MATHEMATICS XII

Solutions Chapter 4 – Determinants

Page No 108:

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) = 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

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 (C1) 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 R1 and R3 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→ C3 + C2, we have:

Here, two columns C1 and Care proportional.

Δ = 0.

Question 5:

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

Answer:

Applying R2 → R2 − R3, we have:

Applying R1 ↔R3 and R2 ↔R3, we have:

Applying R→ 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.

Page No 120:

Question 6:

By using properties of determinants, show that:

Answer:

We have,

Here, the two rows R1 and Rare identical.

∴Δ = 0.

Question 7:

By using properties of determinants, show that:

Answer:

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

Question 8:

By using properties of determinants, show that:

(i) 

(ii) 

Answer:

(i) 

Applying R1 → R1 − Rand 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 − Cand C2 → C2 − C3, we have:

Applying C1 → C1 + C2, we have:

Expanding along C1, we have:

Hence, the given result is proved.

Question 9:

By using properties of determinants, show that:

Answer:

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

Applying R3 → R3 + R2, we have:

Expanding along R3, we have:

Hence, the given result is proved.

Question 10:

By using properties of determinants, show that:

(i) 

(ii) 

Answer:

(i) 

Applying R1 → R1 + R+ 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 + R+ R3, we have:

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

Expanding along C3, we have:

Hence, the given result is proved.

Question 11:

By using properties of determinants, show that:

(i) 

(ii) 

Answer:

(i) 

Applying R1 → R1 + R+ 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 + C+ C3, we have:

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

Expanding along R3, we have:

Hence, the given result is proved.

Page No 121:

Question 12:

By using properties of determinants, show that:

Answer:

Applying R1 → R1 + R+ R3, we have:

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

Expanding along R1, we have:

Hence, the given result is proved.

Question 13:

By using properties of determinants, show that:

Answer:

Applying R1 → R1 + bRand R2 → R2 − aR3, we have:

Expanding along R1, we have:

Question 14:

By using properties of determinants, show that:

Answer:

Taking out common factors ab, and c from R1, R2, and Rrespectively, we have:

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

Applying C1 → aC1, C→ bC2, and C3 → cC3, we have:

Expanding along R3, 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 (x1y1), (x2y2), and

(x3y3) 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 (xy) 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 (xy) 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.

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 aij is Mij.

∴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 aij is Aij = (−1)i + j Mij.

∴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 aij is Mij.

∴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 aij is Aij = (−1)i + j Mij.

∴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

Question 2:

(i)  (ii) 

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

Question 3:

Using Cofactors of elements of second row, evaluate.

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

Question 4:

Using Cofactors of elements of third column, evaluate

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, 

Question 5:

If  and Aij is Cofactors of aij, 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 = a11A11 + a21A21 + a31A31

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 AA = I .

Answer:

Question 4:

Verify A (adj A) = (adj AA = 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−1A−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 A2 + 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 A3 − 6A2 + 5A + 11 I = O. Hence, find A−1.

Answer:

From equation (1), we have:

Question 16:

If verify that A3 − 6A2 + 9A − 4I = 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 (AB.  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.

+ 2= 2

2x + 3= 3

Answer:

The given system of equations is:

+ 2= 2

2x + 3= 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− y = 5

x + = 4

Answer:

The given system of equations is:

2− y = 5

x + = 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 + 3y = 5

2x + 6y = 8

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.

Question 4:

Examine the consistency of the system of equations.

x + y z = 1

2x + 3y + 2z = 2

ax + ay + 2az = 4

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.

Question 5:

Examine the consistency of the system of equations.

3x − y − 2z = 2

2y − z = −1

3x − 5y = 3

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.

Question 6:

Examine the consistency of the system of equations.

5x − y + 4z = 5

2x + 3y + 5z = 2

5x − 2y + 6z = −1

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.

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.

5x + 2y = 3

3x + 2y = 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

2x + y − 3z = 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.

2x + 3y + 3z = 5

x − 2y + z = −4

3x − y − 2z = 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 + 2z = 7

3x + 4y − 5z = −5

2x − y + 3z = 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, Δ 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 C3, we have:

Question 4:

If ab and are real numbers, and,

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

Answer:

Expanding along R1, 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 R3, 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 R1, we have:

Question 10:

Evaluate 

Answer:

Expanding along C1, we have:

Question 11:

Using properties of determinants, prove that:

Answer:

Expanding along R3, we have:

Hence, the given result is proved.

Question 12:

Using properties of determinants, prove that:

Answer:

Expanding along R3, we have:

Hence, the given result is proved.

Question 13:

Using properties of determinants, prove that:

Answer:

Expanding along C1, we have:

Hence, the given result is proved.

Question 14:

Using properties of determinants, prove that:

Answer:

Expanding along C1, 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,

A11 = 75, A12 = 110, A13 = 72

A21 = 150, A22 = −100, A23 = 0

A31 = 75, A32 = 30, A33 = − 24

Page No 143:

Question 17:

Choose the correct answer.

If abc, are in A.P., then the determinant

A. 0 B. 1 C. D. 2x

Answer:

Answer: A

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

Hence, we have Δ = 0.

The correct answer is A.

Question 18:

Choose the correct answer.

If xyz 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.

Solutions Chapter 3 – Matrices (Prev Lesson)
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