Is Vector w in the Range of Matrix A?

[newtomath]

L: R^3 -> R^3 is a linear transformation defined by L(v) =A(v)
A is given as -1 2 0 and w= 1
1 1 1 2
2 -1 1 -1

is w in the range of L?

My understanding is that if a vector exists such that the product of A and this vector = W then it is in the range. I then have the following equations:

-X +2Y = 1
X+Y+Z= 2
2X -Y+Z= -1

no solutions exists, so w is not in the range. Can someone confirm this or offer any other insight?

thanks

 

[lanedance]

bit hard to read is A given as:
A = \begin{bmatrix} -1 & 2 & 0\\ 1 & 1 & 1 \\ 2 & -1 & 1 \end{bmatrix}

 

[lanedance]

what you've done sounds reasonable, though i haven't checked the math,

note so what you are trying to solve is:
A \textbf{x} = \textbf{x}

If A was invertible a solution must exist given by:
\textbf{x} = A^{-1} \textbf{x}
so you could check that det(A) = 0

Now, consider the action of A on a unit vector:
\textbf{e}_1 = (1,0,0)^T
A\textbf{e}_1 = \textbf{a}_1
where \textbf{a}_1 is the first column vector of A

Now every vector can be wirtten in terms of the basis of unit vectors, so the range of A is in fact the column space of A. So in effect you are checking w is not in the column space of A.

 

[HallsofIvy]

Internet readers do not respect spaces so better to use LaTex.

If
A= \begin{bmatrix}-1 & 2 & 0 \\ 1 & 1 & 1 \\ 2 & - 1 & 1\end{bmatrix}
and
w= \begin{bmatrix} 1 \\ 2 \\ -1\end{bmatrix}

is w in the range of A?
That will be true if and only if there exist x, y, z such that
\begin{bmatrix}-1 &amp; 2 &amp; 0 \\ 1 &amp; 1 &amp; 1 \\ 2 &amp; - 1 &amp; 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix} 1 \\ 2 \\ -1\end{bmatrix}[/itex]<br /> (Click on those to see the code.)<br /> <br /> That can be written as the three equations, -x+ 2y= 1, x+ y+ z= 2, 2x- y+ z= -1.<br /> <br /> IF those three equations have a solution (not necessarily unique) then w is in the range of A.<br /> <br /> It might be simpler to check the determinant of A. If the determinant is not 0, the range of A is all of R^3 and so <b>any</b> w is in it.<br /> <br /> If the determinant of a is 0, then you would have to check if w is in the &quot;columns space&quot;. In that case, it is probably simpler to solve the three equations above.