Question

Is b a linear combination of a1, a2, and a3?
a1= (1, -2, 0); a2= (0,1,2); a3=(5, -6, 8); b= (2,-1,6)

Answer 1

form a matrix with the a's as vectors, and b as the last vector
use elimination on this augmented matrix.

Answer 2

the last matrix i Have is
row 1: 1 0 5 2
row 2: 0 1 4 3
row 3: 0 0 0 0

I don't know what to do from here. The answer says "No, b is not a linear combination of a1, a2, and a3"

Answer 3

you have 2 pivot columns. those are your basis
the last column says b= 2*1st pivot col + 3*2nd pivot col
btw, a3 is a linear combination of a1 and a2

Answer 4

You can check
2*a1 + 3*a2 should = b

Answer 5

Also, if b were not a combination, it would have been a pivot column

Answer 6

I see that the back of the book is wrong, but I'm still a little lost as to your explanation. Can you explain "you have 2 pivot...." in a little more detail

Answer 7

The pivot is the leading (left most) non-zero entry in a row
So the 1 in location (1,1) is a pivot.
Also, the 1 in location (2,2).
Notice there is only 1 pivot in the 1st column.
and one in the 2nd column
that means those two vectors are independent

Answer 8

the columns that do not have a pivot are dependent.
their entries tell the dependence. for example
the 2 in location 1,3 tells you that 2 times the first pivot column is part of vector b
the 3 in location 2,3 tells you that 3 times the 2nd pivot column is part of b

Answer 9

the same with a3. a3= 5*a1 + 4*a2

Answer 10

location 1,3? isn't that the 5?
and location 2,3 is a 4 i believe. Unless I dont understand how the location thing works.

The last matrix tells me:
x1+ 5x3=2
x2+4x3=3

sorry, I'm still stuck

Answer 11

sorry, I can't count. change 3 to 4

Answer 12

the 3rd column is a3 and the 4th column is b

Answer 13

a1= (1, -2, 0); a2= (0,1,2); a3=(5, -6, 8); b= (2,-1,6)
2*a1= (2, -4, 0)
3*a2= (0, 3, 6)
--------------
sum = (2, -1,6) = b

Answer 14

btw, normally I would write the vectors vertically, but it's a pain to do so here.

Answer 15

I guess here is the part that I'm stuck. I see that they can add up to b. But...

how I understand it is
x1a1+x2a2+x3a3=b
so if I solve for x1, x2, and x3. I should have some answer.

The two pivot columns tell me that x1 and x2 are basic variables. I still don't see how I can go from my final matrix

row 1: 1 0 5 2
row 2: 0 1 4 3
row 3: 0 0 0 0

to solving x1, and x2.

I get here

x1+ 5x3=2
x2+4x3=3

where x3 is a free variable

Answer 16

That is a valid way to think of it. But there is another way, as linear combination of vectors.
so the first column (1,0,0) means a1 depends on a1 and 0 times on a2 or a3
the 2nd column (0,1,0) means a2 depends only on a2 (0*a1+1*a2+0*a3)
the 3rd column means that a3= 5*a1+4*a2+0*a3 (it is dependent)
It's another way to interpret the matrix, but explaining why it works is not so easy for me

Answer 17

To actually solve your system, set your free variable to something convenient like zero

Answer 18

and to be complete you would add in the null space

\[x=\left(\begin{matrix}2 \\ 3 \\ 0\end{matrix}\right)+c\left(\begin{matrix}-5 \\ -4 \\ 1\end{matrix}\right)\]

Answer 19

But that's for solving the system. Determining if b is in the column space of A is a simpler matter.