Chapter 4 – Determinants

Class 12 Mathematics · 54 questions · 0 with answers

Solved examples

example-1Short answer

If = , then find x.

example-2Short answer

If ∆ = 1 y y , ∆1 = yz zx xy , then prove that ∆ + ∆1 = 0. 1 z z2 x y z 1 1 1 Solution We have ∆1 = yz zx xy x y z Interchanging rows and columns, we get 1 yz x x xyz x2 ∆1 = 1 zx y = y xyz y2 1 xy z z xyz z2 70 MATHEMATICS x 1 x2 y 1 y2 = xyz Interchanging C1 and C2 z 1 z2 1 x x2 (–1) 1 y y2 = – ∆ = 1 z z2 ⇒ ∆1 + ∆ = 0

example-3Short answer

Without expanding, show that cosec 2θ cot 2θ 1 ∆ = cot θ 2 cosec θ −1 = 0. 42 40 2 Solution Applying C1 → C1 – C2 – C3, we have cosec 2θ – cot 2θ – 1 cot 2θ 1 0 cot 2 θ 1 ∆ = cot θ – cosec θ + 1 cosec θ −1 2 2 2 0 cosec θ −1 = 0 = 0 40 2 0 40 2 x p q

example-4Short answer

Show that ∆ = p x q = (x – p) (x2 + px – 2q2) q q x Solution Applying C1 → C1 – C2, we have x− p p q 1 p q ∆= p − x x q = ( x − p) −1 x q 0 q x 0 q x DETERMINANTS 71 0 p + x 2q = ( x − p) −1 x q Applying R1 → R1 + R2 0 q x Expanding along C1, we have ∆ = ( x − p) ( px + x 2 − 2q 2 ) = ( x − p ) ( x 2 + px − 2q 2 ) 0 b−a c−a

example-5Short answer

If ∆ = a − b 0 c − b , then show that ∆ is equal to zero. a−c b−c 0 0 a −b a −c Solution Interchanging rows and columns, we get ∆ = b − a 0 b−c c − a c −b 0 Taking ‘–1’ common from R1, R2 and R3, we get 0 b−a c−a ∆ = (–1) a − b 0 c −b = – ∆ a−c b−c 0 ⇒ 2∆ = 0 or ∆ =0

example-6Short answer

Prove that (A–1)′ = (A′)–1, where A is an invertible matrix. Solution Since A is an invertible matrix, so it is non-singular. We know that |A| = |A′|. But |A| ≠ 0. So |A′| ≠ 0 i.e. A′ is invertible matrix. Now we know that AA–1 = A–1 A = I. Taking transpose on both sides, we get (A–1)′ A′ = A′ (A–1)′ = (I)′ = I Hence (A–1)′ is inverse of A′, i.e., (A′)–1 = (A–1)′

example-7Long answer

If x = – 4 is a root of ∆ = 1 x 1 = 0, then find the other two roots. 3 2 x 72 MATHEMATICS Solution Applying R1 → (R1 + R2 + R3), we get x+4 x+4 x+4 1 x 1 . 3 2 x Taking (x + 4) common from R1, we get 1 1 1 ∆ = ( x + 4) 1 x 1 3 2 x Applying C2 → C2 – C1, C3 → C3 – C1, we get 1 0 0 ∆ = ( x + 4) 1 x − 1 0 . 3 −1 x − 3 Expanding along R1, ∆ = (x + 4) [(x – 1) (x – 3) – 0]. Thus, ∆ = 0 implies x = – 4, 1, 3

example-8Long answer

In a triangle ABC, if 1 1 1 1 + sin A 1 + sin B 1 + sin C =0 , 2 2 2 sinA +sin A sinB+sin B sinC+sin C then prove that ∆ABC is an isoceles triangle. 1 1 1 1 + sin A 1 + sin B 1 + sin C Solution Let ∆ = sinA +sin 2 A sinB+sin 2 B sinC+sin 2 C DETERMINANTS 73 1 1 1 1 + sin A 1 + sin B 1 + sin C = R3 → R3 – R2 − cos 2 A − cos 2 B − cos 2 C 1 0 0 1 + sin A sin B − sin A sin C − sin B = . (C3 → C3 – C2 and C2 → C2 – C1) − cos 2 A cos 2 A − cos 2 B cos 2 B − cos 2 C Expanding along R1, we get ∆ = (sinB – sinA) (sin2C – sin2B) – (sinC – sin B) (sin2B – sin2A) = (sinB – sinA) (sinC – sinB) (sinC – sin A) = 0 ⇒ either sinB – sinA = 0 or sinC – sinB or sinC – sinA = 0 ⇒ A = B or B = C or C = A i.e. triangle ABC is isoceles. 3 −2 sin 3θ

example-9Long answer

Show that if the determinant ∆ = −7 8 cos 2θ = 0 , then sinθ = 0 or . −11 14 2 Solution Applying R2 → R2 + 4R1 and R3 → R3 + 7R1, we get 3 −2 sin 3θ 5 0 cos 2θ + 4sin 3θ = 0 10 0 2 + 7sin3θ or 2 [5 (2 + 7 sin3θ) – 10 (cos2θ + 4sin3θ)] = 0 or 2 + 7sin3θ – 2cos2θ – 8sin3θ = 0 or 2 – 2cos 2θ – sin 3θ = 0 sinθ (4sin2θ + 4sinθ – 3) = 0 74 MATHEMATICS or sinθ = 0 or (2sinθ – 1) = 0 or (2sinθ + 3) = 0 or sinθ = 0 or sinθ = (Why ?).

example-10Multiple choice

Let ∆ = By y 2 1 and ∆1 = x y z , then Cz z2 1 zy zx xy

(A)∆1 = – ∆
(B)∆ ≠ ∆1
(C)∆ – ∆1 = 0
(D)None of these A B C A x yz Solution (C) is the correct answer since ∆1 = x y z =B y zx zy zx xy C z xy Ax x2 xyz Ax x2 1 1 By y2 1 = By y2 xyz = xyz =∆ xyz 2 Cz z2 1 Cz z xyz cos x − sin x 1

example-11Multiple choice

If x, y ∈ R, then the determinant ∆ = sin x cos x 1 lies cos( x + y ) − sin( x + y ) 0 in the interval

(A) − 2, 2 
(B)[–1, 1]
(C) − 2,1
(D) −1, − 2, Solution The correct choice is A. Indeed applying R3→ R3 – cosyR1 + sinyR2, we get DETERMINANTS 75 cos x − sin x 1 ∆ = sin x cos x 1 . 0 0 sin y − cos y Expanding along R3, we have ∆ = (siny – cosy) (cos2x + sin2x)  1 1  = (siny – cosy) = 2 sin y − cos y   2 2   π π  π = 2 cos sin y − sin cos y  = 2 sin (y – 4 )  4 4  Hence – 2 ≤∆≤ 2. Fill in the blanks in each of the Examples 12 to 14.

example-12Short answer

If A, B, C are the angles of a triangle, then sin 2 A cotA 1 ∆ = sin 2 B cotB 1 = ................ sin C cotC 1 Solution Answer is 0. Apply R2 → R2 – R1, R3 → R3 – R1. 23 + 3 5 5

example-13Short answer

The determinant ∆ = 15 + 46 5 10 is equal to ............... 3 + 115 15 5 Solution Answer is 0.Taking 5 common from C 2 and C 3 and applying C1 → C3 – 3 C2, we get the desired result.

example-14True / False

The value of the determinant 76 MATHEMATICS sin 2 23° sin 2 67° cos180° ∆ = − sin 67° − sin 23° cos 2 180° = .......... 2 2 cos180° sin 2 23° sin 2 67° Solution ∆ = 0. Apply C1 → C1 + C2 + C3. State whether the statements in the Examples 15 to 18 is True or False.

example-15Short answer

The determinant cos ( x + y ) − sin ( x + y ) cos 2 y ∆ = sin x cos x sin y − cos x sin x cos y is independent of x only. Solution True. Apply R1 → R1 + sinyR2 + cosy R3, and expand

example-16Short answer

The value of 1 1 1 n n+2 n+4 C1 C1 C1 is 8. n n+2 n+4 C2 C2 C2 Solution True  x 5 2

example-17Short answer

If A =  2 y 3 , xyz = 80, 3x + 2y + 10z = 20, then  1 1 z  81 0 0  A adj. A =  0 81 0  .  0 0 81

Show solution

Solution : False. DETERMINANTS 77  1 5   2 −4 0 1 3 2    A = 1 2 x  , A –1 =  − 1 3 − 

example-18Short answer

If    2 2  2 3 1     1 y 1   2 2  then x = 1, y = – 1. Solution True

Questions

Q8Short answer

x 8 3 2x 5 6 5 Solution We have = . This gives 8 x 8 3 2x2 – 40 = 18 – 40 ⇒ x2 = 9 ⇒ x = ± 3. 1 x x2 1 1 1

Q1Short answer

2. x +1 x +1 x y a+ z 0 xy 2 xz 2 3x − x + y − x + z x− y z−y

Q2Short answer

2 x y 0 yz 3y

Q3Short answer

4. x z zy 2 0 x−z y−z 3z x+4 x x a−b−c 2a 2a

Q5Short answer

x x+4 x 6. 2b b−c−a 2b x x x+4 2c 2c c−a−b Using the proprties of determinants in Exercises 7 to 9, prove that: y2 z2 yz y+z y+z z y 2 2 z x zx z+x = 0 z z+x x = 4 xyz

Q7Short answer

8. x y 2 xy x+ y y x x+ y 78 MATHEMATICS a 2 + 2a 2a + 1 1 2a + 1 a + 2 1 = (a − 1)3 9. 3 3 1 1 cos C cos B

Q10Short answer

If A + B + C = 0, then prove that cos C 1 cos A = 0 cos B cos A 1

Q11Short answer

If the co-ordinates of the vertices of an equilateral triangle with sides of length x1 y1 1 3a 4 ‘a’ are (x1, y1), (x2, y2), (x3, y3), then x2 y2 1 = 4 . x3 y3 1  1 1 sin 3θ 

Q12Short answer

Find the value of θ satisfying  −4 3 cos 2θ  = 0 .    7 −7 −2  4 − x 4 + x 4 + x

Q13Short answer

If  4 + x 4 − x 4 + x  = 0 , then find values of x.  4 + x 4 + x 4 − x 

Q14Short answer

If a 1 , a 2 , a 3 , ..., a r are in G.P., then prove that the determinant ar +1 ar + 5 ar + 9 ar + 7 ar +11 ar +15 is independent of r. ar +11 ar +17 ar + 21

Q15Short answer

Show that the points (a + 5, a – 4), (a – 2, a + 3) and (a, a) do not lie on a straight line for any value of a.

Q16Short answer

Show that the ∆ABC is an isosceles triangle if the determinant DETERMINANTS 79  1 1 1    ∆ =  1 + cos A 1 + cos B 1 + cos C  = 0 . cos 2 A + cos A cos 2 B + cos B cos 2 C + cos C    0 1 1 A 2 − 3I Find A if A = 1 0 1 and show that A = –1 –1

Q17Short answer

. 1 1 0

Q18Long answer

If A =  −2 −1 −2  , find A–1.  0 −1 1  Using A –1 , solve the system of linear equations x – 2y = 10 , 2x – y – z = 8 , –2y + z = 7.

Q19Long answer

Using matrix method, solve the system of equations 3x + 2y – 2z = 3, x + 2y + 3z = 6, 2x – y + z = 2 .  2 2 −4 1 −1 0

Q20Long answer

Given A =  −4 2 −4 , B =  2 3 4 , find BA and use this to solve the    2 −1 5   0 1 2 system of equations y + 2z = 7, x – y = 3, 2x + 3y + 4z = 17.

Q21Long answer

If a + b + c ≠ 0 and b c a = 0 , then prove that a = b = c. bc − a 2 ca − b 2 ab − c 2

Q22Long answer

Prove that ca − b 2 ab − c 2 bc − a 2 is divisible by a + b + c and find the ab − c 2 bc − a 2 ca − b 2 quotient. 80 MATHEMATICS xa yb zc a b c

Q23Long answer

If x + y + z = 0, prove that yc za xb = xyz c a b zb xc ya b c a

Q24Multiple choice

If = , then value of x is 8 x 7 3

(A)3
(B)±3
(C)±6
(D)6 a −b b + c a b−a c+a b

Q25Multiple choice

The value of determinant c−a a +b c

(A)a3 + b3 + c3
(B)3 bc
(C)a3 + b3 + c3 – 3abc
(D)none of these

Q26Multiple choice

The area of a triangle with vertices (–3, 0), (3, 0) and (0, k) is 9 sq. units. The value of k will be

(A)9
(B)3
(C)–9
(D)6 b 2 − ab b − c bc − ac The determinant ab − a a − b b 2 − ab equals 27. bc − ac c − a ab − a 2 (A) abc (b–c) (c – a) (a – b) (B) (b–c) (c – a) (a – b) (C) (a + b + c) (b – c) (c – a) (a – b) (D) None of these DETERMINANTS 81 sin x cos x cos x

Q28Multiple choice

The number of distinct real roots of cos x sin x cos x = 0 in the interval cos x cos x sin x π π − ≤ x ≤ is 4 4

(A)0
(B)2
(C)1
(D)3

Q29Multiple choice

If A, B and C are angles of a triangle, then the determinant −1 cos C cos B cos C −1 cos A is equal to cos B cos A −1

(A)0
(B)–1
(C)1
(D)None of these cos t t 1 f (t )

Q30Multiple choice

Let f (t) = 2sin t t 2t , then lim 2 is equal to t →0 t sin t t t

(A)0
(B)–1
(C)2
(D)3 1 1 1

Q31Multiple choice

The maximum value of ∆ = 1 1 + sin θ 1 is (θ is real number) 1 + cos θ 1 1 1 3

(A)
(B)2 2 2 3
(C)2
(D)82 MATHEMATICS 0 x−a x−b

Q32Multiple choice

If f (x) = x + a 0 x − c , then x+b x+c 0

(A)f (a) = 0
(B)f (b) = 0
(C)f (0) = 0
(D)f (1) = 0  2 λ −3  

Q33Multiple choice

If A =  0 2 5  , then A–1 exists if 1 1 3 

(A)λ=2
(B)λ≠ 2
(C)λ≠–2
(D)None of these

Q34Multiple choice

If A and B are invertible matrices, then which of the following is not correct?

(A)adj A = |A|. A–1
(B)det(A)–1 = [det (A)]–1
(C)(AB)–1 = B–1 A–1
(D)(A + B)–1 = B–1 + A–1 1+ x 1 1

Q35Multiple choice

If x, y, z are all different from zero and 1 1+ y 1 = 0 , then value of 1 1 1+ z x–1 + y–1 + z–1 is

(A)xyz
(B)x–1 y–1 z–1
(C)–x –y –z
(D)–1 x x+ y x+ 2y

Q36Multiple choice

The value of the determinant x + 2 y x x + y is x+ y x+ 2y x

(A)9x2 (x + y)
(B)9y2 (x + y)
(C)3y2 (x + y)
(D)7x2 (x + y) DETERMINANTS 83 1 –2 5

Q37Multiple choice

There are two values of a which makes determinant, ∆ = 2 a −1 = 86, then 0 4 2a sum of these number is

(A)4
(B)5
(C)–4
(D)9 Fill in the blanks

Q38Fill in the blanks

If A is a matrix of order 3 × 3, then |3A| = _______ .

Q39Fill in the blanks

If A is invertible matrix of order 3 × 3, then |A–1 | _______ . (2 + 2 ) (2 − 2 ) x –x 2 x –x 2 If x, y, z ∈ R, then the value of determinant ( 3 + 3 ) (3 − 3 ) x –x 2 x –x 2

Q40Fill in the blanks

1 is (4 + 4 ) (4 − 4 ) x –x 2 x –x 2 equal to _______. 0 cos θ sin θ

Q41Fill in the blanks

If cos2θ = 0, then cos θ sin θ 0 = _________. sin θ 0 cos θ

Q42Fill in the blanks

If A is a matrix of order 3 × 3, then (A2)–1 = ________.

Q43Fill in the blanks

If A is a matrix of order 3 × 3, then number of minors in determinant of A are ________.

Q44Fill in the blanks

The sum of the products of elements of any row with the co-factors of corresponding elements is equal to _________. x 3 7

Q45Fill in the blanks

If x = – 9 is a root of 2 x 2 = 0, then other two roots are __________. 7 6 x 0 xyz x−z

Q46Fill in the blanks

y−x 0 y−z = __________. z−x z− y 0 84 MATHEMATICS (1 + x)17 (1 + x)19 (1 + x) 23

Q47Fill in the blanks

If f (x) = (1 + x) 23 (1 + x) 29 (1 + x)34 = A + Bx + Cx 2 + ..., then (1 + x) 41 (1 + x) 43 (1 + x) 47 A = ________. State True or False for the statements of the following Exercises:

Q48Multiple choice

( A ) = ( A ) , where A is a square matrix and |A| ≠ 0. 3 –1 −1 3 1 –1

Q49Multiple choice

(aA)–1 = A , where a is any real number and A is a square matrix.

Q50Multiple choice

|A–1| ≠ |A|–1 , where A is non-singular matrix.

Q51Multiple choice

If A and B are matrices of order 3 and |A| = 5, |B| = 3, then |3AB| = 27 × 5 × 3 = 405.

Q52Multiple choice

If the value of a third order determinant is 12, then the value of the determinant formed by replacing each element by its co-factor will be 144. x +1 x + 2 x + a

Q53Multiple choice

x + 2 x + 3 x + b = 0 , where a, b, c are in A.P.. x + 3 x + 4 x+c

Q54Multiple choice

|adj. A| = |A|2 , where A is a square matrix of order two. sin A cos A sin A + cos B

Q55Multiple choice

The determinant sin B cos A sin B+ cos B is equal to zero. sin C cos A sin C + cos B x+a p+ u l + f

Q56Multiple choice

If the determinant y + b q + v m + g splits into exactly K determinants of z+c r +w n+h order 3, each element of which contains only one term, then the value of K is 8. DETERMINANTS 85 a p x p+ x a+x a+ p

Q57Multiple choice

Let ∆ = b q y = 16 , then ∆1 = q + y b + y b + q = 32 . c r z r+ z c+ z c+r 1 1 1

Q58Multiple choice

The maximum value of 1 (1+ sin θ) 1 is . 1 1 1 + cos θ