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#### Page No 108:

#### Question 1:

Evaluate the determinants in Exercises 1 and 2.

#### Answer:

= 2(−1) − 4(−5) = − 2 + 20 = 18

#### Page No 108:

#### 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

#### Page No 108:

#### Question 3:

If, then show that

#### Answer:

The given matrix is.

#### Page No 108:

#### 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.

#### Page No 108:

#### 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:

#### Page No 109:

#### Question 7:

Find values of *x*, if

(i) $\left|\begin{array}{cc}2& 4\\ 5& 1\end{array}\right|=\left|\begin{array}{cc}2x& 4\\ 6& x\end{array}\right|$

(ii) $\left|\begin{array}{cc}2& 3\\ 4& 5\end{array}\right|=\left|\begin{array}{cc}x& 3\\ 2x& 5\end{array}\right|$

#### Answer:

(i)

(ii)

#### Page No 109:

#### 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:

#### Page No 119:

#### 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.

#### Page No 119:

#### Question 3:

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

#### Answer:

#### Page No 119:

#### 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.

#### Page No 119:

#### 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.

#### Page No 120:

#### 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:

#### Page No 120:

#### 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.

#### Page No 120:

#### 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.

#### Page No 120:

#### 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.

#### Page No 120:

#### 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.

#### Page No 121:

#### 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:

#### Page No 121:

#### 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.

#### Page No 121:

#### 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.

#### Page No 121:

#### 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.

#### Page No 123:

#### 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.

#### Page No 123:

#### 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.

#### Page No 123:

#### 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*

#### Page No 126:

#### 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

#### Page No 126:

#### 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

#### Page No 126:

#### 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,

#### Page No 126:

#### 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:

#### Page No 131:

#### Question 2:

Find adjoint of each of the matrices.

#### Answer:

#### Page No 131:

#### Question 3:

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

#### Answer:

#### Page No 131:

#### 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:

#### Page No 132:

#### Question 6:

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

#### Answer:

#### Page No 132:

#### Question 7:

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

#### Answer:

#### Page No 132:

#### Question 8:

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

#### Answer:

#### Page No 132:

#### Question 9:

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

#### Answer:

#### Page No 132:

#### Question 10:

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

.

#### Answer:

#### Page No 132:

#### Question 11:

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

#### Answer:

#### Page No 132:

#### 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.

#### Page No 132:

#### Question 13:

If, show that. Hence find.

#### Answer:

#### Page No 132:

#### 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.

#### Page No 132:

#### 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:

#### Page No 132:

#### 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:

#### Page No 132:

#### 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.

#### Page No 132:

#### 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.

#### Page No 136:

#### 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.

#### Page No 136:

#### 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.

#### Page No 136:

#### 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.

#### Page No 136:

#### 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.

#### Page No 136:

#### 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.

#### Page No 136:

#### 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.

#### Page No 136:

#### 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.

#### Page No 136:

#### 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.

#### Page No 136:

#### 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.

#### Page No 136:

#### 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.

#### Page No 136:

#### 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.

#### Page No 136:

#### 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.

#### Page No 136:

#### 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

#### Page No 137:

#### 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, Δ is independent of *θ*.

#### Page No 141:

#### Question 2:

Without expanding the determinant, prove that

#### Answer:

Hence, the given result is proved.

#### Page No 141:

#### Question 3:

Evaluate

#### Answer:

Expanding
along C_{3}, we have:

#### Page No 141:

#### 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*.

#### Page No 141:

#### Question 5:

Solve the equations

#### Answer:

#### Page No 141:

#### Question 6:

Prove that

#### Answer:

Expanding along R_{3}, we have:

Hence, the given result is proved.

#### Page No 141:

#### Question 7:

If

#### Answer:

We know that.

#### Page No 142:

#### Question 8:

Let verify that

(i)

(ii)

#### Answer:

(i)

We have,

(ii)

#### Page No 142:

#### Question 9:

Evaluate

#### Answer:

Expanding
along R_{1}, we have:

#### Page No 142:

#### Question 10:

Evaluate

#### Answer:

Expanding
along C_{1}, we have:

#### Page No 142:

#### Question 11:

Using properties of determinants, prove that:

#### Answer:

Expanding
along R_{3}, we have:

Hence, the given result is proved.

#### Page No 142:

#### Question 12:

Using properties of determinants, prove that:

#### Answer:

Expanding
along R_{3}, we have:

Hence, the given result is proved.

#### Page No 142:

#### Question 13:

Using properties of determinants, prove that:

#### Answer:

Expanding
along C_{1}, we have:

Hence, the given result is proved.

#### Page No 142:

#### Question 14:

Using properties of determinants, prove that:

#### Answer:

Expanding
along C_{1}, we have:

Hence, the given result is proved.

#### Page No 142:

#### Question 15:

Using properties of determinants, prove that:

#### Answer:

Hence, the given result is proved.

#### Page No 142:

#### 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.

#### Page No 143:

#### 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.

#### Page No 143:

#### 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, $0\le \theta \le 2\mathrm{\pi}$

$\Rightarrow -1\le \mathrm{sin}\theta \le 1$

The correct answer is D.

View NCERT Solutions for all chapters of Class 15