Determinants Class 12th Mathematics Part I CBSE Solution

Class 12^th Mathematics Part I CBSE Solution

Exercise 4.1

1. | cc 2&4 -5&-1 | Evaluate the determinants:
2. | cc costheta &-sintegrate heta sintegrate heta | Evaluate the determinants:…
3. | cc x^2 - x+1 x+1+1 | Evaluate the determinants:
4. If a = [ll 1&2 4&2] then show that |2A| = 4|A|.
5. If a = [lll 1&0&1 0&1&2 0&0&4] then show that |3A| = 27|A|
6. Evaluate the determinants | ccc 3&-1&-2 0&1&-1 3&-5&0 |
7. | ccc 3&-4&5 1&1&-2 2&3&1 | Evaluate the determinants
8. | ccc 0&1&2 -1&0&-3 -2&3&0 | Evaluate the determinants
9. | ccc 2&-1&-2 0&2&-1 3&-5&0 | Evaluate the determinants
10. If A = [lll 1&1&-2 2&1&-3 5&4&-9] , find |A|.
11. Find values of x, if | ll 2&4 5&1 | = | cc 2x&4 6 | | ll 2&4 5&1 | = | cc 2x&4…
12. | ll 2&3 4&5 | = | ll x&3 2x&5 | Evaluate the determinants
13. If , then x is equal toA. 6 B. 6 C. 6 D. 0

Exercise 4.2

1. | lll x+a y+b x+c | = 0 Using the property of determinants and without expanding…
2. | lll a-b b-c c-a | = 0 Using the property of determinants and without expanding…
3. | lll 2&7&65 3&8&75 5&9&86 | = 0 Using the property of determinants and without…
4. Using the property of determinants and without expanding in prove that:…
5. | ccc b+c+r+z c+a+p+x a+b+q+y | = 2 | ccc a b c | Using the property of…
6. | ccc 0&-b -a&0&-c b&0 | = 0 Using the property of determinants and without…
7. | ccc - a^2 ba& - b^2 ca& - c^2 | = 4a^2b^2c^2 Using the property of…
8. | lll 1& a^2 1& b^2 1& c^2 | = (a-b) (b-c) (c-a) By using properties of…
9. | lll 1&1&1 a a^3 & b^3 & c^3 | = (a-b) (b-c) (c-a) (a+b+c) By using properties…
10. | lll x& x^2 y& y^2 z& z^2 | = (x-y) (y-z) (z-x) (xy+yz+zx) By using properties…
11. | ccc x+4&2x&2x 2x+4&2x 2x&2x+4 | = (5x+4) (4-x)^2 By using properties of…
12. | ccc y+k y+k y+k | = k^2 (3y+k) By using properties of determinants, show…
13. | ccc a-b-c&2a&2a 2b&2b 2c&2c | = (a+b+c)^3 By using properties of…
14. | ccc x+y+2z z+z+2x z+x+2y | = 2 (x+y+z)^3 By using properties of…
15. | ccc 1& x^2 x^2 &1 x& x^2 &1 | = (1-x^3)^2 By using properties of…
16. | ccc 1+a^2 - b^2 &2ab&-2b 2ab& 1-a^2 + b^2 &2a 2b&-2a& 1-a^2 - b^2 | = (1+a^2…
17. | ccc a^2 + 1 ab& b^2 + 1 ca& c^2 + 1 | = 1+a^2 + b^2 + c^2 By using properties…
18. Let A be a square matrix of order 3 × 3, then | kA| is equal toA. k|A| B. k^2…
19. Which of the following is correctA. Determinant is a square matrix. B.…

Exercise 4.3

1. Find area of the triangle with vertices at the point given in each of the…
2. (2, 7), (1, 1), (10, 8) Find area of the triangle with vertices at the point…
3. (-2, -3), (3, 2), (-1, -8) Find area of the triangle with vertices at the point…
4. Show that points A (a, b + c), B (b, c + a), C (c, a + b) are collinear.…
5. (k, 0), (4, 0), (0, 2) Find values of k if area of triangle is 4 sq. units and…
6. (-2, 0), (0, 4), (0, k) Find values of k if area of triangle is 4 sq. units and…
7. Find equation of line joining (1, 2) and (3, 6) using determinants.…
8. Find equation of line joining (3, 1) and (9, 3) using determinants.…
9. If area of triangle is 35 sq units with vertices (2, -6), (5, 4) and (k, 4).…

Exercise 4.4

1. | cc 2&-4 0&3 | Write Minors and Cofactors of the elements of following…
2. | ll a b | Write Minors and Cofactors of the elements of following…
3. | lll 1&0&0 0&1&0 0&0&1 | Write Minors and Cofactors of the elements of…
4. | ccc 1&0&4 3&5&-1 0&1&2 | Write Minors and Cofactors of the elements of…
5. Using Cofactors of elements of second row, evaluate delta = | lll 5&3&8 2&0&1…
6. Using Cofactors of elements of third column, evaluate delta = | lll 1 1 1 | .…
7. If delta = | lll a_11_12_13 a_21_22_23 a_31_32_33 | and Aij is Cofactors of aij,…

Exercise 4.5

1. Find adjoint of each of the matrices. [ll 1&2 3&4]
2. Find adjoint of each of the matrices. [ccc 1&-1&2 2&3&5 -2&0&1]
3. [cc 2&3 -4&-6] Verify A (adj A) = (adj A) A = |A|
4. [ccc 1&-1&2 3&0&-2 1&0&3] Verify A (adj A) = (adj A) A = |A|
5. [cc 2&-2 4&3] Find the inverse of each of the matrices (if it exists)…
6. [ll -1&5 -3&2] Find the inverse of each of the matrices (if it exists)…
7. [lll 1&2&3 0&2&4 0&0&5] Find the inverse of each of the matrices (if it exists)…
8. [ccc 1&0&0 3&3&0 5&2&-1] Find the inverse of each of the matrices (if it exists)…
9. [ccc 2&1&3 4&-1&0 -7&2&1] Find the inverse of each of the matrices (if it…
10. [ccc 1&-1&2 0&2&-3 3&-2&4] Find the inverse of each of the matrices (if it…
11. [ccc 1&0&0 0 0 &-cosalpha] Find the inverse of each of the matrices (if it…
12. Let a = [ll 3&7 2&5] b = [ll 6&8 7&9] . Verify that (AB)-1 = B-1 A-1.…
13. If a = [cc 3&1 -1&2] , show that A^2 - 5A + 7I = O. Hence find A-1.…
14. For the matrix a = [ll 3&1 1&2] , find the numbers a and b such that A^2 + aA +…
15. For the matrix a = [ccc 1&1&1 1&2&-3 2&-1&3] Show that A^3 - 6A^2 + 5A + 11 I =…
16. If a = [ccc 2&-1&1 -1&2&-1 1&-1&2] Verify that A^3 - 6A^2 + 9A - 4I = O and…
17. Let A be a non-singular square matrix of order 3 × 3. Then |adj A| is equal…
18. If A is an invertible matrix of order 2, then det (A-1) is equal toA. det (A)…

Exercise 4.6

1. x + 2y = 2 2x + 3y = 3 Examine the consistency of the system of equations.…
2. 2x - y = 5 x + y = 4 Examine the consistency of the system of equations.…
3. x + 3y = 5 2x + 6y = 8 Examine the consistency of the system of equations.…
4. x + y + z = 1 2x + 3y + 2z = 2 ax + ay + 2az = 4 Examine the consistency of the…
5. 3x-y - 2z = 2 2y - z = -1 3x - 5y = 3 Examine the consistency of the system of…
6. 5x - y + 4z = 5 2x + 3y + 5z = 2 5x - 2y + 6z = -1 Examine the consistency of…
7. 5x + 2y = 4 7x + 3y = 5 Solve system of linear equations, using matrix method.…
8. 2x - y = -2 3x + 4y = 3 Solve system of linear equations, using matrix method.…
9. 4x - 3y = 3 3x - 5y = 7 Solve system of linear equations, using matrix method.…
10. 5x + 2y = 3 3x + 2y = 5 Solve system of linear equations, using matrix method.…
11. 2x+y+z = 1 x-2y-z = 3/2 3y-5z = 9 Solve system of linear equations, using…
12. x - y + z = 4 2x + y - 3z = 0 x + y + z = 2 Solve system of linear equations,…
13. 2x + 3y +3 z = 5 x - 2y + z = -4 3x - y - 2z = 3 Solve system of linear…
14. x - y + 2z = 7 3x + 4y - 5z = -5 2x - y + 3z = 12 Solve system of linear…
15. If a = [ccc 2&-3&5 3&2&-4 1&1&-2] , find A-1. Using A-1 solve the system of…
16. The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs. 60. The cost of 2 kg…

Miscellaneous Exercise

1. Prove that the determinant | ccc x heta -sintegrate heta &-x&1 costheta &1 | is…
2. Without expanding the determinant, prove that | ccc a& a^2 b& b^2 c& c^2 | = |…
3. Evaluate | ccc cosalpha cosbeta sinbeta &-sinalpha -sinbeta &0 sinalpha cosbeta…
4. If a, b and c are real numbers, and delta = | lll b+c+a+b c+a+b+c a+b+c+a | = 0…
5. Solve the equation | ccc x+a x+a x+a | = 0 , a not equal 0
6. Prove that | ccc a^2 & ac+c^2 a^2 + ab & b^2 ab& b^2 + bc & c^2 | = 4a^2b^2c^2…
7. If a^-1 = [ccc 3&-1&1 -15&6&-5 5&-2&2] b = [ccc 1&2&-2 -1&3&0 0&-2&1] , find…
8. Let a = [ccc 1&-2&1 -2&3&1 1&1&5] . Verify that [adj A]-1 = adj (A-1)…
9. Let a = [ccc 1&-2&1 -2&3&1 1&1&5] . Verify that (A-1)-1 = A
10. Evaluate | ccc x+y y+y x+y |
11. Evaluate | ccc 1 1+y 1+y |
12. Prove that | lll alpha & alpha^2 & beta + gamma beta & beta^2 & gamma + alpha…
13. Prove that | lll x& x^2 & 1+px^3 y& y^2 & 1+py^3 z& z^2 & 1+pz^3 | = (1 = pxyz)…
14. Prove that | ccc 3a&-a+b&-a+c -b+a&3b&-b+c -c+a&-c+b&3c | = 3 (a+b+c)…
15. Prove that | ccc 1&1+p&1+p+q 2&3+2p&4+3p+2q 3&6+3p&10+6p+3q | = 1…
16. Prove that | ccc sinalpha & cos (alpha + delta) sinbeta & cos (beta + delta)…
17. Solve the system of equations 2/x + 3/y + 10/z = 4 4/x - 6/y + 5/z = 1 6/x +…
18. If a, b, c, are in A.P, then the determinant | lll x+2+3+2a x+3+4+20 x+4+5+2c |…
19. If x, y, z are nonzero real numbers, then the inverse of matrix a = [lll x&0&0…
20. Let a = [ccc 1 heta &1 -sintegrate heta &1 heta -1&-sintegrate heta &1] , where…

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Exercise 4.1

Question 1.

Evaluate the determinants:

Answer:

We know that determinant of A is calculated as 

Now,

= 2(-1) – 4(-5)

= -2 – (-20)

= -2 + 20

= 18

The determinant of the above matrix is 18.

Question 2.

Evaluate the determinants:

Answer:

We know that determinant of A is calculated as 

Now, 

= cosθ(cosθ) - (-sinθ)(sinθ)

= cos²θ + sin²θ

= 1 [∵ cos²θ + sin²θ = 1]

The determinant of the above matrix is 1.

Question 3.

Evaluate the determinants:

Answer:

We know that determinant of A is calculated as 

Now, 

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

= (x³ - x² + x + x² – x + 1) - (x² - 1)

= x³ + 1 - x² + 1

= x³ - x² + 2

Ans. The determinant of the above matrix is x³ - x² + 2.

Question 4.

If  then show that |2A| = 4|A|.

Answer:

|A| = 

We know that determinant of A is calculated as 

= 1(2) - 2(4)

= 2 - 8

|A| = -6

LHS: |2A|

= 2(4) - 4(8)

= 8 - 32 = -24

|2A| = -24 …LHS

RHS: 4|A|

4|A|= 4(-6)

= -24

4|A| = -24 …RHS

LHS = RHS

Hence proved.

Question 5.

If  then show that |3A| = 27|A|

Answer:

We know that a determinant of a 3 x 3 matrix is calculated as

= 1[1(4) - 2(0)] – 0 + 1[0-0]

= 1[4 - 0] – 0 + 0

= 4

|A|= 4

LHS: |3A|

= 3[3(12) - 0(6)] – 0 + 3[0 – 0]

= 3(36) – 0 + 0

= 108

|3A| = 108 ----LHS

RHS: 27|A|

27|A| = 27(4)

= 108

27|A| = 108 ----RHS

LHS=RHS

Hence proved.

Question 6.

Evaluate the determinants

Answer:

Now,

We know that a determinant of a 3x3 matrix is calculated as

= 3[0 – (-1)(-5)] +1[0 – (-1)(3)] – 2[0 – 0]

= 3(-5) + 1(3) – 0

= -15 + 3

= -12

The determinant of the above matrix is -12

Question 7.

Evaluate the determinants

Answer:

Now, 

We know that a determinant of a 3 x 3 matrix is calculated as

= 3[1 – (-2)(3)] + 4[1 – (-2)(2)] + 5[3-2]

= 3[1 + 6] + 4[1 + 4] + 5[1]

= 3[7] + 4[5] + 5

= 21 + 20 + 5

= 46

The determinant of the above matrix is 46.

Question 8.

Evaluate the determinants

Answer:

Now, 

We know that a determinant of a 3 x 3 matrix is calculated as

= 0 – 1[0 – (-3)(-2)] + 2[(-1)(3) – 0]

= 0 – 1[0 - 6] + 2[-3 – 0]

= 0 – 1[-6] + 2[-3]

= 0 + 6 – 6

= 0

The determinant of the above matrix is 0.

Question 9.

Evaluate the determinants

Answer:

Now, 

We know that a determinant of a 3 x 3 matrix is calculated as

= 2[0 – (-1)(-5)] + 1[0 – (-1)(3)] – 2[0 – 3(2)]

= 2[0 – 5] + 1[0 + 3] – 2[-6]

= 2[-5] + 1[3] -2[-6]

= -10 + 3 + 12

= 5

The determinant of the above matrix is 5.

Question 10.

If A = , find |A|.

Answer:

GIVEN: 

Now, 

We know that a determinant of a 3 x 3 matrix is calculated as

.

= 1[-9 – (-3)(4)] – 1[2(-9) – (-3)(5)] – 2[2(4) – 1(5)]

= 1[-9 + 12] – 1[-18 + 15] – 2[8 – 5]

= 1[3] – 1[-3] – 2[3]

= 3 + 3 – 6

= 0

Ans. |A| = 0

Question 11.

Find values of x, if

Answer:

We have

We know that determinant of A is calculated as 

⇒ 2(1) – 4(5) = 2x(x) – 4(6)

⇒ 2 – 20 = 2x² - 24

⇒ -18 = 2x² – 24

⇒ 2x² = -24 + 18

⇒ 2x² = 6

⇒ x² = 6/2

⇒ x² = 3

x = �√3

Ans. The value of x is �√3

Question 12.

Evaluate the determinants

Answer:

We know that determinant of A is calculated as 

We have 

⇒ 2(5) – 3(4) = x(5) – 3(2x)

⇒ 10 – 12 = 5x – 6x

⇒ -2 = -x

⇒ x = 2

Ans. The value of x is 2.

Question 13.

If , then x is equal to
A. 6
B. ±6

C. –6

D. 0

Answer:

We have 

We know that determinant of A is calculated as 

⇒ x(x) – 2(18) = 6(6) – 2(18)

⇒ x² - 36 = 36 – 36

⇒ x² =36 – 36 + 36

⇒ x² = 36

⇒ x = ±6

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Exercise 4.2

Question 1.

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

Answer:

Applying Operations C₁→ C₁ + C₂ (i.e. Replacing 1^st column by addition of 1^st and 2^nd column)

C₁→ C₁ - C₃ (i.e. Replacing 1^st column by subtraction of 1^st and 3^rd column)

If any one of the rows or columns of a determinant is 0 then the value of that determinant is 0.

∴ LHS = 0 = RHS

Question 2.

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

Answer:

Applying Operation C₁→ C₁ + C₂ (i.e. Replacing 1^st column by addition of 1^st and 2^nd column)

C₁→ C₁ + C₃ (i.e. Replacing 1^st column by addition of 1^st and 3^rd column)

If any one of the rows or columns of a determinant is 0 then the value of that determinant is 0.

∴ LHS = 0 = RHS

Question 3.

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

Answer:

Applying Operation C₁→ C₁ + 9C₂ (i.e. Replacing 1^st column by addition of 1^st column and 9 times second column)

C₁→ C₁ - C₃ (i.e. Replacing 1^st column by subtraction of 1^st and 3rd column)

If any one of the rows or columns of a determinant is 0 then the value of that determinant is 0.

∴ LHS = 0 = RHS

Question 4.

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

Answer:

C₃→ C₂ + C₃ (i.e. replace 3^rd column by addition of 2^nd and 3^rd column)

Taking ab + bc + ac outside determinant

C₁→ C₁ - C₃ (i.e. replace 1^st column by subtraction of 1^st and 3^rd column)

If any one of the rows or columns of a determinant is 0 then the value of that determinant is 0.

∴ LHS = 0 = RHS

∴

Question 5.

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

Answer:

When two determinants are added each of the corresponding elements gets added.

Here we can split the LHS determinant as

For the determinant, u perform the following transformation R₁↔ R₃ (i.e. interchange 1^st row with 3^rd row)

When two particular rows/columns of a determinant are interchanged the value becomes negative 1 times the original value.

R₁↔ R₂ (i.e. interchange 1^st row with 2^nd row)

R1 ↔ R3 (i.e. interchange 1^st row with 3^rd row)

R1 ↔ R2 (i.e. interchange 1^st row with 2^nd row)

∴ LHS = RHS

∴

Question 6.

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

Answer:

To Prove:



R₁→ cR₁ (i.e. replace 1^st row by multiplying it with c)

As we are multiplying we should also divide c so that the original given determinant is not changed

R₁→ R₁ - bR₂ (i.e. replace 1^st row by subtraction of 1^st row and b times 2^nd row)

Taking a outside the determinant from 1^st row

If any two rows or columns of a determinant are identical then the value of that determinant is 0 because we get a row or column with all elements 0 when we when we subtract those particular rows/columns here the transformation is R1 → R1 - R3

∴LHS = 0 = RHS

∴

Question 7.

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

Answer:

Taking ‘a’, ‘b’ and ‘c’ outside the determinant from 1^st,2^nd and 3^rd column respectively

Taking ‘a’, ‘b’ and ‘c’ outside the determinant from 1^st,2^nd and 3^rd row respectively

.

R1 → R1 + R2 (i.e. Replacing 1^st row by addition of 1^st and 2^nd row)

panding the determinant along 1^st row

∴ LHS = a² b² c² × 2(1 - (-1)) = 4a²b²c² = RHS

∴

Question 8.

By using properties of determinants, show that:

Answer:

R1 → R1 - R2 (i.e. Replacing 1^st row by subtraction of 1^st and 2^nd row)

R2 → R2 - R3 (i.e. Replacing 2^nd row by subtraction of 2^nd and 3^rd row)

Since we know a² - b² = (a + b)(a - b)

Therefore taking (a - b) and (b - c) outside the determinant from 1^st and 2^nd row respectively

R1 → R1 - R2 (i.e. Replacing 1^st row by subtraction of 1^st and 2^nd row)

Expanding the determinant along 1^st column

∴

∴LHS = (a - b)(b - c)(0 - (a - c))

∴LHS = (a - b)(b - c)(c - a) = RHS

∴

Question 9.

By using properties of determinants, show that:

Answer:

C1 → C1 - C2 (i.e. Replacing 1^st column by subtraction of 1^st and 2^nd column)

C2 → C2 - C3 (i.e. Replacing 2^nd column by subtraction of 2^nd and 3^rd column)

We have a³ - b³ = (a - b)(a² + ab + b²) and b³ - c³ = (b - c)(b² + bc + c²)

Therefore taking (a - b) and (b - c) outside the determinant from 1^st and 2^nd column respectively

C1 → C1 - C2 (i.e. Replacing 1^st column by subtraction of 1^st and 2^nd column)

As a² - c²) = (a + c)(a - c) therefore taking (a - c) outside the determinant from 1^st column we get

.

Expanding the determinant along 1^st row

∴ LHS = (a - b)(b - c)(a - c)( - (a + b + c))

Adjusting the minus sign with (a - c)

∴LHS = (a - b)(b - c)(c - a)(a + b + c) = RHS

∴

Question 10.

By using properties of determinants, show that:

Answer:

R1 → R1 - R2 (i.e. Replacing 1^st row by subtraction of 1^st and 2^nd row)

R2 → R2 - R3 (i.e. Replacing 2^nd row by subtraction of 2^nd and 3^rd row)

We know x² - y² = (x + y)(x - y) and y² - z² = (y + z)(y - z)

Therefore taking (x - y) and (y - z) outside the determinant from 1^st and 2^nd row respectively

R1 → R1 - R2 (i.e. Replacing 1^st row by subtraction of 1^st and 2^nd row)

Taking (x - z) outside determinant from 1^st row

C2 → C2 - C3 (i.e. Replacing 2^nd column by subtraction of 2^nd and 3^rd column)

Expanding the determinant along 1^st row

∴ LHS = (x - y)(y - z)(x - z)(z² - xy - xz - yz - z²)

Cancelling z² and adjusting the negative sign with (x - z)

∴LHS = (x - y)(y - z)(z - x)(xy + yz + zx) = RHS

∴

Question 11.

By using properties of determinants, show that:

Answer:

R1 → R1 + R2 + R3 (i.e. replace 1^st row by addition of 1^st, 2^nd and 3^rd row)

Taking 5x + 4 outside the determinant from 1^st row

C2 → C2 - C1 (i.e. replace 2^nd column by subtraction of 2^nd and 1^st column)

C3 → C3 - C1 (i.e. replace 3^rd column by subtraction of 3^rd and 1^st column)

Expanding the determinant along 1^st row

∴ LHS = (5x - 4) (4 - x)² = RHS

∴

Question 12.

By using properties of determinants, show that:

Answer:

R1 → R1 + R2 + R3 (i.e. replace 1^st row by addition of 1^st, 2^nd and 3^rd row)

Taking 3y + k outside the determinant from 1^st row

C2 → C2 - C1 (i.e. replace 2^nd column by subtraction of 2^nd and 1^st column)

C3 → C3 - C1 (i.e. replace 3^rd column by subtraction of 3^rd and 1^st column)

Expanding the determinant along 1^st row

∴ LHS = (3y + k) k² = RHS

∴

Question 13.

By using properties of determinants, show that:

Answer:

R1 → R1 + R2 + R3 (i.e. replace 1^st row by addition of 1^st, 2^nd and 3^rd row)

Taking a + b + c outside the determinant from 1^st row

C2 → C2 - C1 (i.e. replace 2^nd column by subtraction of 2^nd and 1^st column)

C3 → C3 - C1 (i.e. replace 3^rd column by subtraction of 3^rd and 1^st column)

Expanding the determinant along 1^st row

∴ LHS = (a + b + c)³ = RHS

∴

Question 14.

By using properties of determinants, show that:

Answer:

C1 → C1 + C2 + C3 (i.e. replace 1^st column by addition of 1^st, 2^nd and 3^rd column)

Taking 2(x + y + z) outside the determinant from 1^st column

R2 → R2 - R1 (i.e. replace 2^nd row by subtraction of 2^nd and 1^st row)

R3 → R3 - R1 (i.e. replace 3^rd row by subtraction of 3^rd and 1^st row)

Expanding the determinant along 1^st column

∴ LHS = 2(x + y + z)³ = RHS

∴

Question 15.

By using properties of determinants, show that:

Answer:

R1 → R1 - xR2 (i.e. replace 1^st row by subtraction of 1^st row and ‘x’ times 2^nd row)

Taking (1 - x³) outside the determinant from 1^st row

.

Expanding the determinant along 1^st row

HS = (1 - x³)² = RHS

∴

Question 16.

By using properties of determinants, show that:

Answer:

R1 → R1 + bR3 (i.e. replace 1^st row by addition of 1^st row and b times 3^rd row)

R2 → R2 - aR3 (i.e. replace 2^nd row by subtraction of 2^nd row and a times 3^rd row)

.

Taking both (1 + a² + b²) outside the determinant from 1^st and 2^nd row

Expanding the determinant along 1^st row

∴ LHS = (1 + a² + b²)² [1 + a² - b² - (-2b²)]

∴ LHS = (1 + a² + b² )² (1 + a² + b²)

∴ LHS = (1 + a² + b² )³ = RHS

Question 17.

By using properties of determinants, show that:

Answer:

Taking out a, b and c from the determinant from 1^st, 2^nd and 3^rd row respectively.

R2 → R2 - R1 (i.e. replace 2^nd row by subtraction of 2^nd and 1^st row)

R3 → R3 - R1 (i.e. replace 3^rd row by subtraction of 3^rd and 1^st row)

.

C1 → aC1 (i.e. replace 1^st column by ‘a’ times 1^st column)

C2 → bC2 (i.e. replace 2^nd column by ‘b’ times 2^nd column)

→ cC3 (i.e. replace 3^rd column by ‘c’ times 3^rd column)

As we are multiplying by a, b and c we should also divide by a, b and c to keep the original determinant value unchanged.

.

C1 → C1 + C2 + C3 (i.e. replace 1^st column by addition of 1^st, 2^nd and 3^rd column)

Expanding determinant along 1^st column

∴ LHS = (1 + a² + b² + c²) = RHS

∴

Question 18.

Let A be a square matrix of order 3 × 3, then | kA| is equal to
A. k|A|

B. k² |A|

C. k³ |A|

D. 3k |A|

Answer:

Let A be any 3×3 matrix

∴ 

Taking out k from the determinant from 1^st, 2^nd and 3^rd row

∴ 

∴ |kA| = k³|A|

Therefore, answer is option (C) k³|A|

Question 19.

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:

Determinant is an operation which we perform on arranged numbers. A square matrix is set of arranged numbers. We perform some operations on a matrix and we get a value that value is called as determinant of that matrix hence determinant is a number associated to square matrix.

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Exercise 4.3

Question 1.

Find area of the triangle with vertices at the point given in each of the following:

(1, 0), (6, 0), (4, 3)

Answer:

Given vertices of the triangle are (1, 0), (6, 0), (4, 3)

Let the vertices of the triangle be given by (x₁, y₁), (x₂, y₂), (x₃, y₃)

Area of triangle is given by Δ = 

Area of triangle = Δ = 

Expanding the determinant along Row 1

Δ = 1/2 × [1 × (0 × 1 – 3 × 1) – 0 × (6 × 1 – 4 × 1) + 1 × (6 × 3 – 4 × 0)]

Δ = 1/2 × [1 × (0 – 3) – 0 + 1 × (18 – 0)]

Δ = 1/2 × (-3 + 18) = 15/2 sq. units

∴ Δ = 15/2 sq. units = 7.5 sq. units

Question 2.

Find area of the triangle with vertices at the point given in each of the following:

(2, 7), (1, 1), (10, 8)

Answer:

Given vertices of the triangle are (2, 7), (1, 1), (10, 8)

Let the vertices of the triangle be given by (x₁, y₁), (x₂, y₂), (x₃, y₃)

Area of triangle is given by Δ = 

Area of triangle = Δ = 

Expanding the determinant along Row 1

Δ = 1/2 × [2 × (1 × 1 – 8 × 1) – 7 × (1 × 1 – 10 × 1) + 1 × (8 × 1 – 1 × 10)]

Δ = 1/2 × [2 × (1 – 8) – 7 × (1 – 10) + 1 × (8 – 10)]

Δ = 1/2 × [2 × (-7) – 7 × (-9) + 1 × (-2)] = 1/2 × (-14 + 63 – 2) sq. units

Δ = 1/2 × 47 sq. units = 47/2 sq. units

∴ Δ = 47/2 sq. units = 23.5 sq. units

Question 3.

Find area of the triangle with vertices at the point given in each of the following:

(–2, –3), (3, 2), (–1, –8)

Answer:

Given vertices of the triangle are (–2, –3), (3, 2), (–1, –8)

Let the vertices of the triangle be given by (x₁, y₁), (x₂, y₂), (x₃, y₃)

Area of triangle is given by Δ = 

Area of triangle = Δ = 

Expanding the determinant along Row 1

Δ = 1/2 × |[(-2) × (2 × 1 – (-8) × 1) – (-3) × (3 × 1 – (-1) × 1) + 1 × (3 × (-8) – 2 × (-1))]|

Δ = 1/2 × |[(-2) × (2 + 8) + 3 × (3 + 1) + 1 × (-24 + 2)]|

Δ = 1/2 × |[(-2) × 10 + 3 × 4 + 1 × (-22)]| = 1/2 ×| (-20 + 12 – 22)| sq. units

Δ = 1/2 × |-30| sq. units = 30/2 sq. units

∴ Δ = 30/2 sq. units = 15 sq. units

Question 4.

Show that points

A (a, b + c), B (b, c + a), C (c, a + b) are collinear.

Answer:

Given vertices of the triangle are A (a, b + c), B (b, c + a), C (c, a + b)

Let the vertices of the triangle be given by (x₁, y₁), (x₂, y₂), (x₃, y₃)

Area of triangle is given by Δ = 

For points to be collinear area of triangle = Δ = 0

So, we have to show that area of triangle formed by ABC is 0

Area of triangle = Δ = 

Expanding the determinant along Row 1

Δ = 1/2 × [a × {(c + a) × 1 – (a + b) × 1) – (b + c) × {b × 1 – c × 1} + 1 × {b × (a + b) – c × (c +a)}]

Δ = 1/2 × [a × (c + a – a – b) – (b + c) × (b – c) + 1 × (ab + b² – c² - ca)]

Δ = 1/2 × [a × (c – b) – (b² – c²) + 1 × (ab + b² – c² – ca)]

Δ = 1/2 × (ac – ab – b² + c² + ab + b² – c² – ca) sq units

Δ = 1/2 × 0 sq units

∴ Δ = 0

∴ Given vertices of the triangle are A (a, b + c), B (b, c + a), C (c, a + b) are collinear

Question 5.

Find values of k if area of triangle is 4 sq. units and vertices are

(k, 0), (4, 0), (0, 2)

Answer:

Given vertices of the triangle are (k, 0), (4, 0), (0, 2)

Let the vertices of the triangle be given by (x₁, y₁), (x₂, y₂), (x₃, y₃)

Area of triangle is given by Δ = 

Given, Area of triangle = Δ = 4 sq. units

 = 4

⇒ 4 = 1/2 |[k × (0 × 1 – 2 × 1) – 0 × (4 × 1 – 0 × 1) + 1 × (4 × 2 – 0 × 0)]|

⇒ 4 = 1/2 × |[k × (0 – 2) – 0 + 1 × (8 – 0)]|

⇒ � 4 × 2 = -2k + 8

⇒ 8 = -2k + 8 and -8 = -2k + 8

⇒ 8 – 8 = -2k and 8 + 8 = 2k

⇒ 2k = 0 and 16 = 2k

⇒ k = 0 and k = 8

Question 6.

Find values of k if area of triangle is 4 sq. units and vertices are

(–2, 0), (0, 4), (0, k)

Answer:

Given vertices of the triangle are (–2, 0), (0, 4), (0, k)

Let the vertices of the triangle be given by (x₁, y₁), (x₂, y₂), (x₃, y₃)

Area of triangle is given by Δ = 

Given, Area of triangle = Δ = 4 sq. units

 = 4

⇒ 4 = 1/2 |[(-2) × (4 × 1 – k × 1) – 0 × (0 × 1 – 0 × 1) + 1 × (0 × k – 0 × 4)]|

⇒ 4 = 1/2 × |[(-2) × (4 – k) – 0 + 1 × (0 – 0)]|

⇒ 4 × 2 = |(-8 + 2k)

⇒ � 8= 2k - 8

⇒ 8 = 2k - 8 and ⇒ -8 = 2k - 8

⇒ 8 + 8 = 2k and ⇒ 8 - 8 = 2k

⇒ k = 16/2 and ⇒ k = 0/2

⇒ k = 8 and ⇒ k = 0

Question 7.

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

Answer:

Equation of line joining points (x₁, y₁) & (x₂, y₂) is given by = 0

Given points are (1, 2) and (3, 6)

Equation of line is given by  = 0

⇒ 1/2 × [1 × (6 × 1 – y × 1) – 2 × (3 × 1 – x × 1) + 1 × (3 × y – x × 6)] = 0

⇒ [(6 – y) – 2 × (3 – x) + (3y – 6x)] = 0 × 2

⇒ (6 – y – 6 + 2x + 3y – 6x) = 0

⇒ 2y – 4x = 0

⇒ y – 2x = 0 ⇒ y = 2x

∴ Required Equation of line is y = 2x

Question 8.

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

Answer:

Equation of line joining points (x₁, y₁) & (x₂, y₂) is given by = 0

Given points are (3, 1) and (9, 3)

Equation of line is given by  = 0

⇒ 1/2 × [3 × (3 × 1 – y × 1) – 1 × (9 × 1 – x × 1) + 1 × (9 × y – x × 3)] = 0

⇒ [3 × (3 – y) – 1 × (9 – x) + (9y – 3x)] = 0 × 2

⇒ (9 – 3y – 9 + x + 9y – 3x) = 0

⇒ 6y – 2x = 0

⇒ 2x – 6y = 0 ⇒ x – 3y = 0

∴ Required Equation of line is x – 3y = 0

Question 9.

If area of triangle is 35 sq units with vertices (2, –6), (5, 4) and (k, 4). Then k is
A. 12

B. –2

C. –12, –2

D. 12, –2

Answer:

Given vertices of the triangle are (2, – 6), (5, 4) and (k, 4).

Let the vertices of the triangle be given by (x₁, y₁), (x₂, y₂), (x₃, y₃)

Area of triangle is given by Δ = 

Given, Area of triangle = Δ = 35 sq. units

 = 35

⇒ � 35 = 1/2 × [2 × (4 × 1 – 4 × 1) – (-6) × (5 × 1 – k × 1) + 1 × (5 × 4 – k × 4)]

⇒ � 35 = 1/2 × [2 × (4 – 4) + 6 × (5 – k) + 1 × (20 – 4k)]

⇒ � 35 × 2 = (2 × 0 + 30 – 6k + 20 – 4k)

⇒ � 70 = 30 – 6k + 20 – 4k

⇒ � 70 = 50 – 10k

⇒ 70 – 50 = -10k and ⇒ -70 – 50 = -10k

⇒ 20 = -10k and ⇒ -120 = -10k

⇒ k = -20/10 and ⇒ k = 120/10

⇒ k = -2 and ⇒ k = 12

---

Exercise 4.4

Question 1.

Write Minors and Cofactors of the elements of following determinants:

Answer:

Minor: Minor of an element a_ij of a determinant is the determinant obtained by removing i^th row and j^th column in which element a_ij lies. It is denoted by M_ij.

Cofactor: Cofactor of an element a_ij, A_ij = (-1)^i+j M_ij.

Minor of element a_ij = M_ij

a₁₁ = 2, Minor of element a₁₁ = M₁₁ = 3

Here removing 1^st row and 1^st column from the determinant we are left out with 3 so M₁₁ = 3.

Similarly, finding other Minors of the determinant

a₁₂ = -4, Minor of element a₁₂ = M₁₂ = 0

a₂₁ = 0, Minor of element a₂₁ = M₂₁ = -4

a₂₂ = 3, Minor of element a₂₂ = M₂₂ = 2

Cofactor of an element a_ij, A_ij = (-1)^i+j × M_ij

A₁₁ = (-1)¹⁺¹ × M₁₁ = 1 × 3 = 3

A₁₂ = (-1)¹⁺² × M₁₂ = (-1) × 0 = 0

A₂₁ = (-1)²⁺¹ × M₁₁ = (-1) × (-4) = 4

A₂₂ = (-1)²⁺² × M₂₂ = 1 × 2 = 2

Question 2.

Write Minors and Cofactors of the elements of following determinants:

Answer:

Minor of an element a_ij = M_ij

a₁₁ = a, Minor of element a₁₁ = M₁₁ = d

Here removing 1^st row and 1^st column from the determinant we are left out with d so M₁₁ = d.

Similarly, finding other Minors of the determinant

a₁₂ = c, Minor of element a₁₂ = M₁₂ = b

a₂₁ = b, Minor of element a₂₁ = M₂₁ = c

a₂₂ = d, Minor of element a₂₂ = M₂₂ = a

Cofactor of an element a_ij, A_ij = (-1)^i+j × M_ij

A₁₁ = (-1)¹⁺¹ × M₁₁ = 1 × d = d

A₁₂ = (-1)¹⁺² × M₁₂ = (-1) × b = -b

A₂₁ = (-1)²⁺¹ × M₁₁ = (-1) × c = -c

A₂₂ = (-1)²⁺² × M₂₂ = 1 × a = a

Question 3.

Write Minors and Cofactors of the elements of following determinants:

Answer:

Minor of an element a_ij = M_ij

a₁₁ = 1, Minor of element a₁₁ = M₁₁ =  = (1 × 1) – (0 × 0) = 1

Here removing 1^st row and 1^st column from the determinant we are left out with the determinant. Solving this we get M₁₁ = 1

Similarly, finding other Minors of the determinant

a₁₂ = 0, Minor of element a₁₂ = M₁₂ =  = (0 × 1) – (0 × 0) = 0

a₁₃ = 0, Minor of element a₁₃ = M₁₃ =  = (0 × 0) - (1 × 0) = 0

a₂₁ = 0, Minor of element a₂₁ = M₂₁ =  = (0 × 1) – (0 × 0) = 0

a₂₂ = 1, Minor of element a₂₂ = M₂₂ =  = (1 × 1) – (0 × 0) = 1

a₂₃ = 0, Minor of element a₂₃ = M₂₃ =  = (1 × 0) – (0 × 0) = 0

a₃₁ = 0, Minor of element a₃₁ = M₃₁ =  = (0 × 0) – (0 × 1) = 0

a₃₂ = 0, Minor of element a₃₂ = M₃₂ =  = (1 × 0) – (0 × 0) = 0

a₃₃ = 1, Minor of element a₃₃ = M₃₃ =  = (1 × 1) – (0 × 0) = 1

Cofactor of an element a_ij, A_ij = (-1)^i+j × M_ij

A₁₁ = (-1)¹⁺¹ × M₁₁ = 1 × 1 = 1

A₁₂ = (-1)¹⁺² × M₁₂ = (-1) × 0 = 0

A₁₃ = (-1)¹⁺³ × M₁₃ = 1 × 0 = 0

A₂₁ = (-1)²⁺¹ × M₂₁ = (-1) × 0 = 0

A₂₂ = (-1)²⁺² × M₂₂ = 1 × 1 = 1

A₂₃ = (-1)²⁺³ × M₂₃ = (-1) × 0 = 0

A₃₁ = (-1)³⁺¹ × M₃₁ = 1 × 0 = 0

A₃₂ = (-1)³⁺² × M₃₂ = (-1) × 0 = 0

A₃₃ = (-1)³⁺³ × M₃₃ = 1 × 1 = 1

Question 4.

Write Minors and Cofactors of the elements of following determinants:

Answer:

Minor of an element a_ij = M_ij

a₁₁ = 1, Minor of element a₁₁ = M₁₁ =  = (5 × 2) – ((-1) × 1) = 10 + 1 = 11

Here removing 1^st row and 1^st column from the determinant we are left out with the determinant. Solving this we get M₁₁ = 11

Similarly, finding other Minors of the determinant

a₁₂ = 0, Minor of element a₁₂ = M₁₂ = = (3 × 2) – ((-1) × 0) = (6 - 0) = 6

a₁₃ = 4, Minor of element a₁₃ = M₁₃ =  = (3 × 1) – (5 × 0) = 3 - 0 = 3

a₂₁ = 3, Minor of element a₂₁ = M₂₁ =  = (0 × 2) – (4 × 1) = 0 – 4 = -4

a₂₂ = 5, Minor of element a₂₂ = M₂₂ =  = (1 × 2) – (4 × 0) = 2 – 0 = 2

a₂₃ = -1, Minor of element a₂₃ = M₂₃ =  = (1 × 1) – (0 × 0) = 1

a₃₁ = 0, Minor of element a₃₁ = M₃₁ =  = (0 × (-1)) – (4 × 5) = 0 – 20 = -20

a₃₂ = 1, Minor of element a₃₂ = M₃₂ =  = (1 × (-1)) – (4 × 3) = -1 – 12 = -13

a₃₃ = 2, Minor of element a₃₃ = M₃₃ = = (1 × 5) – (0 × 3) = (5 – 0) = 5

Cofactor of an element a_ij, A_ij = (-1)^i+j × M_ij

A₁₁ = (-1)¹⁺¹ × M₁₁ = 1 × 11 = 11

A₁₂ = (-1)¹⁺² × M₁₂ = (-1) × 6 = -6

A₁₃ = (-1)¹⁺³ × M₁₃ = 1 × 3 = 3

A₂₁ = (-1)²⁺¹ × M₂₁ = (-1) × (-4) = 4

A₂₂ = (-1)²⁺² × M₂₂ = 1 × 2 = 2

A₂₃ = (-1)²⁺³ × M₂₃ = (-1) × 1 = -1

A₃₁ = (-1)³⁺¹ × M₃₁ = 1 × (-20) = -20

A₃₂ = (-1)³⁺² × M₃₂ = (-1) × (-13) = 13

A₃₃ = (-1)³⁺³ × M₃₃ = 1 × 5 = 5

Question 5.

Using Cofactors of elements of second row, evaluate 

Answer:

To evaluate a determinant using cofactors, Let

B = 

Expanding along Row 1

B = 

B = a₁₁ A₁₁ + a₁₂ A₁₂ + a₁₃ A₁₃

[Where A_ij represents cofactors of a_ij of determinant B.]

B = Sum of product of elements of R₁ with their corresponding cofactors

Similarly, the determinant can be solved by expanding along column

So, B = sum of product of elements of any row or column with their corresponding cofactors

Cofactors of second row

A₂₁ = (-1)²⁺¹ × M₂₁ = (-1) ×  = (-1) × (3 × 3 – 8 × 2) = (-1) × (-7) = 7

A₂₂ = (-1)²⁺² × M₂₂ = 1 × = (5 × 3 – 8 × 1) = 7

A₂₃ = (-1)²⁺³ × M₂₃ = (-1) ×  = (-1) × (5 × 2 – 3 × 1) = (-1) × 7 = -7

[Where A_ij = (-1)^i+j × M_ij, M_ij = Minor of i^th row & j^th column]

Therefore,

Δ = a₂₁A₂₁ + a₂₂A₂₂ + a₂₃A₂₃

Δ = 2 × 7 + 1 × (-7) = 14 - 7 = 7

Ans: Δ = 7

Question 6.

Using Cofactors of elements of third column, evaluate.

Answer:

To evaluate a determinant using cofactors, Let

B = 

Expanding along Row 1

B = 

B = a₁₁ A₁₁ + a₁₂ A₁₂ + a₁₃ A₁₃

[Where A_ij represents cofactors of a_ij of determinant B.]

B = Sum of product of elements of R₁ with their corresponding cofactors

Similarly, the determinant can be solved by expanding along column

So, B = sum of product of elements of any row or column with their corresponding cofactors

Cofactors of third column

A₁₃ = (-1)¹⁺³ × M₁₃ = 1 ×  = 1 × (1 × z – 1 × y) = (z – y)

A₂₃ = (-1)²⁺³ × M₂₃ = (-1) ×  = (-1) × (1 × z – 1 × x) = - (z - x) = (x - z)

A₃₃ = (-1)³⁺³ × M₃₃ = 1 ×  = 1 × (1 × y – 1 × x) = (y – x)

[Where A_ij = (-1)^i+j × M_ij, M_ij = Minor of i^th row & j^th column]

Therefore,

Δ = a₁₃A₁₃ + a₂₃A₂₃ + a₃₃A₃₃

Δ = yz (z - y) + zx (x - z) + xy (y - x) = z [y (z - y) + x (x - z)] + xy (y - x)

Δ = z (yz - y² + x² - xz) + xy (y - x) = z [(yz - xz) + (x² - y²)] + xy (y - x)

Δ = z [z × (y - x) + (x + y) × (x - y)] + xy (y - x)

Δ = z × (y - x) × (z – x - y) + xy (y - x)

Δ = (y - x) × (z² – xz – yz + xy)

Δ = (y - x) × [z (z - x) – y (z - x)] = (y - x) × (z - y) × (z - x)

Δ = (x - y) (y - z) (z - x)

Ans: Δ = (x - y) (y - z) (z - x)

Question 7.

If  and A_ij is Cofactors of a_ij, then value of Δ is given by
A. a₁₁ A₃₁+ a₁₂ A₃₂ + a₁₃ A₃₃

B. a₁₁ A₁₁+ a₁₂ A₂₁ + a₁₃ A₃₁

C. a₂₁ A₁₁+ a₂₂ A₁₂ + a₂₃ A₁₃

D. a₁₁ A₁₁+ a₂₁ A₂₁ + a₃₁ A₃₁

Answer:

Δ = 

Expanding along Column 1

Δ = 

Δ = a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁

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Exercise 4.5

Question 1.

Find adjoint of each of the matrices.

Answer:

Adjoint of the matrix A = [a_ij]_n×n is defined as the transpose of the matrix [A_ij]_n×n where A_ij is the co-factor of the element a_ij.

Let’s find the cofactors for all the positions first-

Here, A₁₁ = 4, A₁₂ = -3, A₂₁ = -2, A₂₂ = 1.

∴ Adj A = 

= 

Question 2.

Find adjoint of each of the matrices.

Answer:

Adjoint of the matrix A = [a_ij]_n×n is defined as the transpose of the matrix [A_ij]_n×n where A_ij is the co-factor of the element a_ij.

Let’s find the cofactors for all the positions first-

Here, A₁₁ = 1{(3×1-0×5)} = 3

Similarly,

A₁₂ = -12, A₁₃ = 6, A₂₁ = 1, A₂₂ = 5, A₂₃ = 2, A₃₁ = -11, A₃₂ = -1, A₃₃ = 5.

Question 3.

Verify A (adj A) = (adj A) A = |A|

Answer:

Adjoint of the matrix A = [a_ij]_n×n is defined as the transpose of the matrix [A_ij]_n×n where A_ij is the co-factor of the element a_ij.

Let’s find the cofactors for all the positions first-

Here, A₁₁ = -6, A₁₂ = 4, A₂₁ = -3, A₂₂ = 2.

∴ Adj A = 

= 

So LHS = A(AdjA) = 

Also AdjA(A) = 

Determinant of A = |A| = 2(-6)-(3)(-4) = 0

So RHS = |A|I = 0

Hence A(AdjA) = AdjA(A) = |A|I = 0 {hence proved}

Question 4.

Verify A (adj A) = (adj A) A = |A|

Answer:

Adjoint of the matrix A = [a_ij]_n×n is defined as the transpose of the matrix [A_ij]_n×n where A_ij is the co-factor of the element a_ij.

Let’s find the cofactors for all the positions first-

Here, A₁₁ = 0, A₁₂ = -11, A₁₃ = 0, A₂₁ = 3, A₂₂ = 1, A₂₃ = -1, A₃₁ = 2, A₃₂ = 8, A₃₃ = 3.

∴ Adj A = 

= 

So, LHS = A(AdjA) = 

Also AdjA(A) = 

Determinant of A = |A| = 11

So RHS = |A|I = .

Hence A(AdjA) = AdjA(A) = |A|I =  {hence proved}

Question 5.

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

Answer:

We know that 

Adjoint of the matrix A = [a_ij]_n×n is defined as the transpose of the matrix [A_ij]_n×n where A_ij is the co-factor of the element a_ij.

Let’s find the cofactors for all the positions first-

Here, A₁₁ = 3, A₁₂ = -4, A₂₁ = 2, A₂₂ = 2.

∴ Adj A = 

= 

And |A| = 2(3)-(-2)(4) = 14

So .

Question 6.

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

Answer:

We know that 

Adjoint of the matrix A = [a_ij]_n×n is defined as the transpose of the matrix [A_ij]_n×n where A_ij is the co-factor of the element a_ij.

Let’s find the cofactors for all the positions first-

Here, A₁₁ = 2, A₁₂ = 3, A₂₁ = -5, A₂₂ = -1.

∴ Adj A = 

= 

And |A| = -1(2)-(-3)(5) = 13

So 

Question 7.

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

Answer:

Adjoint of the matrix A = [a_ij]_n×n is defined as the transpose of the matrix [A_ij]_n×n where A_ij is the co-factor of the element a_ij.

Let’s find the cofactors for all the positions first-

Here, A₁₁ = 10, A₁₂ = 0, A₁₃ = 0, A₂₁ = -10, A₂₂ = 5, A₂₃ = 0, A₃₁ = 2, A₃₂ = -4, A₃₃ = 2.

∴ Adj A = 

And |A| = 10.

Question 8.

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

Answer:

Adjoint of the matrix A = [a_ij]_n×n is defined as the transpose of the matrix [A_ij]_n×n where A_ij is the co-factor of the element a_ij.

Let’s find the cofactors for all the positions first-

Here, A₁₁ = -3, A₁₂ = 3, A₁₃ = -9, A₂₁ = 0, A₂₂ = -1, A₂₃ = -2, A₃₁ = 0, A₃₂ = 0, A₃₃ = 3.

∴ Adj A = 

And |A| = -3.

.

Question 9.

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

Answer:

Adjoint of the matrix A = [a_ij]_n×n is defined as the transpose of the matrix [A_ij]_n×n where A_ij is the co-factor of the element a_ij.

Let’s find the cofactors for all the positions first-

Here, A₁₁ = -1, A₁₂ = -4, A₁₃ = 1, A₂₁ = 5, A₂₂ = 23, A₂₃ = -11, A₃₁ = 3, A₃₂ = 12, A₃₃ = -6.

∴ Adj A = 

= .

And |A| = -3.

.

Question 10.

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

Answer:

Adjoint of the matrix A = [a_ij]_n×n is defined as the transpose of the matrix [A_ij]_n×n where A_ij is the co-factor of the element a_ij.

Let’s find the cofactors for all the positions first-

Here, A₁₁ = 2, A₁₂ = -9, A₁₃ = -6, A₂₁ = 0, A₂₂ = -2, A₂₃ = -1, A₃₁ = -1, A₃₂ = 3, A₃₃ = 2.

∴ Adj A = 

= 

And |A| = -1.

Question 11.

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

Answer:

Adjoint of the matrix A = [a_ij]_n×n is defined as the transpose of the matrix [A_ij]_n×n where A_ij is the co-factor of the element a_ij.

Let’s find the cofactors for all the positions first-

Here, A₁₁ = -1, A₁₂ = 0, A₁₃ = 0, A₂₁ = 0, A₂₂ = -cosα, A₂₃ = -sinα, A₃₁ = 0, A₃₂ = -sinα, A₃₃ = cosα.

∴ Adj A = 

And |A| = 1.

.

Question 12.

Let . Verify that (AB)^–1 = B^–1 A^–1.

Answer:

We have AB =  = (61)(67)-(47)(87) = -2

Here determinant of matrix = |AB|≠ 0 hence (AB)^-1 exists.

Also |A| = 1 ≠ 0 and |B| = -2 ≠ 0.

∴ A^-1 and B^-1 will also exist and are given by-

And hence,

{Hence proved}

Question 13.

If , show that A² – 5A + 7I = O. Hence find A^–1.

Answer:

We have A² = A.A = .

So A² – 5A + 7I = 

Hence A² – 5A + 7I = 0

∴ A.A – 5A = -7I

Now post multiply with A^-1

So A.A.A^-1-5A.A^-1 = -7I.A^-1

→ A.I – 5I = -7I.A^-1 {since A.A^-1 = I}

A – 5I = -7A^-1 {since X.I = X}

Question 14.

For the matrix , find the numbers a and b such that A² + aA + bI = O.

Answer:

We have A² = A.A = 

Since A² + aA + bI = 

So A² + aA + bI = 

Hence 10+3a+b = 0 …(i)

5+a = 0 …(ii)

5+2a+b = 0 …(iii)

From (ii) a = -5

Putting a in (iii) we get b = 5

So a = -5 and b = 5 satisfy the equation.

Question 15.

For the matrix 

Show that A³– 6A² + 5A + 11 I = O. Hence, find A^–1.

Answer:

Here A² = A.A =

And hence A³ = A. A² =

∴ A³– 6A² + 5A + 11 I =

Thus, A³– 6A² + 5A + 11 I = 0

Now, A³– 6A² + 5A + 11 I = 0,

→ (A.A.A)- 6 (A.A) +5A = -11I

Post-multiply with A^-1 on both sides-

→ (A.A.A.A^-1)- 6 (A.A.A^-1) +5A.A^-1 = -11I. A^-1

→ (A.A.I) – 6(A.I) + 5I = -11I. A^-1 {since A.A^-1 = I}

→ (A.A) – 6A +5I = -11A^-1 {since X.I = X}

Hence 

Question 16.

If Verify that A³ – 6A² + 9A – 4I = O and hence find A^-1.

Answer:

Here A² = A.A = 

And hence A³ = A. A² = 

∴ A³– 6A² + 9A -4I

Thus, A³– 6A² + 9A -4I = 0

Now, A³– 6A² + 9A -4I = 0,

→ (A.A.A)- 6 (A.A) +9A = 4I

Post-multiply with A^-1 on both sides-

→ (A.A.A.A^-1)- 6 (A.A.A^-1) +9A.A^-1 = 4I. A^-1

→ (A.A.I) – 6(A.I) + 9I = 4I. A^-1 {since A.A^-1 = I}

→ (A.A) – 6A +9I = 4A^-1 {since X.I = X}

Hence 

Question 17.

Let A be a non-singular square matrix of order 3 × 3. Then |adj A| is equal to
A. |A |

B. | A|²

C. | A|³

D. 3|A|

Answer:

For a square matrix of order n×n,

We know that |AdjA| = |A|^n-1

So, |AdjA| = |A|^(3-1) = |A|²

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 adj(A) is of order n and |Adj(A)| = |A|^n-1}

Alternative-

We know that AA^-1 = I

So |A||A^-1| = |I| = 1

---

Exercise 4.6

Question 1.

Examine the consistency of the system of equations.

x + 2y = 2

2x + 3y = 3

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

Now |A| = 3(1)-2(2) = -1 ≠ 0.

∴ A is a non-singular matrix and hence A-1 exists.

So, The system of equations will be consistent.

Question 2.

Examine the consistency of the system of equations.

2x – y = 5

x + y = 4

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

Now |A| = 2(1)-1(-1) = 3 ≠ 0.

∴ A is a non-singular matrix and hence A^-1 exists.

So, The system of equations will be 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

Now |A| = 1(6)-3(2) = 0

∴ A is a singular matrix and hence A^-1 doesn’t exist.

So, The system of equations will be 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

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

Now |A| = 1(6a-2a)-1(4a-2a) +1(2a-3a) = a ≠ 0

∴ A is a non-singular matrix and hence A^-1 exists.

So the system of equations will be 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

0x + 2y - z = -1

3x - 5y +0z = 3

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

Now |A| = 3(-5) + 3(5) = 0

∴ A is a singular matrix and hence A^-1 doesn’t exist.

So the system of equations will be 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

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

Now |A| = 5(18+10) + 1(12-25) + 4(-4-15) = 51 ≠ 0

∴ A is a non-singular matrix and hence A^-1 exists.

So, The system of equations will be consistent.

Question 7.

Solve system of linear equations, using matrix method.

5x + 2y = 4

7x + 3y = 5

Answer:

The given system of equations is:

5x + 2y = 4

7x + 3y = 5

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

Now |A| = 3(5)-2(7) = 1 ≠ 0

∴ A is a non-singular matrix and hence A^-1 exists.

Now 

So 

And 

So 

Hence x = 2 and y = -3.

Question 8.

Solve system of linear equations, using matrix method.

2x – y = –2

3x + 4y = 3

Answer:

The given system of equations is:

2x - y = -2

3x + 4y = 3

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

Now |A| = 2(4)-3(-1) = 11 ≠ 0.

∴ A is a non-singular matrix and hence A^-1 exists.

Now 

So 

And 

So 

Hence 

Question 9.

Solve system of linear equations, using matrix method.

4x – 3y = 3

3x – 5y = 7

Answer:

The given system of equations is:

4x - 3y = 3

3x - 5y = 7

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

Now |A| = 4(-5)-3(-3) = -11 ≠ 0

∴ A is a non-singular matrix and hence A^-1 exists.

Now 

AdjA = 

So 

And 

So 

Hence 

Question 10.

Solve system of linear equations, using matrix method.

5x + 2y = 3

3x + 2y = 5

Answer:

The given system of equations is:

5x + 2y = 3

3x + 2y = 5

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

Now |A| = 5(2)-2(3) = 4 ≠ 0

∴ A is a non-singular matrix and hence A^-1 exists.

Now 

AdjA = 

So 

And 

So 

Hence x = -1 and y = 4

Question 11.

Solve system of linear equations, using matrix method.

Answer:

The given system of equations is:

2x + y + z = 1

0x + 3y - 5z = 9

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

Now |A| = 2(10+3)-1(-5) +1(3) = 34 ≠ 0

∴ A is a non-singular matrix and hence A^-1 exists.

Now A₁₁ = 13, A₁₂ = 5, A₁₃ = 3, A₂₁ = 8, A₂₂ = -10, A₂₃ = -6, A₃₁ = 1, A₃₂ = 3, A₃₃ = -5

So AdjA = 

∴ 

And hence X = A^-1B

So 

Hence 

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 is:

x - y + z = 4

2x + y -3z = 0

x + y + z = 2

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

Now |A| = 1(1+3) + 1(5) + 1(1) = 10 ≠ 0

∴ A is a non-singular matrix and hence A^-1 exists.

Now A₁₁ = 4, A₁₂ = -5, A₁₃ = 1, A₂₁ = 2, A₂₂ = 0, A₂₃ = -2, A₃₁ = 1, A₃₂ = -2, A₃₃ = 3

So AdjA = 

And hence X = A^-1B

So 

Hence x = 2, y = -1 and z = 1.

Question 13.

Solve system of linear equations, using matrix method.

2x + 3y +3 z = 5

x – 2y + z = –4

3x – y – 2z = 3

Answer:

The given system of equations is:

2x + 3y + 3z = 5

x – 2y + z = -4

3x - y - 2z = 3

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

Now |A| = 2(4+1)-3(-5) +3(5) = 40 ≠ 0

∴ A is a non-singular matrix and hence A^-1 exists.

Now A₁₁ = 5, A₁₂ = 5, A₁₃ = 5, A₂₁ = 3, A₂₂ = -13, A₂₃ = 11, A₃₁ = 9, A₃₂ = 1, A₃₃ = -7

So AdjA = 

∴ 

And hence X = A^-1B

So 

Hence x = 1, y = 2 and z = -1

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 is:

x - y + 2z = 7

3x + 4y - 5z = -5

2x – y + 3z = 12

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

Now |A| = 1(12-5) + 1(9+10) + 2(-3-8) = 4 ≠ 0

∴ A is a non-singular matrix and hence A^-1 exists.

Now A₁₁ = 7, A₁₂ = -19, A₁₃ = -11, A₂₁ = 1, A₂₂ = -1, A₂₃ = -1, A₃₁ = -3, A₃₂ = 11, A₃₃ = 7

So AdjA = 

∴ 

So 

Hence 

Question 15.

If , find A^–1. Using A^–1 solve the system of equations

2x – 3y + 5z = 11

3x + 2y – 4z = – 5

x + y – 2z = – 3

Answer:

∴ |A| = 2(-4+4) + 3(-6+4) + 5(3-2)


= -1 ≠ 0

So inverse of A exists,

Now A₁₁ = 0, A₁₂ = 2, A₁₃ = 1, A₂₁ = -1, A₂₂ = -9, A₂₃ = -5, A₃₁ = 2, A₃₂ = 23, A₃₃ = 13

So AdjA = 

∴ 

So .

Hence x = 1, y = 2 and z = 3

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 according to the giving situation, it can be represented by a system of equations as-

4x + 3y + 2z = 60

2x + 4y + 6z = 90

6x + 2y + 3z = 70

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

Here |A| = 4(12-12)-3(6-36) + 2(4-24) = 50 ≠ 0

Hence A^-1 will exist.

Now, A₁₁ = 0, A₁₂ = 30, A₁₃ = -20, A₂₁ = -5, A₂₂ = 0, A₂₃ = 10, A₃₁ = 10, A₃₂ = -20, A₃₃ = 10

Hence x = 5, y = 8 & z = 8.

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

---

Miscellaneous Exercise

Question 1.

Prove that the determinant  is independent of θ.

Answer:

Let Δ = 

Expanding the above determinant along R₁ i.e. Row 1

Δ = x (-x² – 1) – sinθ (-x × sinθ – cosθ × 1) + cosθ (-sinθ × 1 + x cosθ)

Δ = -x³ – x + xsin²θ + sinθ × cos θ – sinθ × cosθ + x cos²θ

Δ = -x³ – x + x (sin²θ + cos²θ)

Since, sin²θ + cos²θ = 1

∴ Δ = -x³ – x + x

Hence Δ is independent of θ.

Question 2.

Without expanding the determinant, prove that 

Answer:

Multiplying row 1 with a, row 2 with b and row 3 with c

[R₁→ aR₁, R₂→ bR₂, R₃→ cR₃]

 [Taking common abc from C₃]

 [Applying C₁↔ C₃]

=  [Applying C₂ ↔ C₃]

= RHS

Since, LHS = RHS

∴ the given result is proved.

Question 3.

Evaluate 

Answer:

Let Δ = 

Expanding along C₃ we get

Δ = -sinα (-sinβ × sinα sinβ – cosβ × sinα cosβ) – 0 (sinα cosβ × cosα sinβ – cosα cosβ × sinα sinβ) + cosα (cosα cosβ × cosβ – cosα sinβ × (-sinβ))

Δ = -sinα (-sinα sin²β – sinα cos²β) – 0 + cosα (cosα cos²β + cosα sin²β)

Δ = sinα × sinα (sin²β + cos²β) + cosα × cosα (cos²β + sin²β)

[Taking –sinα and cosα common]

Since, sin²β + cos²β = 1

∴ Δ = sin²α + cos²α

Δ = 1 [sin²α + cos²α = 1]

Question 4.

If a, b and c are real numbers, and



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

Answer:

Given, Δ = 

Applying Elementary transformations, we get

R₁→ R₁ + R₂ + R₃ we have,

Δ = 2 (a + b + c) 

Now applying C₂→ C₂ – C₁ and C₃→ C₃ – C₁

Δ = 2 (a + b + c) 

Expanding along R₁

Δ = 2 (a + b + c) [1 {(b – c)(c – b) – (b – a)(c – a)} – 0 + 0]

Δ = 2 (a + b + c) [- b² – c² + 2bc – bc + ba + ac – a²]

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

Given that Δ = 0

∴ 2 (a + b + c) [ab + bc + ca – a² – b² – c²] = 0

⇒ Either a + b + c = 0, or ab + bc + ca – a² – b² – c² = 0

Now

ab + bc + ca – a² – b² – c² = 0

Multiplying both sides by -2

⇒ - 2ab – 2bc – 2ca + 2a² + 2b² + 2c² = 0

⇒ a² – 2ab + b² + b² – 2bc + c² + c² – 2ca + a² = 0

⇒ (a – b)² + (b – c)² + (c – a)² = 0

Since, (a – b)², (b – c)², (c – a)² are non-negative

∴ (a – b)² = (b – c)² = (c – a)²

⇒ (a – b) = (b – c) = (c – a)

⇒ a = b = c

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

Question 5.

Solve the equation 

Answer:

Given, 

Applying elementary transformations we get,

R₁→ R₁ + R₂ + R₃

⇒ (3x + a)  = 0

C₂→ C₂ – C₁ and C₃→ C₃ – C₁, we get

(3x + a)  = 0

Expanding along R₁ we have

(3x + a) [1 (a × a – 0 × 0) – 0 (x × a – 0 × a) + 0 (x × 0 – a × x)] = 0

⇒ (3x + a) [1 × a²] = 0

⇒ a² (3x + a) = 0

Since, a ≠ 0

∴ 3x + a = 0

⇒ x = -a/3

Question 6.

Prove that 

Answer:

Given,

LHS = 

RHS = 4a²b²c²

LHS = Δ = 

Taking out common factors a, b and c from C₁, C₂ and C₃, we have

Δ = abc 

Applying Elementary Transformations

R₂→ R₂ – R₁ and R₃→ R₃ – R₁

Δ = abc 

R₂→ R₂ + R₁

Δ = abc 

R₃→ R₃ + R₂

Δ = abc 

Δ = 2ab²c 

C₂→ C₂ – C₁

Δ = 2ab²c 

Expanding along R₃ we get,

Δ = 2ab²c [a (c – a) + a (a + c)]

= 2ab²c [ac – a² + a² + ac]

= 2ab²c (2ac)

= 4a²b²c²

Δ = RHS

∴ LHS = RHS

Hence, Proved

Question 7.

If , find (AB)^-1

Answer:

B = 

We need to find B^-1

To find the inverse of a matrix we need to find the Adjoint of that matrix

For finding the adjoint of the matrix we need to find its cofactors

Let B_ij denote the cofactors of Matrix B

Minor of an element b_ij = M_{ij �}

b₁₁ = 1, Minor of element b₁₁ = M₁₁ =  = (3 × 1) – ((-2) × 0) = 3

b₁₂ = 2, Minor of element b₁₂ = M₁₂ = = ((-1) × 1) – (0 × 0) = -1

b₁₃ = -2, Minor of element b₁₃ = M₁₃ =  = ((-1) × (-2)) – (3 × 0) = 2

b₂₁ = -1, Minor of element b₂₁ = M₂₁ = = (2 × 1) – ((-2) × (-2)) = -2

b₂₂ = 3, Minor of element b₂₂ = M₂₂ =  = (1 × 1) – ((-2) × 0) = 1

b₂₃ = 0, Minor of element b₂₃ = M₂₃ =  = (1 × (-2)) – (2 × 0) = -2

b₃₁ = 0, Minor of element b₃₁ = M₃₁ =  = (2 × 0) – ((-2) × 3) = 6

b₃₂ = -2, Minor of element b₃₂ = M₃₂ =  = (1 × 0) – ((-2) × (-1)) = -2

b₃₃ = 1, Minor of element b₃₃ = M₃₃ = = (1 × 3) – (2 × (-1)) = 5

Cofactor of an element b_ij, B_ij = (-1)^i+j × M_ij

B₁₁ = (-1)¹⁺¹× M₁₁ = 1 × 3 = 3

B₁₂ = (-1)¹⁺²× M₁₂ = (-1) × (-1) = 1

B₁₃ = (-1)¹⁺³× M₁₃ = 1 × 2 = 2

B₂₁ = (-1)²⁺¹× M₂₁ = (-1) × (-2) = 2

B₂₂ = (-1)²⁺² × M₂₂ = 1 × 1 = 1

B₂₃ = (-1)²⁺³ × M₂₃ = (-1) × (-2) = 2

B₃₁ = (-1)³⁺¹ × M₃₁ = 1 × 6 = 6

B₃₂ = (-1)³⁺² × M₃₂ = (-1) × (-2) = 2

B₃₃ = (-1)³⁺³ × M₃₃ = 1 × 5 = 5

Adj B =  = 

|B| = 1 (3 × 1 – (-2) × 0) – 2 ((-1) × 1 – 0 × 0) + (-2) ((-1) × (-2) – 3 × 0)

|B| = 3 – 2 ( -1 – 0) – 2 (2 – 0)

|B| = 3 + 2 – 4 = 1

∴ B^-1 = (Adj B)/|B| = /1 = 

We know (AB)^-1 = B^-1A^-1

(AB)^-1 = × 

Solving the above matrix we get

(AB)^-1 = 

Question 8.

Let . Verify that

[adj A]^-1 = adj (A^-1)

Answer:

A = 

|A| = 1 (3 × 5 – 1 × 1) – (-2) ((-2) × 5 – 1 × 1) + 1 ((-2) × 1 – 3 × 1)

|A| = (15 – 1) + 2 (-10 – 1) + (-2 – 3)

|A| = 14 – 22 – 5 = -13

To find the inverse of a matrix we need to find the Adjoint of that matrix

For finding the adjoint of the matrix we need to find its cofactors

Let A_ij denote the cofactors of Matrix A

Minor of an element a_ij = M_{ij �}

a₁₁ = 1, Minor of element a₁₁ = M₁₁ =  = (3 × 5) – (1 × 1) = 14

a₁₂ = -2, Minor of element a₁₂ = M₁₂ = = (-2 × 5) – (1 × 1) = -11

a₁₃ = 1, Minor of element a₁₃ = M₁₃ =  = (-2 × 1) – (3 × 1) = -5

a₂₁ = -2, Minor of element a₂₁ = M₂₁ = = ((-2) × 5) – (1 × 1) = -11

a₂₂ = 3, Minor of element a₂₂ = M₂₂ =  = (1 × 5) – (1 × 1) = 4

a₂₃ = 1, Minor of element a₂₃ = M₂₃ =  = (1 × 1) – ((-2) × 1) = 3

a₃₁ = 1, Minor of element a₃₁ = M₃₁ =  = (-2 × 1) – (3 × 1) = -5

a₃₂ = 1, Minor of element a₃₂ = M₃₂ =  = (1 × 1) – (1 × (-2)) = 3

a₃₃ = 5, Minor of element a₃₃ = M₃₃ = = (1 × 3) – ((-2) × (-2)) = -1

Cofactor of an element a_ij = A_ij

A₁₁ = (-1)¹⁺¹× 14 = 1 × 14 = 14

A₁₂ = (-1)¹⁺²× (-11) = (-1) × (-11) = 11

A₁₃ = (-1)¹⁺³× (-5) = 1 × (-5) = -5

A₂₁ = (-1)²⁺¹× (-11) = (-1) × (-11) = 11

A₂₂ = (-1)²⁺² × 4 = 1 × 4 = 4

A₂₃ = (-1)²⁺³ × 3 = (-1) × 3 = -3

A₃₁ = (-1)³⁺¹ × (-5) = 1 × (-5) = -5

A₃₂ = (-1)³⁺² × 3 = (-1) × 3 = -3

A₃₃ = (-1)³⁺³ × (-1) = 1 × (-1) = -1

Adj A =  = 

A^-1 = (Adj A)/|A|

A^-1 =  = 

(i) |Adj A| = 14(-4 – 9) – 11 (-11 – 15) – 5 (-33 + 20)

= 14 × (-13) – 11 × (-26) – 5 (-13)

= -182 + 286 + 65 = 169

Similarly Finding the Adj (Adj A) as found above

Adj (Adj A) = 

[Adj A]^-1 = Adj (Adj A)/|Adj A|

= 

= 

A^-1 =  = 

Similarly Finding the Adj (A^-1) as found above

Adj (A^-1) =  = 

Hence, [Adj A]^-1 = Adj (A^-1)

Question 9.

Let . Verify that

(A^-1)^-1 = A

Answer:

A = 

|A| = 1 (3 × 5 – 1 × 1) – (-2) ((-2) × 5 – 1 × 1) + 1 ((-2) × 1 – 3 × 1)

|A| = (15 – 1) + 2 (-10 – 1) + (-2 – 3)

|A| = 14 – 22 – 5 = -13

To find the inverse of a matrix we need to find the Adjoint of that matrix

For finding the adjoint of the matrix we need to find its cofactors

Let A_ij denote the cofactors of Matrix A

Minor of an element a_ij = M_{ij �}

a₁₁ = 1, Minor of element a₁₁ = M₁₁ =  = (3 × 5) – (1 × 1) = 14

a₁₂ = -2, Minor of element a₁₂ = M₁₂ = = (-2 × 5) – (1 × 1) = -11

a₁₃ = 1, Minor of element a₁₃ = M₁₃ =  = (-2 × 1) – (3 × 1) = -5

a₂₁ = -2, Minor of element a₂₁ = M₂₁ = = ((-2) × 5) – (1 × 1) = -11

a₂₂ = 3, Minor of element a₂₂ = M₂₂ =  = (1 × 5) – (1 × 1) = 4

a₂₃ = 1, Minor of element a₂₃ = M₂₃ =  = (1 × 1) – ((-2) × 1) = 3

a₃₁ = 1, Minor of element a₃₁ = M₃₁ =  = (-2 × 1) – (3 × 1) = -5

a₃₂ = 1, Minor of element a₃₂ = M₃₂ =  = (1 × 1) – (1 × (-2)) = 3

a₃₃ = 5, Minor of element a₃₃ = M₃₃ = = (1 × 3) – ((-2) × (-2)) = -1

Cofactor of an element a_ij = A_ij

A₁₁ = (-1)¹⁺¹× 14 = 1 × 14 = 14

A₁₂ = (-1)¹⁺²× (-11) = (-1) × (-11) = 11

A₁₃ = (-1)¹⁺³× (-5) = 1 × (-5) = -5

A₂₁ = (-1)²⁺¹× (-11) = (-1) × (-11) = 11

A₂₂ = (-1)²⁺² × 4 = 1 × 4 = 4

A₂₃ = (-1)²⁺³ × 3 = (-1) × 3 = -3

A₃₁ = (-1)³⁺¹ × (-5) = 1 × (-5) = -5

A₃₂ = (-1)³⁺² × 3 = (-1) × 3 = -3

A₃₃ = (-1)³⁺³ × (-1) = 1 × (-1) = -1

Adj A =  = 

A^-1 = (Adj A)/|A|

A^-1 =  = 

(ii) To find (A^-1)^-1 we have to find out Adj(A^-1)

A^-1 = 

|A^-1| = (-1/13)³ [14 (4 × (-1) – (-3) × (-3)) – 11 (11 × (-1) – (-3) × (-5)) + (-5) (11 × (-3) – 4 × (-5))]

|A| = (-1/13)³ [14 (-4 – 9) – 11 (-11 – 15) – 5 (-33 + 20)]

|A| = (-1/13)³ [14 × (-13) – 11 × (-26) – 5 × (-13)]

|A| = (-1/13)³ × 169 = -1/13

Cofactor of an element a_ij = A_ij

A₁₁ = (-1)¹⁺¹× (-1/13) = 1 × (-1/13) = -1/13

A₁₂ = (-1)¹⁺²× (-2/13) = (-1) × (-2/13) = 2/13

A₁₃ = (-1)¹⁺³× (-1/13) = 1 × (-1/13) = -1/13

A₂₁ = (-1)²⁺¹× (-2/13) = (-1) × (-2/13) = 2/13

A₂₂ = (-1)²⁺² × (3/13) = 1 × (-3/13) = -3/13

A₂₃ = (-1)²⁺³ × (1/13) = (-1) × (1/13) = -1/13

A₃₁ = (-1)³⁺¹ × (-1/13) = 1 × (-1/13) = -1/13

A₃₂ = (-1)³⁺² × (1/13) = (-1) × 1/13 = -1/13

A₃₃ = (-1)³⁺³ × (-5/13) = 1 × (-5/13) = -5/13

Adj (A^-1) =  = 

(A^-1)^-1 = Adj(A^-1)/|A^-1|

(A^-1)^-1 =  =  = A

∴ (A^-1)^-1 = A

Question 10.

Evaluate 

Answer:

Let Δ = 

Applying Elementary Transformations.

Applying R₁→ R₁ + R₂ + R₃, we have

Δ = 

Δ = 2 (x + y)

Applying C₂→ C₂ – C₁ and C₃→ C₃ – C₁, we have

Δ = 2 (x + y)

Expanding along R₁, we have

Δ = 2 (x + y) [1 (x × (-x) – (-y) × (x – y)) – 0 + 0]

Δ = 2 (x + y) [-x² + y (x – y)]

Δ = 2 (x + y) [-x² + xy – y²]

Δ = -2 (x + y) [x² – xy + y²]

Δ = -2 (x³ + y³)

Question 11.

Evaluate 

Answer:

Let Δ = 

Applying Row transformations

R₂→ R₂ – R₁; R₃→ R₃ – R₁

Δ = 

Δ = 

Now, Expanding along C₁

Δ = 1 (y × x – 0 × 0) – 0 (x × x – 0 × y) + 0 (x × 0 – y × y)

Δ = 1 (xy) – 0 + 0

Δ = xy

Question 12.

Prove that 

Answer:

Let Δ = 

Applying Row Transformations

R₂→ R₂ – R₁

Δ = 

R₃→ R₃ – R₁

Δ = 

Taking (β – α)(γ – α) from R₂ and R₃ respectively

Δ = (β – α) (γ – α)

Applying R₃→ R₃ – R₂, we have

Δ = (β – α) (γ – α)

Expanding along R₃, we have

Δ = (β – α) (γ – α) [0 (α² × (-1) – (β + γ) × (β + γ) – (γ – β)((-1) × α – 1 × (β + γ) + 0 (α × (β + γ) – 1 × α²)

Δ = (β – α) (γ – α) [0 – (γ – β)( - α - β – γ) + 0]

Δ = (β – α) (γ – α) (γ – β) (α + β + γ)

Δ = (α – β) (β – γ) (γ – α) (α + β + γ)

Question 13.

Prove that  where p is any scalar.

Answer:

Let Δ = 

Applying Elementary Row Transformations

R₂→ R₂ – R₁ and R₃→ R₃ – R₁

Δ = 

Taking (y – x) and (z – x) common from R₂ and R₃ respectively

Δ = (y – x) (z – x)

Applying R₃→ R₃ – R₂

Δ = (y – x) (z – x)

Taking (z – y) common from R₃

Δ = (y – x) (z – x) (z – y)

Expanding along R₃, we have

Δ = (x – y) (y – z) (z – x) [0 – 1 {x × p(y² + x² + xy) – 1 × (1 + px³)} + p (x + y + z) {x × (y + x) – 1 × x²}

Δ = (x – y) (y – z) (z – x) (-px³ – pxy² – px²y + 1 + px³ + px²y + pxy² + pxyz)

Δ = (x – y) (y – z) (z – x) (1 + pxyz)

Hence, the given result is proved.

Question 14.

Prove that 

Answer:

Let Δ = 

Applying Elementary Transformations

C₁→ C₁ + C₂ + C₃, we have

Δ = 

Taking (a + b + c) common from C₁, we get

Δ = (a + b + c)

Applying R₂→ R₂ – R₁ and R₃→ R₃ – R₁

Expanding along C₁

Δ = (a + b + c) [1 × {(2b + a) (2c + a) – (a – b) (a – c)} – 0 + 0]

Δ = (a + b + c) [4bc + 2ab + 2ac + a² – a² + ac + ba – bc]

Δ = (a + b + c) (3ab + 3bc + 3ac)

Δ = 3 (a + b + c) (ab + bc + ca)

Hence, the given result proved.

Question 15.

Prove that 

Answer:

Let Δ = 

Applying Elementary Row Transformations

R₂→ R₂ – 2R₁

Δ = 

R₃→ R₃ – 3R₁

Δ = 

R₃→ R₃ – 3R₂

Δ = 

Expanding Along C₁, we have

Δ = 1 (1 × 1 – 0 × (2 + p)) – 0 + 0

Δ = 1 – 0

Δ = 1

Hence, the given result is proved

Question 16.

Prove that 

Answer:

Let Δ = 

Δ = 

Δ = 

Applying Elementary Column Transformations

C₁→ C₁ + C₃

Δ = 

Since, the two columns are identical

[In a determinant if two columns are identical the the value of determinant is 0]

So, the value of given determinant is 0

∴ Δ = 0

Hence, the given result is proved.

Question 17.

Solve the system of equations

Answer:

Given System of equations are

Let 

∴ Given system of equation becomes

2p + 3q + 10r = 4

4p – 6q + 5r = 1

6p + 9q – 20r = 2

The given System of Equations can be written in the form of AX = B

Here A =  and B = , X = 

Now we need to find |A|

∴ |A| = = 2 × (120 – 45) – 3 × (- 80 – 30) + 10 × (36 + 36)

= 150 + 330 + 720

= 1200

Since, |A| ≠ 0

∴ A is non-singular. So, it’s inverse exists

Cofactor of an element a_ij = A_ij

A₁₁ = (-1)¹⁺¹× 75 = 1 × 75 = 75

A₁₂ = (-1)¹⁺²× (-110) = (-1) × (-110) = 110

A₁₃ = (-1)¹⁺³× 72 = 1 × (72) = 72

A₂₁ = (-1)²⁺¹× (-150) = (-1) × (-150) = 150

A₂₂ = (-1)²⁺² × (-100) = 1 × (-100) = -100

A₂₃ = (-1)²⁺³ × 0 = (-1) × 0 = 0

A₃₁ = (-1)³⁺¹ × 75 = 1 × 75 = 75

A₃₂ = (-1)³⁺² × (-30) = (-1) × (-30) = 30

A₃₃ = (-1)³⁺³ × (-24) = 1 × (-24) = -24

Adj A =  = 

A^-1 = (Adj A)/|A|

A^-1 = 

Now,

Since, AX = B

∴ X = A^-1B

∴ p = �, q = 1/3, r = 1/5

∴ x = 1/p = 2; y = 1/q = 3; z = 1/r = 5

So, x = 2; y = 3; z = 5

Question 18.

If a, b, c, are in A.P, then the determinant  is
A. 0

B. 1

C. x

D. 2x

Answer:

Let Δ = 

Since, a, b, c are in A.P.

∴ 2b = a + c

Δ = 

Applying Elementary Row Transformations

R₁→ R₁ – R₂ and R₃→ R₃ – R₂

Δ = 

R₁→ R₁ + R₃, we have

Δ = 

[In a determinant if all elements of a row is 0 then the value of determinant is 0.]

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

∴ Δ = 0

Question 19.

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

B. 

C. 

D. 

Answer:

A = 

|A| = x × (y × z) = xyz

Since, |A| ≠ 0

A^-1 exists

To find the inverse of a matrix we need to find the Adjoint of that matrix

For finding the adjoint of the matrix we need to find its cofactors

Let A_ij denote the cofactors of Matrix A

Minor of an element a_ij = M_{ij �}

a₁₁ = x, Minor of element a₁₁ = M₁₁ =  = (y × z) – (0 × 0) = yz

a₁₂ = 0, Minor of element a₁₂ = M₁₂ = = (0 × z) – (0 × 0) = 0

a₁₃ = 0, Minor of element a₁₃ = M₁₃ =  = (0 × 0) – (0 × y) = 0

a₂₁ = 0, Minor of element a₂₁ = M₂₁ = = (0 × z) – (0 × 0) = 0

a₂₂ = y, Minor of element a₂₂ = M₂₂ =  = (x × z) – (0 × 0) = xz

a₂₃ = 0, Minor of element a₂₃ = M₂₃ =  = (x × 0) – (0 × 0) = 0

a₃₁ = 0, Minor of element a₃₁ = M₃₁ =  = (z × 0) – (0 × 0) = 0

a₃₂ = 0, Minor of element a₃₂ = M₃₂ =  = (x × 0) – (0 × 0) = 0

a₃₃ = z, Minor of element a₃₃ = M₃₃ = = (x × y) – (0 × 0) = xy

Cofactor of an element a_ij, A_ij = (-1)^i+j × M_ij

A₁₁ = (-1)¹⁺¹× M₁₁ = 1 × yz = yz

A₁₂ = (-1)¹⁺²× M₁₂ = (-1) × 0 = 0

A₁₃ = (-1)¹⁺³× M₁₃ = 1 × 0 = 0

A₂₁ = (-1)²⁺¹× M₂₁ = (-1) × 0 = 0

A₂₂ = (-1)²⁺² × M₂₂ = 1 × xz = xz

A₂₃ = (-1)²⁺³ × M₂₃ = (-1) × 0 = 0

A₃₁ = (-1)³⁺¹ × M₃₁ = 1 × 0 = 0

A₃₂ = (-1)³⁺² × M₃₂ = (-1) × 0 = 0

A₃₃ = (-1)³⁺³ × M₃₃ = 1 × xy = xy

Adj A =  = 

A^-1 = adj A / |A|

A^-1 = /xyz

A^-1 = 

A^-1 =  = 

The correct answer is A

Question 20.

Let , where 0 ≤ θ ≤ 2π. Then
A. Det (A) = 0

B. Det (A) є (2, ∞)

C. Det (A) є (2, 4)

D. Det (A) є [2, 4]

Answer:

A = 

|A| = 1 (1 × 1 – sin θ × (-sin θ)) – sin θ ((-sin θ) × 1 – (-1) × sin θ) + 1 ((-sin θ) × (-sin θ) – (-1) × 1)

|A| = 1 + sin²θ + sin²θ – sin²θ + sin²θ + 1

|A| = 2 + 2sin²θ

|A| = 2(1 + sin²θ)

Now, 0 ≤ θ ≤ 2π

⇒ sin 0 ≤ sin θ ≤ sin 2π

⇒ 0 ≤ sin²θ ≤ 1

⇒ 1 + 0 ≤ 1 + sin²θ ≤ 1 + 1

⇒ 2 ≤ 2(1 + sin²θ) ≤ 4

∴ Det (A) є [2, 4]