Question 1057358
Use row operations to reduce the matrix "a" and determine which columns form a basis
of the column space.


The resulting matrix is:


{{{(
 matrix( 3, 5, 
   1, 0, 0, 4, -1,
   0, 1, 0, 1, 1,
   0, 0, 1, -2, 0
 )
)
}}}


From this, you can see that the first three columns are linearly independent
because each column has a single "1" in it, and the adjacent columns have 
only 0 entries in the same row (looking only at the first three columns).


The fourth and fifth columns are linear combinations of the first three columns of a.


Specifically, the fourth column is a combination of 4 times the first column, 
plus the second column, minus 2 times the third column. (Check this and see 
for yourself.)


Now, if b = (1, 0, 1)^T is in the span of the columns of a, then there has to be 
some linear combination of the first three columns that is equal to (1, 0, 1)^T.


That is:


(1, 2, 1)^T * x + (2, 1, 5)^T * y + (3, 3, 2)^T * z = (1, 0, 1)^T


This gives three equations:


x + 2y + 3z = 1


2x + y + 3z = 0


x + 5y + 2z = 1


If there is a solution for this system of equations, then b = (1, 0, 1)^T is in the 
span of the columns of a.


I'll assume that you can take if from here. :)