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Class 12 Maths NCERT Solutions for Chapter 4 Determinants Miscellaneous Exercise

Class 12 Maths NCERT Solutions for Chapter 4 Determinants Miscellaneous Exercise

Determinants Miscellaneous Exercise Solutions

1. Prove that the determinant  is independent of θ is independent of θ. 

Solution

Δ = 

= x(–x2 – 1) – sinθ(– x sinθ – cos θ) + cos θ(– sinθ + x cosθ)
= –x3 – x + xsin2 θ + sinθ cosθ – sinθ cosθ + x cos2θ
= –x3 – x + x(sin2θ + cos2θ)
= –x3 – x + x
= –x3 (Independent of θ)
Hence, Δ is independent of θ.


2. Without expanding the determinant, prove that

Solution

Hence, the given result is proved.


3. Evaluate 

Solution  Δ = 
Expanding along C3 , we have :
∆ = − sin α(−sin α sin2β – cos2 β sinα) + cos α(cos α cos2 β + cos α sin2 β)
= sin2 α(sin2 β + cos2 β) + cos2 α(cos2 β + sin2 β)
= sin2 α(1) + cos2 α(1)
= 1

4. If a, b and c are real numbers, and triangle  = Show that either a + b + c = 0 or a = b = c. 
Solution
Expanding along R1 , we have : 
Δ = 2 (a + b + c)(1)[(b - c)(c - b) - (b - a)(c - a)] 
= 2(a + b + c)[-b2 - c2 + 2bc - bc + ba + ac - a2 ]
= 2(a + b + c)[ab + bc +  ca - a2  - b2  - c2 ]
It is given that Δ = 0. 
(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 
⇒ -2ab - 2bc - 2ca + 2a2 + 2b2 + 2c2  = 0 
⇒ (a - b)2 + (b - c)2 + (c - a)2 = 0 
⇒ (a - b)2 = (b - c)2 = (c - a)2  = 0 [(a-b)2, (b-c)2, (c-a)2 are non-negative] 
⇒ (a - b) = (b - c) = (c - a) = 0 
⇒ a = b = c
Hence, if Δ = 0 then either a + b + c = 0 or a = b = c. 

5. Solve the equations = 0, a ≠ 0 
Solution

Expanding along R1 , we have : 
(3x + a)[1 × a2 ]= 0 
⇒ a2 (3x + a) = 0 
But a ≠ 0. 
Therefore, we have : 
3x + a = 0 
⇒ x = -a/3 

6. Prove that  = 4a2 b2 c2 
Solution

Expanding along R3 , we have : 
Δ = 2ab2 c[a(c - a) + a(a + c)]
= 2ab2 c[ac - a2 + a2 + ac]
= 2ab2 c(2ac)
= 4a2 b2 c2 
Hence, the given result is proved.

7. If A ^(-1) = find (AB)-1 .
Solution
We know that (AB)-1 = B-1 A-1 .

∴ |B| = 1 × 3 - 2× (-1) - 2(2)
= 3 + 2 - 4 
= 5 - 4 = 1 
Now, A11 = 3, A12 = 1, A13 = 2 
A21 = 2, A22 = 1, A23 = 2 
A31 = 6, A32 = 2, A33 = 5 

 
8. Let A = verify that 
(i) [adj A]-1 = adj(A-1
(ii) (A-1)-1 = A 
Solution
A = 
∴ |A| = 1 (15 - 1) + 2(-10 - 1) + 1(-2 - 3) = 14 - 22 - 5= -13 
Now,  A11 = 14, A12 = 11, A13 = -5
A21 = 11, A22 = 4, A23 = -3 
A31 = -5, A32 = -3, A13 = -1

(i) |adj A| = 14(-4 - 9) - 11(-11 - 15) - 5(-33 + 20)
= 14(-13) - 11(-26) - 5(-13) 
= -182 + 286 + 65 = 169 
we have , 

(ii) We have show that : 

9. Evaluate 
Solution

10. Evaluate  
Solution

Expanding along C1 , we have : 
Δ = 1(xy - 0) = xy

11. Using properties of determinants, prove that : 
= (β - γ) (γ - α)(α - β) (α + β + γ)
Solution

Expanding along R3 , we have : 
Δ =  (β - α) (γ - α)[- (γ - β)( - α - β - γ)]
= (β - α) (γ - α) (γ - β)(α + β + γ)
= (α - β) (β - γ) (γ - α)(α + β + γ)
Hence, the given result is proved. 

12. Using properties of determinants, prove that : 
 = (1 + pxyz)(x - y)(y - z)(z -x), where p is any scalar. 
Solution

13. Using properties of determinants, prove that : 
= 3(a + b + c)(ab + bc + ca)
Solution
Expanding along C1 , we have : 
Δ = (a + b + c) [(2b + a)(2c + a) - (a - b)(a - c)]
= (a + b +c)[4abc + 2ab + 2ac + a2 - a2 + ac + ba - bc]
= (a + b + c)(3ab + 3bc + 3ac)
= 3(a + b + c)(ab + bc + ca)
Hence, the given result is proved. 

14. Using properties of determinants, prove that : 
Solution

15 . Solve the system of the following equations 
2/x + 3/y + 10/z = 4 
4/x - 6/y + 5/z = 1 
6/x + 9/y - 20/x = 2 
Solution
Let 1/x = p, 1/y = q, 1/z = r. 
Then the given system of equations is as follows : 
2p + 3q + 10r = 4 
4p - 6q + 5r = 1 
6p + 9q - 20r = 2 
This system can be written in the form of AX = B, where 

Now, 
|A| = 2(120 - 45) - 3 (-80 - 30) + 10(36 + 36)
= 150 + 330 + 720 
=1200 
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 
∴ A-1 = 1/|A| [adj A]

∴ p = 1/2, q = 1/3, and r = 1/5 
Hence, x = 2, y = 3, and z = 5. 

16. Choose the correct answer. 
If a, b, c, are in A.P., then the determinant 

A. 0
B. 1 
C. x 
D. 2x 
Solution 

Here, all the elements of the first row (R1 ) are zero. 
Hence, we have Δ = 0.
The correct answer is A.

18. If x, y, z are nonzero real numbers, then the inverse of matrix A =  is 
Solution
A = 
∴ |A| = x(yz - 0) = xyz ≠ 0 
Now,  A11 = yz, A12 = 0, A13 = 0 
A21 = 0, A22 = xz, A23 = 0 
A31 = 0, A32 = 0, A33 = xy

The correct answer is A. 

19. Choose the correct answer . 
Let A =  where 0 ≤ θ ≤ 2π, then 

A. Det (A) = 0
B. Det (A) ∈ (2, ∞)
C. Det (A) ∈ (2, 4)
D. Det (A)∈ [2, 4]

Solution

A = 
∴ |A| = 1(1 + sin2 θ) - sinθ( - sinθ + sinθ) + 1 (sin2 θ + 1)
= 1 + sin2 θ + sin2 θ + 1 
= 2 + 2sin2 θ
= 2(1 + sin2 θ) 
Now, 0 ≤ θ ≤ 2π 
⇒ 0 ≤ sinθ  ≤ 1 
⇒ 0 ≤ sin2 θ  ≤ 1 
⇒ 1 ≤ 1 + sin2 θ  ≤ 2
⇒ 2 ≤ 2(1 + sin2 θ)  ≤ 2
∴ Det (A) ∊ [2, 4]
The correct answer is D.

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