors *a*, *b*, and *c* from R_{1}, R_{2}, and R_{3 }respectively, we have:

Applying R_{2} → R_{2} − R_{1 }and R_{3} → R_{3} − R_{1}, we have:

Applying C_{1} → *a*C_{1}, C_{2 }→ *b*C_{2, }and C_{3} → *c*C_{3}, we have:

Expanding along R_{3}, we have:

Hence, the given result is proved.

#### Question 15:

Choose the correct answer.

Let *A* be a square matrix of order 3 × 3, then is equal to

**A. ** **B. ** **C. ** **D.**

#### Answer:

**Answer: C**

*A* is a square matrix of order 3 × 3.

Hence, the correct answer is C.

#### Question 16:

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.

**D. **None of these

#### 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.

#### Page No 122:

#### Question 1:

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)

#### 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.

#### Page No 123:

#### Question 2:

Show that points

are collinear

#### 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.

#### Question 3:

Find values of *k* if area of triangle is 4 square units and vertices are

(i) (*k*, 0), (4, 0), (0, 2) (ii) (−2, 0), (0, 4), (0, *k*)

#### Answer:

We know that the area of a triangle whose vertices are (*x*_{1}, *y*_{1}), (*x*_{2}, *y*_{2}), and

(*x*_{3}, *y*_{3}) 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.

#### Question 4:

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

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

#### 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* = 2*x*.

(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* − 3*y* = 0.

#### Question 5:

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

**A.** 12 **B.** −2 **C.** −12, −2 **D.** 12, −2

#### 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.

#### Page No 126:

#### Question 1:

Write Minors and Cofactors of the elements of following determinants:

(i) (ii)

#### Answer:

(i) The given determinant is.

Minor of element *a*_{ij} is M_{ij.}

∴M_{11} = minor of element *a*_{11 }= 3

M_{12} = minor of element *a*_{12 }= 0

M_{21} = minor of element *a*_{21 }= −4

M_{22} = minor of element *a*_{22 }= 2

Cofactor of *a*_{ij} is A_{ij} = (−1)^{i + j} M_{ij}.

∴A_{11} = (−1)^{1+1} M_{11} = (−1)^{2} (3) = 3

A_{12} = (−1)^{1+2} M_{12} = (−1)^{3} (0) = 0

A_{21} = (−1)^{2+1} M_{21} = (−1)^{3} (−4) = 4

A_{22} = (−1)^{2+2} M_{22} = (−1)^{4} (2) = 2

(ii) The given determinant is.

Minor of element *a*_{ij} is M_{ij}.

∴M_{11} = minor of element *a*_{11 }= *d*

M_{12} = minor of element *a*_{12 }= *b*

M_{21} = minor of element *a*_{21 }= *c*

M_{22} = minor of element *a*_{22 }= *a*

Cofactor of *a*_{ij} is A_{ij} = (−1)^{i + j} M_{ij.}

∴A_{11} = (−1)^{1+1} M_{11} = (−1)^{2} (*d*) = *d*

A_{12} = (−1)^{1+2} M_{12} = (−1)^{3} (*b*) = −*b*

A_{21} = (−1)^{2+1} M_{21} = (−1)^{3} (*c*) = −*c*

A_{22} = (−1)^{2+2} M_{22} = (−1)^{4} (*a*) = *a*

#### Question 2:

(i) (ii)

#### Answer:

(i) The given determinant is.

By the definition of minors and cofactors, we have:

M_{11 }= minor of *a*_{11}=

M_{12 }= minor of *a*_{12}=

M_{13 }= minor of *a*_{13 }=

M_{21 }= minor of *a*_{21 }=

M_{22 }= minor of *a*_{22 }=

M_{23 }= minor of *a*_{23 }=

M_{31 }= minor of *a*_{31}=

M_{32 }= minor of *a*_{32 }=

M_{33 }= minor of *a*_{33 }=

A_{11 }= cofactor of *a*_{11}= (−1)^{1+1} M_{11} = 1

A_{12 }= cofactor of *a*_{12 }= (−1)^{1+2} M_{12} = 0

A_{13 }= cofactor of *a*_{13 }= (−1)^{1+3} M_{13} = 0

A_{21 }= cofactor of *a*_{21 }= (−1)^{2+1} M_{21} = 0

A_{22 }= cofactor of *a*_{22 }= (−1)^{2+2} M_{22} = 1

A_{23 }= cofactor of *a*_{23 }= (−1)^{2+3} M_{23} = 0

A_{31 }= cofactor of *a*_{31 }= (−1)^{3+1} M_{31} = 0

A_{32 }= cofactor of *a*_{32 }= (−1)^{3+2} M_{32} = 0

A_{33 }= cofactor of *a*_{33 }= (−1)^{3+3} M_{33} = 1

(ii) The given determinant is.

By definition of minors and cofactors, we have:

M_{11 }= minor of *a*_{11}=

M_{12 }= minor of *a*_{12}=

M_{13 }= minor of *a*_{13 }=

M_{21 }= minor of *a*_{21 }=

M_{22 }= minor of *a*_{22 }=

M_{23 }= minor of *a*_{23 }=

M_{31 }= minor of *a*_{31}=

M_{32 }= minor of *a*_{32 }=

M_{33 }= minor of *a*_{33 }=

A_{11 }= cofactor of *a*_{11}= (−1)^{1+1} M_{11} = 11

A_{12 }= cofactor of *a*_{12 }= (−1)^{1+2} M_{12} = −6

A_{13 }= cofactor of *a*_{13 }= (−1)^{1+3} M_{13} = 3

A_{21 }= cofactor of *a*_{21 }= (−1)^{2+1} M_{21} = 4

A_{22 }= cofactor of *a*_{22 }= (−1)^{2+2} M_{22} = 2

A_{23 }= cofactor of *a*_{23 }= (−1)^{2+3} M_{23} = −1

A_{31 }= cofactor of *a*_{31 }= (−1)^{3+1} M_{31} = −20

A_{32 }= cofactor of *a*_{32 }= (−1)^{3+2} M_{32} = 13

A_{33 }= cofactor of *a*_{33 }= (−1)^{3+3} M_{33} = 5

#### Question 3:

Using Cofactors of elements of second row, evaluate.

#### Answer:

The given determinant is.

We have:

M_{21 }=

∴A_{21 }= cofactor of *a*_{21 }= (−1)^{2+1} M_{21} = 7

M_{22 }=

∴A_{22 }= cofactor of *a*_{22 }= (−1)^{2+2} M_{22} = 7

M_{23 }=

∴A_{23 }= cofactor of *a*_{23 }= (−1)^{2+3} M_{23} = −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 = *a*_{21}A_{21} + *a*_{22}A_{22} + *a*_{23}A_{23} = 2(7) + 0(7) + 1(−7) = 14 − 7 = 7

#### Question 4:

Using Cofactors of elements of third column, evaluate

#### Answer:

The given determinant is.

We have:

M_{13 }=

M_{23 }=

M_{33 }=

∴A_{13 }= cofactor of *a*_{13 }= (−1)^{1+3} M_{13} = (*z − y*)

A_{23 }= cofactor of *a*_{23 }= (−1)^{2+3} M_{23} = − (*z − x*) = (*x − z*)

A_{33 }= cofactor of *a*_{33 }= (−1)^{3+3} M_{33} = (*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,

#### Question 5:

If and A_{ij} is Cofactors of *a*_{ij}, then value of Δ is given by

#### 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 = *a*_{11}A_{11} + *a*_{21}A_{21} + *a*_{31}A_{31}

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

The correct answer is D.

#### Page No 131:

#### Question 1:

Find adjoint of each of the matrices.

#### Answer:

#### Question 2:

Find adjoint of each of the matrices.

#### Answer:

#### Question 3:

Verify *A* (*adj A*) = (*adj A*) *A* = *I* .

#### Answer:

#### Question 4:

Verify *A* (*adj A*) = (*adj A*) *A* = *I* .

#### Answer:

#### Page No 132:

#### Question 5:

Find the inverse of each of the matrices (if it exists).

#### Answer:

#### Question 6:

Find the inverse of each of the matrices (if it exists).

#### Answer:

#### Question 7:

Find the inverse of each of the matrices (if it exists).

#### Answer:

#### Question 8:

Find the inverse of each of the matrices (if it exists).

#### Answer:

#### Question 9:

Find the inverse of each of the matrices (if it exists).

#### Answer:

#### Question 10:

Find the inverse of each of the matrices (if it exists).

.

#### Answer:

#### Question 11:

Find the inverse of each of the matrices (if it exists).

#### Answer:

#### Question 12:

Let and. Verify that

#### Answer:

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

(*AB*)^{−1} = *B*^{−1}*A*^{−1}

Hence, the given result is proved.

#### Question 13:

If, show that. Hence find.

#### Answer:

#### Question 14:

For the matrix, find the numbers *a* and *b* such that *A*^{2} + *aA* + *bI *= *O*.

#### 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.

#### Question 15:

For the matrixshow that *A*^{3} − 6*A*^{2} + 5*A* + 11 *I* = O. Hence, find *A*^{−1.}

#### Answer:

From equation (1), we have:

#### Question 16:

If verify that *A*^{3} − 6*A*^{2} + 9*A* − 4*I* = *O* and hence find *A*^{−1}

#### Answer:

From equation (1), we have:

#### Question 17:

Let *A* be a nonsingular square matrix of order 3 × 3. Then is equal to

**A.** **B.** **C.** **D. **

#### Answer:

**Answer: B**

We know that,

Hence, the correct answer is B.

#### Question 18:

If *A* is an invertible matrix of order 2, then det (*A*^{−1}) is equal to

**A.** det (*A*) ** B.** **C.** 1 **D. **0

#### Answer:

Since *A* is an invertible matrix,

Hence, the correct answer is B.

#### Page No 136:

#### Question 1:

Examine the consistency of the system of equations.

*x *+ 2*y *= 2

2*x* + 3*y *= 3

#### Answer:

The given system of equations is:

*x *+ 2*y *= 2

2*x* + 3*y *= 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.

#### Question 2:

Examine the consistency of the system of equations.

2*x *− *y* = 5

*x* + *y *= 4

#### Answer:

The given system of equations is:

2*x *− *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.

#### Question 3:

Examine the consistency of the system of equations.

*x* + 3*y* = 5

2*x* + 6*y* = 8

#### Answer:

The given system of equations is:

*x* + 3*y* = 5

2*x* + 6*y* = 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.

#### Question 4:

Examine the consistency of the system of equations.

*x* +* y *+ *z* = 1

2*x* + 3*y* + 2*z* = 2

*ax* + *ay* + 2*az* = 4

#### Answer:

The given system of equations is:

*x* +* y *+ *z* = 1

2*x* + 3*y* + 2*z* = 2

*ax* + *ay* + 2*az* = 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.

#### Question 5:

Examine the consistency of the system of equations.

3*x* −* y *− 2z = 2

2*y* − *z* = −1

3*x* − 5*y* = 3

#### Answer:

The given system of equations is:

3*x* −* y *− 2z = 2

2*y* − *z* = −1

3*x* − 5*y* = 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.

#### Question 6:

Examine the consistency of the system of equations.

5*x* −* y *+ 4*z* = 5

2*x* + 3*y* + 5*z* = 2

5*x* − 2*y* + 6*z* = −1

#### Answer:

The given system of equations is:

5*x* −* y *+ 4*z* = 5

2*x* + 3*y* + 5*z* = 2

5*x* − 2*y* + 6*z* = −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.

#### Question 7:

Solve system of linear equations, using matrix method.

#### 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.

#### Question 8:

Solve system of linear equations, using matrix method.

#### 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.

#### Question 9:

Solve system of linear equations, using matrix method.

#### 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.

#### Question 10:

Solve system of linear equations, using matrix method.

5*x* + 2*y* = 3

3*x* + 2*y* = 5

#### 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.

#### Question 11:

Solve system of linear equations, using matrix method.

#### 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.

#### Question 12:

Solve system of linear equations, using matrix method.

*x* − *y* + *z* = 4

2*x* + *y* − 3*z* = 0

*x* + *y* + *z* = 2

#### 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.

#### Question 13:

Solve system of linear equations, using matrix method.

2*x* + 3*y* + 3*z* = 5

*x* − 2*y* + *z* = −4

3*x* − *y* − 2*z* = 3

#### Answer:

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

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

#### Question 14:

Solve system of linear equations, using matrix method.

*x* − *y* + 2*z* = 7

3*x* + 4*y* − 5*z* = −5

2*x* −* y* + 3*z* = 12

#### 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.

#### Page No 137:

#### Question 15:

If, find *A*^{−1}. Using A^{−1} solve the system of equations

#### Answer:

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

#### Question 16:

The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60. The cost of 2 kg onion, 4 kg

wheat and 6 kg rice is Rs 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70.

Find cost of each item per kg by matrix method.

#### 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.

#### Page No 141:

#### Question 1:

Prove that the determinant is independent of *θ*.

#### Answer:

Hence, Î” is independent of *Î¸*.

#### Question 2:

Without expanding the determinant, prove that

#### Answer:

Hence, the given result is proved.

#### Question 3:

Evaluate

#### Answer:

Expanding along C_{3}, we have:

#### Question 4:

If *a*, *b* and *c *are real numbers, and,

Show that either *a* + *b* + *c* = 0 or *a* = *b* = *c*.

#### Answer:

Expanding along R_{1}, we have:

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

#### Question 5:

Solve the equations

#### Answer:

#### Question 6:

Prove that

#### Answer:

Expanding along R_{3}, we have:

Hence, the given result is proved.

#### Question 7:

If

#### Answer:

We know that.

#### Page No 142:

#### Question 8:

Let verify that

(i)

(ii)

#### Answer:

(i)

We have,

(ii)

#### Question 9:

Evaluate

#### Answer:

Expanding along R_{1}, we have:

#### Question 10:

Evaluate

#### Answer:

Expanding along C_{1}, we have:

#### Question 11:

Using properties of determinants, prove that:

#### Answer:

Expanding along R_{3}, we have:

Hence, the given result is proved.

#### Question 12:

Using properties of determinants, prove that:

#### Answer:

Expanding along R_{3}, we have:

Hence, the given result is proved.

#### Question 13:

Using properties of determinants, prove that:

#### Answer:

Expanding along C_{1}, we have:

Hence, the given result is proved.

#### Question 14:

Using properties of determinants, prove that:

#### Answer:

Expanding along C_{1}, we have:

Hence, the given result is proved.

#### Question 15:

Using properties of determinants, prove that:

#### Answer:

Hence, the given result is proved.

#### Question 16:

Solve the system of the following equations

#### 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,

*A*_{11} = 75, *A*_{12} = 110, *A*_{13} = 72

*A*_{21} = 150, *A*_{22} = −100, *A*_{23} = 0

*A*_{31} = 75, *A*_{32} = 30, *A*_{33} = − 24

#### Page No 143:

#### Question 17:

Choose the correct answer.

If *a*, *b*, *c*, are in A.P., then the determinant

**A.** 0 **B.** 1 **C.** *x ***D. **2*x*

#### Answer:

**Answer:** **A**

Here, all the elements of the first row (R_{1}) are zero.

Hence, we have Δ = 0.

The correct answer is A.

#### Question 18:

Choose the correct answer.

If *x*, *y*, *z* are nonzero real numbers, then the inverse of matrix is

**A.** **B.**

**C.** **D. **

#### Answer:

**Answer: A**

The correct answer is A.

#### Question 19:

Choose the correct answer.

Let, where 0 ≤ *θ*≤ 2π, then

**A.** Det (A) = 0

**B.** Det (A) ∈ (2, ∞)

**C.** Det (A) ∈ (2, 4)

**D. **Det (A)∈ [2, 4]

#### Answer:

Answer: D

Now,

0≤θ≤2π

⇒-1≤sinθ≤1 The correct answer is D.