CHAPTER 3: DETERMINANTS

Transcription

1 CHAPER 3: DEERMINANS Previous Years Board Exam (Important Questions & Answers) MARKS WEIGHAGE 0 marks. Let A be a square matrix of order 3 3. Write the value of A, where A = 4. Since A = n A where n is order of matrix A. Here A = 4 and n = 3 A = 3 4 = 3. Write the value of the following determinant: Given that Appling R R 6R (Since R is zero) If A is a square matrix and A =, then write the value of AA', where A' is the transpose of matrix A. AA ' = A. A' = A. A = A = x = 4. [since, AB = A. B and A = A', where A and B are square matrices.] 4. If A is a 3 3 matrix, A 0 and 3A = k A, then write the value of k. Here, 3A = k A 3 3 A = k A [ ka = kn A where n is order of A] 7 A = k A k = 7 a ib c id 5. Evaluate: c id a ib a ib c id ( a ib)( a ib) ( c id)( c id) c id a ib ( a ib)( a ib) ( c id)( c id) a i b c i d a b c d x 3 6. If 3, find the value of x. x 5 4 Prepared b: M. S. KumarSwam, G(Maths) Page - -

2 x 3 Given that 3 x 5 4 4x + 8 3x 5 = 3 x 7 = 3 x = 0 7. If = , write the minor of the element a 3. 3 Minor of a 3 = 5 3 = 0 3 = Evaluate: cos5 sin 75 sin5 0 0 cos Expanding the determinant, we get cos 5. cos 75 - sin 5. sin 75 = cos ( ) = cos 90 = 0 [since cos (A + B) = cos A. cos B sina. sin B] 9. Using properties of determinants, prove the following: a a b a b Let a b a a b a b a b a Appling R R + R + R 3, we have 3( a b) 3( a b) 3( a b) a b a a b a b a b a aking out 3(a + b) from st row, we have 3( a b) a b a a b a b a b a a a b a b a b a a b b a b 9 ( ) a b a b a Appling C C C and C C C ( a b) b b a b b b a Expanding along first row, we have D = 3(a + b) [. (4b b )] = 3 (a + b) 3b = 9b (a + b) Prepared b: M. S. KumarSwam, G(Maths) Page - -

3 0. Write the value of the determinant Given determinant A = x = 0 ( R = R 3 ) 3 4 6x 9x x x 9x x. wo schools P and Q want to award their selected students on the values of olerance, Kindness and Leadership. he school P wants to award Rs. x each, Rs. each and Rs. z each for the three respective values to 3, and students respectivel with a total award mone of Rs.,00. School Q wants to spend Rs. 3,00 to award its 4, and 3 students on the respective values (b giving the same award mone to the three values as school P). If the total amount of award for one prize on each value is Rs.,00, using matrices, find the award mone for each value. Apart from these three values, suggest one more value that should be considered for award. According to question, 3x + + z = 00 4x + + 3z = 300 x + + z = 00 he above sstem of equation ma be written in matrix form as AX = B 3 x 00 A 4 3, X and B 300 z 00 3 Here, A 4 3 3( 3) (4 3) (4 ) A exists. Now, A = ( 3) =, A = (4 3) =, A 3 = (4 ) = 3, A = ( ) =, A = (3 ) =, A 3 = (3 ) = A 3 = (6 ) = 5, A 3 = (9 4) = 5, A 33 = (3 8) = adj( A) A ( adja) 5 5 A Prepared b: M. S. KumarSwam, G(Maths) Page - 3 -

4 x 5 00 Now, X A B z x z x = 300, = 400, z = 500 i.e., Rs. 300 for tolerance, Rs. 400 for kindness and Rs. 500 for leadership are awarded. One more value like punctualit, honest etc ma be awarded.. Using properties of determinants, prove that a x z LHS x a z x a z Appling C C + C + C 3, we get a x z z a x z a z a x z a z Appl R R R, we get z ( a x z) a z a z 0 a 0 ( a x z) a z a z Expanding along R, we get = (a + x + + z) {0 + a (a + z z)} = a (a + x + + z) = RHS 3. 0 students were selected from a school on the basis of values for giving awards and were divided into three groups. he first group comprises hard workers, the second group has honest and law abiding students and the third group contains vigilant and obedient students. Double the number of students of the first group added to the number in the second group gives 3, while the combined strength of first and second group is four times that of the third group. Using matrix method, find the number of students in each group. Apart from the values, hard work, honest and respect for law, vigilance and obedience, suggest one more value, which in our opinion, the school should consider for awards. Let no. of students in Ist, nd and 3rd group to x,, z respectivel. From the statement we have x + + z = 0 x + =3 x + 4z = 0 he above sstem of linear equations ma be written in matrix form as AX = B where x 0 A 0, X and B 3 4 z 0 Prepared b: M. S. KumarSwam, G(Maths) Page - 4 -

5 Here, A 0 ( 4 0) ( 8 0) ( ) A exists. Now, A = 4 0 = 4 A = ( 8 0) = 8 A 3 = = A = ( 4 ) = 5 A = 4 = 5 A 3 = ( ) = 0 A 3 = 0 = A 3 = (0 ) = A 33 = = adj( A) A ( adja) 8 5 A 5 0 x Now, X A B z 0 0 x z 0 0 x = 5, = 3, z = 4. he management committee of a residential colon decided to award some of its members (sa x) for honest, some (sa ) for helping others and some others (sa z) for supervising the workers to keep the colon neat and clean. he sum of all the awardees is. hree times the sum of awardees for cooperation and supervision added to two times the number of awardees for honest is 33. If the sum of the number of awardees for honest and supervision is twice the number of awardees for helping others, using matrix method, find the number of awardees of each categor. Apart from these values, namel, honest, cooperation and supervision, suggest one more value which the management of the colon must include for awards. According to question x + + z = x z = 33 x + z = 0 he above sstem of linear equation can be written in matrix form as AX = B where x A 3 3, X and B 33 z 0 Here, A 3 3 Prepared b: M. S. KumarSwam, G(Maths) Page - 5 -

6 = (3 + 6) ( 3) + ( 4 3) = = 3 A exists. A = 9, A =, A 3 = 7 A = 3, A = 0, A 3 = 3 A 3 = 0, A 3 =, A 33 = adj( A) A ( adja) 0 A x Now, X A B z x z x = 3, = 4, z = 5 No. of awards for honest = 3 No. of awards for helping others = 4 No. of awards for supervising = 5. he persons, who work in the field of health and hgiene should also be awarded. 5. Using properties of determinants, prove the following: 3x x x z x 3 z 3( x z)( x z zx) x z z 3z 3x x x z LHS x 3 z x z z 3z Appling C C + C + C 3 x z x x z x z 3 z x z z 3z aking out (x + + z) along C, we get x x z ( x z) 3 z z 3z Appling R R R ; R 3 R 3 R x x z ( x z) 0 x x 0 x z x z Appling C C C 3 Prepared b: M. S. KumarSwam, G(Maths) Page - 6 -

7 z x z ( x z) 0 3 x 0 3z x z Expanding along I column, we get D = (x + + z)[(3 (x + z) + 3z (x )] = 3(x + + z)[x + z + z + xz z] = 3(x + + z)(x + z + zx) = R.H.S. 6. A school wants to award its students for the values of Honest, Regularit and Hard work with a total cash award of Rs. 6,000. hree times the award mone for Hardwork added to that given for honest amounts to `,000. he award mone given for Honest and Hardwork together is double the one given for Regularit. Represent the above situation algebraicall and find the award mone for each value, using matrix method. Apart from these values, namel, Honest, Regularit and Hardwork, suggest one more value which the school must include for awards. Let x, and z be the awarded mone for honest, Regularit and hardwork. From the statement x + +z = 6000 (i) x + 3z =000 (ii) x +z = x +z = 0 (iii) he above sstem of three equations ma be written in matrix form as AX = B, where x 6000 A 0 3, X and B z 0 Here, A 0 3 (0 6) ( 3) ( 0) Hence A exist If Aij is co-factor of aij then A = = 6 A = ( 3) = ; A 3 = ( 0) = A = ( + ) = 3 A = 0 A 3 = ( ) = 3 A 3 = 3 0 = 3 A 3 = (3 ) = ; A 33 = 0 = adj( A) A ( adja) 0 A 6 3 x Now, X A B z 3 0 Prepared b: M. S. KumarSwam, G(Maths) Page - 7 -

8 x z x =500, = 000, z = 3500 Except above three values, school must include discipline for award as discipline has great importance in student s life. x x 4 7. If, then write the value of x. x 3 x 3 x x 4 Given that x 3 x 3 (x +) (x + ) (x )(x 3) = + x + x + x + x + 3x + x 3 =3 7x =3 7x =4 x = 8. Using properties of determinants, prove that a a b a b c LHS a 3a b 4a 3b c 3a 6a 3b 0a 6b 3c a a a b c a b a b c a 3a 4a 3b c a b 4a 3b c 3a 6a 0a 6b 3c 3a 3b 0a 6b 3c a b c a b c a 3 4a 3b c ab 4a 3b c 3 6 0a 6b 3c 3 3 0a 6b 3c a b c a 3 4a 3b c ab a 6b 3c a b c a 3 4a 3b c 3 6 0a 6b 3c a b c a 3 4a 3 3b 3 c 3 6 0a 3 6 6b 3 6 3c a. a 3 4 b 3 3 c a a b a b c a 3a b 4a 3b c a 3a 6a 3b 0a 6b 3c [since C = C in second determinant] Prepared b: M. S. KumarSwam, G(Maths) Page - 8-3

9 a. a 3 4 b.0 c.0 [since C = C 3 in second determinant and C = C 3 in third determinant] a Appling C C C and C 3 C 3 C we get 0 0 a Expanding along R we get = a 3.(7 6) = a Using matrices, solve the following sstem of equations: x + z = 4; x + 3z = 0; x + + z = Given equations x + z = 4 x + 3z = 0 x + + z = We can write this sstem of equations as AX = B where x 4 A 3, X and B 0 z Here, A 3 = ( + 3) - (- ) ( + 3) + ( - ) = = 0 A exists. A = 4, A = 5, A 3 = A =, A = 0, A 3 = A 3 =, A 3 = 5, A 33 = adj( A) A ( adja) A 0 3 x 4 4 Now, X A B z 3 Prepared b: M. S. KumarSwam, G(Maths) Page - 9 -

10 x z he required solution is x =, = -, z = 3 0. If A = and B = 3 0, find (AB). 5 0 For B B i.e., B is invertible matrix B exist. A = 3, A =, A 3 = A =, A =, A 3 = A 3 = 6, A 3 =, A 33 = 5 = (3 0) ( 0) ( 0 ) = = adj( B) B ( adjb) B 5 5 Now (AB) = B. A Prepared b: M. S. KumarSwam, G(Maths) Page - 0 -