# Gate maths

date post

15-Aug-2015Category

## Engineering

view

62download

6

Embed Size (px)

### Transcript of Gate maths

- 1. 3. Eigen Values and Eigen Vectors EC All GATE Questions 1. Given the matrix 4 2 , 4 3 the eigenvector is [EC: GATE-2005] (a) 3 2 (b) 4 3 (c) 2 1 (d) 2 1 1. (c) Characteristic equation 2A I 0 = 4 2 0 4 3 = 5,4 = Take 5, then AX X becomes = = = 1 1 2 2 x 5x4 2 4 3 x 5x 1 2 1 1 2 2 4x 2x 5x 4x 3x 5x + = =+ + = = + = 1 2 1 1 2 1 2 2 4x 2x 5x x 2x 4x 3x 5x 2 1if x 1 then x 2 = = 2 is eigen vector corrosponding to 5. 1 = 2. The eigen values and the corresponding eigen vectors of a 2 2 matrix are given by [EC: GATE-2006] Eigenvalue Eigenvector 1 = 8 1v = 1 1 2 = 4 2v = 1 1 The matrix is 20/02/2014 sreekantha.292@gmail.com Page 1 of 177 During the moment of darkness only the stars will be born.
- 2. (a) 6 2 2 6 (b) 4 6 6 4 (c) 2 4 4 2 (d) 4 8 8 4 2. (a) We know, sum of eigen values = trace (A). = Sum of diagonal element of A. Therefore 1 2 8 4 12 + = + = Option = + =(a)gives, trace(A) 6 6 12. 3. For the matrix 4 2 , 2 4 the eigen value corresponding to the eigenvector 101 101 is [EC: GATE-2006] (a) 2 (b) 4 (c) 6 (d) 8 3. (c) 4 2 101 101 2 4 101 101 = 606 101 101 606 6606 101 = = = = 6. All the four entries of the 2 2 matrix P = 11 12 21 22 p p p p are nonzero, and one of its eigen values is zero. Which of the following statements is true? [EC: GATE-2008] (a) P11P22 P12P21 = 1 (b) P11P22 P12P21 = 1 (c) P11P22 P12P21 = 0 (d) P11P22 + 12P21 = 0 6.(c) One eigen value is zero det P 0 = 11 22 12 21P P P P 0 = 7. The eigen values of the following matrix are [EC: GATE-2009] 1 3 5 3 1 6 0 0 3 (a) 3, 3 + 5j, 6 j (b) 6 + 5j, 3 + j, 3 j 20/02/2014 sreekantha.292@gmail.com Page 2 of 177
- 3. (c) 3+ j, 3 j, 5 + j (d) 3, 1 + 3j, 1 3j 7. (d) Let the matrix be A. We know, Trace (A)=sum of eigen values. ME 20 Years GATE Questions 8. Find the eigen value of the matrix 1 0 0 A 2 3 1 0 2 4 = for any one of the eigen values, find out the corresponding eigenvector. [ME: GATE-1994] 8. Same as Q.1 9. The eigen values of the matrix [ME: GATE-1999] 5 3 3 -3 (a) 6 (b) 5 (c) -3 (d) -4 9. (a), (d). 10. The three characteristic roots of the following matrix A [ME: GATE-2000] are 1 2 3 A 0 2 3 0 0 2 = (a) 2,3 (b) 1,2,2 (c) 1,0,0 (d) 0,2,3 10.(b) A is lower triangular matrix. So eigen values are only the diagonal elements. 20/02/2014 sreekantha.292@gmail.com Page 3 of 177
- 4. 11. 4 1 For the matrix the eigen value are 1 4 [ME: GATE-2003] (a) 3 and -3 (b) 3 and -5 (c) 3 and 5 (d) 5 and 0 11. (c) 12. The sum of the eigen values of the matrix given below is [ME: GATE-2004] 1 2 3 1 5 1 3 1 1 (a) 5 (b) 7 (c) 9 (d) 18 12.(b) Sum of eigen values of A= trace (A) 13. For which value of x will the matrix given below become singular? [ME:GATE-2004] 8 x 0 4 0 2 12 6 0 (a) 4 (b) 6 (c) 8 (d) 12 13. (a) Let the given matrix be A. A is singular. det A 0 = 8 x 0 4 0 2 0 12 6 0 = x 4. = 14. Which one of the following is an eigenvector of the matrix [ME: GATE-2005] 20/02/2014 sreekantha.292@gmail.com Page 4 of 177
- 5. 5 0 0 0 0 5 5 0 0 0 2 1 0 0 3 1 1 0 1 1 -2 0 0 -1 (a) (b) (c) (d) 0 1 0 2 0 0 -2 1 14. (a) Let the given matrix be A. Eigen values of A are. 5, 5, Take 5, then AX X gives. = = = 1 1 2 2 3 3 4 4 x 5x5 0 0 0 x 5x0 5 5 0 x 5x0 0 2 1 0 0 3 1 x 5x =1 15x 5x 2 3 2 35x 5x 5x x 0+ = = 3 4 3 4 32x x 5x x 0 x 0+ = = = 3 4 43x x 5x+ = Thus the system of four equation has solution in the form ( )1 2K ,K ,0,0 where 1 2K ,K any real numbers. If we take 1 2K K 2= = than (a) is ture. 15. Eigen values of a matrix 3 2 2 3 S = are 5 and 1. What are the eigen values of the matrix S2 = SS? [ME: GATE-2006] (a) 1 and 25 (b) 6 and 4 (c) 5 and 1 (d) 2 and 10 15. (a) We know If be the eigen value of A 2 2 is an eigen value of A . 16. If a square matrix A is real and symmetric, then the eigenvaluesn [ME: GATE-2007] (a) Are always real (b) Are always real and positive (c) Are always real and non-negative (d) Occur in complex conjugate pairs 20/02/2014 sreekantha.292@gmail.com Page 5 of 177
- 6. 16. (a) 17. The number of linearly independent eigenvectors of 2 1 0 2 is [ME: GATE-2007] (a) 0 (b) 1 (c) 2 (d) Infinite 17. (d) Here 2,2 = For 2, AX X gives, = = 1 1 2 2 x 2x2 1 0 2 x 2x = 1 2 1 2 2 2 2x x 2x x 0 2x 2x + = = = k is the form of eigen vector corrosponding to =2. where 0 k R. 18. The matrix 1 2 4 3 0 6 1 1 p has one eigenvalue equal to 3. The sum of the other two eigenvalues is [ME: GATE-2008] (a) p (b) p-1 (c) p-2 (d) p-3 18.(c) Let the given matrix be A. we know we know i trace(A). = = = + + = +1Here 3 and trace(A) 1 0 P P 1 2 3 P 1 3 P 2 + = + = 19. The eigenvectors of the matrix 1 2 0 2 are written in the form 1 1 and a b . What is a + b? [ME: GATE-2008] (a) 0 (b) (c) 1 (d) 2 19.(b) 1 2 1 2 1 1 Here 1, 2, Given X and X a b = = = = 1 1 1 1For 1, AX X gives = = 20/02/2014 sreekantha.292@gmail.com Page 6 of 177
- 7. 1 2 1 1 0 2 a a = 1 2a 1 a 0 2a a + = = = 2 2 2For 2, AX X gives = = 1 2 1 2 0 2 b 2b = 1 2b 2 b 1 2 2b 2b + = = = 1a b 2 + = 20. For a matrix 3 4 5 5 [ ] , 3 5 M x = the transpose of the matrix is equal to the inverse of the matrix, [M]T = [M]-1. The value of x is given by [ME: GATE-2009] 4 3 3 4 (a) - (b) - (c) (d) 5 5 5 5 20. (a) T 1 Given M M = M is orthogonal matrix T 2MM I = T 2 3 4 3 3x 12 x 1 5 5 5 5 25 Now, MM 3 4 3 3x 12 9 x x 5 5 5 5 25 25 + = = + + =T 2MM I 2 3x 12 1 12 5 45 25 x 25 3 53x 12 9 x 5 25 25 + = = = + + 21. One of the Eigen vectors of the matrix A = 2 1 1 3 is [ME: GATE-2010] 2 2 4 1 (a) (b) (c) (d) 1 1 1 1 20/02/2014 sreekantha.292@gmail.com Page 7 of 177
- 8. 21. (a) The eigen vectors of A are given by AX= X So we can check by multiplication. 2 2 2 2 2 1 1 3 1 1 1 = = = 2 is an eigen vactor of A. corrosponding to 1 1 CE 10 Years GATE Questions 22. The eigen values of the matrix 4 2 2 1 [CE: GATE 2004] (a) are 1 and 4 (b) are 1 and 2 (c) are 0 and 5 (d) cannot be determined 22. (c) 23. Consider the system of equations (n n) (n t)A x = (n )l where, is a scalar. Let i i( , x ) be an eigen-pair of an eigen value and its corresponding eigen vector for real matrix A. Let l be a (n n) unit matrix. Which one of the following statement is NOT correct? (a) For a homogeneous n n system of linear equations, (A ) x = 0 having a nontrivial solution, the rank of (A ) is less than n. [CE: GATE 2005] (b) For matrix Am , m being a positive integer, m m i i( , x ) will be the eigen-pair for all i. (c) If AT = A1 , then i| | = 1 for all i. (d) If AT = A, hen i is real for all i. 23. (b) If be the eigen value of A. then m be the eigen value of m m A .X is no the eigen vector of m A 24. For a given matrix A = 2 2 3 2 1 6 , 1 2 0 one of the eigenvalues is 3. [CE: GATE 2006] The other two eigenvalues are (a) 2, 5 (b) 3, 5 (c) 2, 5 (d) 3, 5 24(b). 20/02/2014 sreekantha.292@gmail.com Page 8 of 177
- 9. 1 2 3we know trace(A). + + = 2 33 2 1 0 1 + + = + = 2 3 2 + = Only choice (b) is possible. 25. The minimum and the maximum eigen values of the matrix 1 1 3 1 5 1 3 1 1 are 2 and 6, respectively. What is the other eigen value? [CE: GATE 2007] (a) 5 (b) 3 (c) 1 (d) 1 25. (b) 1 2 3We know trace(A) + + = + + =3by the condition, 2 6 7 3 3 = 26. The Eigen values of the matrix [P] = 4 5 2 5 are [CE: GATE 2008] (a) 7 and 8 (b) 6 and 5 (c) 3 and 4 (d) 1 and 2 26. (b). EE All GATE Questions 29. The state variable description of a linear autonomous system is, X= AX, Where X is the two dimensional state vector and A is the system matrix given by A = 0 2 2 0 The roots of the characteristic equation are [EE: GATE-2004] (a) -2 and +2 (b)-j2 and +j2 (c)-2 and -2 (d) +2 and +2 29. (a) 20/02/2014 sreekantha.292@gmail.com Page 9 of 177
- 10. 30. In the matrix equation Px = q which of the following is a necessary condition for the existence of at least one solution for the unknown vector x: [EE: GATE-2005] (a) Augmented matrix [Pq] must have the same rank as matrix P (b) Vector q must have only non-zero elements (c) Matrix P must be singular (d) Matrix P must be square 30. (a). 31. For the matrix P= 3 2 2 0 2 1 , 0 0 1 s one of the eigen values is equal to -2. Which of the following is an eigen vector? (a) 3 2 1 (b) 3 2 1 (c) 1 2 3 (d) 2 5 0 31.(d). AX 2X= 1 1 2 2 3 3 x 2x3 2 2 0 2 1 x 2x 0 0 1 x 2x 1 2 3 1 2 3 2 3 3 3x 2x 2x 2x (i) 2x x 2x (ii) x 2x (iii) + = + = = From (ii)and (iii) we get 2 3x 0 and x 0= = 1 2 3From(i)5x 2x 2x (iv)= only choice (d) satisfies equation (iv). 32. If 1 0 1 2 1 1 , 2 3 2 R = then top row of R-1 is [EE: GATE-2005] (a) [ ]5 6 4 (b) [ ]5 3 1 (c) [ ]2 0 -1 (d) [ ]2 1 1/ 2 32(b). 20/02/2014 sreekantha.292@gmail.com Page 10 of 177
- 11. 1 1 R adj R det R = Now, det R = 1 t 5 6 4 5 3 1 adj R 3 4 3 6 4 1 1 1 1 4 3 1 = = = = 1 . top row of R 5 3 1 as det R 1. 35. x=[x1x2..xn]T is an n-tuple nonzero vector. The nn matrix V=xxT [EE: GATE-2007] (a) has rank zero (b) has rank l (c) is orthogonal (d) has rank n 35 (b). As every minor of order 2 is zero. Statement for Linked Answer Questions 37 & 38 Cayley - Hamiltion Theorem states that square matrix satisfies its own characteristic equation, Consider a matrix 3 2 1 0 A = 37. A satisfies the relation [EE: GATE-2007] (a) A +3I + 2A -2 =0 (b) A2+2A+2I=0 (c) (A+I)(A+2I)=0 (d) exp(A)=0 37. (c) Characteristic equation of A is 2A I 0 = 2 3 2 0

Recommended

*View more*