Question

given the following vector X, find a non-zero sq matrix A such that Ax=0.

X= [2 10 3] <-- this is in a column, not row

Answer 1

people would say the vector X is in the Null space of the matrix A

one way to figure this out is notice that multiplying a matrix times a vector can be done this way:

\[\left[\begin{matrix}c1 & c2 \end{matrix}\right]\left[\begin{matrix}a1 \\ a2\end{matrix}\right]= a1*c1+a2*c2\]
where c1 and c2 are columns

for your problem you want the sum of 2 times the first column plus 1 times the 2nd column + 3 times the 4th column add up to a vector with all zeros

Answer 2

can you explain further why you would times 2 with the first column and 1 and 3 times to the 2nd and 4th column respectively?

Answer 3

ohhh ok, I see now, you would want everything to equal to the main 0 as depicted in the formula AX=0

Answer 4

But yeah, please explain how would you come to 2, 1, and 3 so quickly?

Answer 5

See http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/lecture-3-multiplication-and-inverse-matrices/

in the mean time, one matrix that would work is
c1= 1 0 0 0
c2= 0 1 0 0
c3 = 1 2 3 4 (can be anything)
c4= -2/3 -1/3 0 0

Answer 6

where each c1, c2, etc is a column

Answer 7

right on, Dot product then. But I am still not convince on thats how you achieve the answer. The answer is \[\left[\begin{matrix}3 & 3&3 \\2 & 1 &-3\\ 3 & 2 &-2\end{matrix}\right]\]

Answer 8

ooh, I thought the vector was 2 1 0 3
and it's really 2 ten 3

2,10, and 3

but it's the same idea

Answer 9

haha no sorry, its \[\left(\begin{matrix}-8 \\ 10\\-2\end{matrix}\right)\]

Answer 10

that's different from
X= [2 10 3] <-- this is in a column, not row

Answer 11

oops sorry, yeah thats what I meant \[\left(\begin{matrix}2 \\10\\3 \end{matrix}\right)\]

Answer 12

you can also look for rows that give zero when dot product with your X

for example row1= 2 -1 2
row 2= 0 3 -10
row3 = -3 0 2

Answer 13

hmmm thats a bit tedious wont it not? it doesnt really give the answer as I posted above, \[\left[\begin{matrix}3 & 3 & 3 \\ 2 & 1 & -3 \\ 3 & 2 & -2\end{matrix}\right]\]

Answer 14

that matrix works for X= -8 10 -2

another matrix that would work for that X:
1 1 1
0 1 5
1 0 -4

Answer 15

lol, im super sorry, but I am super stumped on just getting that answer from vector X from dot product.