Problem 2 a Is in span b If possible write as a lin
Solution
a) The easiest method is to check the determinant value of the three span vectors, if the determinant is zero, that means they don\'t the span R^3
Determinant
= 1(8 + 12) + 0(0+16) + 5(-4)
= 20 - 20
= 0
Since determinant is zero hence these vector doesn\'t span the complete R^3 space
2 = a + 5c
-1 = 2a + b - 6c
6 = 2b + 8c
Hence it is in the span of the given three vectors
b)
2 = a + 5c
-1 = 2a + b - 6c
6 = 2b + 8c
solving the linear equations we get
Matlab solution
Hence the given vector can be written as a sub-vector
c) Yes there are more solution possible like x=7,y=7 and z=-1
Problem 3
The first matrix is 3X3 and second matrix is 3X1, hence resultant matrix will be 3X1
First element = 1 * 7 + 0 * 7 + 5 * -1 = 2
Second element = -2 * 7 + 1 * 7 + -6 * -1 = -1
Third element = 0 * 7 + 2 * 7 + 8 * -1 = 6
Hence correct answer is [2,-1,6]
The given problem multiplies the given matrix span with the solution to obtained the resultant vector
| 1 | 0 | 5 |
| -2 | 1 | -6 |
| 0 | 2 | 8 |
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