As no zero row occurs in the echelon form of the matrix, the set of vectors
xxx
123
,,
is
linearly independent.
(2) If the given vector lies in the subspace spanned by the vectors
xxx
123
,, then there will
exist scalars c
1
, c
2
, c
3
not all zero such that
(4, -5, 9, -7) =c
1
(-1, 2, 5, 2) +c
2
(3, 0, 4, -1) +c
3
(1, 1, -2, 1)
This gives the following system of equations:
-c
1
+ 3c
2
+c
3
= 4 (i)
2c
1
+c
3
=-5 (ii)
5c
1
+ 4c
2
- 2c
3
= 9 (iii)
2c
1
-c
2
+c
3
=-7 (iv)
Its solution is c
1
=-1, c
2
= 2, c
3
=-3, which shows that the vector (4, -5, 9, -7) lies in the
subspace of R
4
spanned by the given vectors. ...
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