In: Math
Consider the following homogeneous linear system: x1 + 2x2 + 7x3 − 9x4 + 31x5 = 0 2x1 + 4x2 + 7x3 − 11x4 + 34x5 = 0 3x1 + 6x2 + 5x3 − 11x4 + 29x5 = 0 [10p] a) Find the rank of the coefficient matrix. [5p] b) Use part (a) to determine the dimension of the solution space. [10p] c) Find a basis for the solution space.
augmented matrix is
1 | 2 | 7 | -9 | 31 |
2 | 4 | 7 | -11 | 34 |
3 | 6 | 5 | -11 | 29 |
convert into Reduced Row Eschelon Form...
Add (-2 * row1) to row2
1 | 2 | 7 | -9 | 31 |
0 | 0 | -7 | 7 | -28 |
3 | 6 | 5 | -11 | 29 |
Add (-3 * row1) to row3
1 | 2 | 7 | -9 | 31 |
0 | 0 | -7 | 7 | -28 |
0 | 0 | -16 | 16 | -64 |
Divide row2 by -7
1 | 2 | 7 | -9 | 31 |
0 | 0 | 1 | -1 | 4 |
0 | 0 | -16 | 16 | -64 |
Add (16 * row2) to row3
1 | 2 | 7 | -9 | 31 |
0 | 0 | 1 | -1 | 4 |
0 | 0 | 0 | 0 | 0 |
Add (-7 * row2) to row1
1 | 2 | 0 | -2 | 3 |
0 | 0 | 1 | -1 | 4 |
0 | 0 | 0 | 0 | 0 |
reduced system is
there is 2 pivot entry at first and third column hence rank of matrix is 2
there is no pivot entry at second , fourth and fifth column
hence dimension of solution space is 3
.
find l solution
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..........free variable
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.........free variable
.........free variable
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