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Solution to Previous Year Exam Questions for Engineering Mathematics 2 - EM-2 of rgpv by salman rashid

  • Engineering Mathematics 2 - EM-2
  • 2017
  • PYQ Solution
  • Rajiv Gandhi Proudyogiki Vishwavidyalaya Bhopal - rgpv
  • Computer Science Engineering
  • B.Tech
  • 252 Views
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RGPV SOLUTION MA-111 ENGINEERING MATHEMATICS 2 DEC-2017 1. (a) Find rank and nullity of the following matrix by reducing it to the normal form 1 3 3  2 4 10   3 8 4  Solution: Given the matrix is 1 3 3  A  2 4 10 3 8 4  Applying, R2  R2  2R1 , R3  R3  3R1 3 1 3  A ~ 0  2 4  0  1  5 Applying, C2  C2  3C1 , C3  RC3  3C1 0 1 0  A ~ 0  2 4  0  1  5 Applying, R2  R3 1 0 0 A ~ 0 1 5 0  2 4 Applying, R3  R3  2R2 1 0 0  A ~ 0 1 5  0 0 14 Applying, C3  C3  5C2 1 0 0  A ~ 0 1 0  0 0 14 Applying, C3  C3 / 14 www.rgpvonline.com download rgpv online app for more solutions

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1 0 0  A ~ 0 1 0   I 3  0 0 14 Clearly this is required normal form and rank  ( A)  3 Nullity Answer N(A) = order of square matrix-Rank of Matrix N(A) = 3 – 3 = 0 (b) Answer Examine whether the following equations are consistent and solve them if they are consistent. 2 x  y  3z  0 3x  2 y  z  0 x  4 y  5z  0 Solution : Given the system of equation is 2  1 3  x  0 3 2 1  y   0      1  4 5  z  0 AX  B  This is Homogeneous system of equation, and then augmented matrix is  2  1 3 0 C  A : B  3 2 1 0 1  4 5 0 Applying, R1  R3 1  4 5 0 C ~ 3 2 1 0 2  1 3 0 Applying, R2  R2  3R1 , R3  R3  2R1 1  4 5 0 C ~ 0 14  14 0 0 7  7 0 Applying, R2  R2 / 14 1  4 5 0 C ~ 0 1  1 0 0 7  7 0 www.rgpvonline.com download rgpv online app for more solutions

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Applying, R1  R1  4R2 , R3  R3  7 R2 1 0 1 0 C ~ 0 1  1 0 0 0 0 0 Clearly  ( A)  2 and  (C )  2   ( A)   (C )  2  3 (No. of unknown variables)  The system is consistent and having infinite many solutions. 1 0 1   x  0 0 1  1  y   0      0 0 0   z  0 xz 0 yz0 Taking, z = k, then we get x = k and y = k Hence the required solution is x  k , y  k and z  k Answer 2. (a) Find Eigen values and Eigen vectors of the matrix  8 6 2  A   6 7  4  2  4 3  Solution : The characteristic equation is | A  l | 0 8 6 2  6 2 7 4 0 4 3  8   7  3     16  6 63     8  224  2(7     0  8   21  10  2  16 6 18  6  8  224  14  2   0  8   2  10  5  6 10  6   210  2   0  82  80  40  3  102  5  60  36  20  4  0   3  182  45  0    2  18  45  0      15  3  0    0,3,15 www.rgpvonline.com download rgpv online app for more solutions

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 x Case 1 : Suppose X 1   y  be Eigen vector corresponding to Eigen value   0, then    z  A  0, l X1  0   2  8  0  6  6 70 4     4 3  0  2  x  y  0    z   8  6 2   x   6 7  4  y   0      2  4 3   z  Since all rows are non identical then taking R1 and R2 rows as Taking, 8.x  6. y  2.z  0  6.x  7. y  4.z  0  x y z   24  14  12  32 56  36  x y z   10 20 20 i.e.,  Case 2 : x y z   1 2 2 1  X 1  2 2  x Suppose X 1   y  be Eigen vector corresponding to Eigen value   3, then    z  A  3, l X1  0   2  8  3  6  6 73 4     2  4 3  3  x  y  0    z   5  6 2   x   6 4  4  y   0     2  4 0   z  Since all rows are non identical then taking R1 and R2 rows as www.rgpvonline.com download rgpv online app for more solutions

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