# linear algebra linear transformation two.

Let

$$A=\begin{bmatrix}1&1&-2\\0&1&1\\-2&-1&5\end{bmatrix}\quad\text{ and }\quad b=\begin{bmatrix}3\\-1\\-7\end{bmatrix}\;.$$

Define the linear transformation $T: {\mathbb R}^3 \rightarrow {\mathbb R}^3$ as $T(x) = Ax$. Find a vector $x$ whose image under $T$ is $b$.

$x =$

Is the vector $x$ unique? (enter YES or NO)

-
This post needs heavy editing. –  Vedran Šego Jul 4 '13 at 11:13
Please see whether I interpreted the post correctly; I’m simply guessing that the entries in the two matrices appeared in their natural order in the very complicated $\LaTeX$ code that you wrote. –  Brian M. Scott Jul 4 '13 at 11:39
What have you already tried? –  TZakrevskiy Jul 4 '13 at 12:21
@BrianM.Scott I admire you. I tried, but got discouraged, due to the mess it was in. However, if you look at the first version, the last component of $b$ might be $-7$. I don't know about the rest. –  Vedran Šego Jul 5 '13 at 13:03
@Vedran: You’re absolutely right: I misread it. Thanks very much. –  Brian M. Scott Jul 5 '13 at 18:53

so we want a vector $(x,y,z)$ such that when premultiplied by A gives us b.

so therefore $x+y-2z=3$, $y+z=-1$and $-2x-y-5z=-1$. We are allowed multiply these equations by constants and add these equations. (left side with left and right with right. As you can see, (if we ignore the variables and just look at the coefficients this is the same as adding one row of the matric to another and multiplying one row with another. Therefore we can work with matrices. Where each row contains the coefficients of the variables on the first three columns and the constant to which they add up on the right side.

So at the start we have.

$$\begin{bmatrix}1&1&-2&3\\0&1&1&-1\\-2&-1&5&-7\end{bmatrix}$$

Then we switch the second and third rows

$$\begin{bmatrix}1&1&-2&3\\-2&-1&5&-7\\0&1&1&-1\end{bmatrix}$$

add twice the first one to the second one

$$\begin{bmatrix}1&1&-2&3\\0&1&1&-1\\0&1&1&-1\end{bmatrix}$$

and then subtract the second one from the third one to get

$$\begin{bmatrix}1&1&-2&3\\0&1&1&-1\\0&0&0&0\end{bmatrix}\;.$$

We can simplify a bit more by subtracting the second row from the first:

$$\begin{bmatrix}1&0&-3&4\\0&1&1&-1\\0&0&0&0\end{bmatrix}\;.$$

This says that $x_1-3x_3=4$ and $x_2+x_3=-1$, or

\left\{\begin{align*} x_1&=4+3x_3\\ x_2&=-1-x_3\;, \end{align*}\right.

where $x_3$ can be any real number whatsoever. Taking $x_3=0$ makes the calculation very easy and gives us one solution:

$$x=\begin{bmatrix}4\\-1\\0\end{bmatrix}\;.$$

But this is clearly not unique: we get infinitely many other solutions by changing the value of $x_3$.

-
Since I made an error in transcribing the question into readable form, I took the liberty of changing this answer to match the corrected version of the question and save you the trouble. –  Brian M. Scott Jul 5 '13 at 19:02
Ok ,thanks for your work Proffesor. –  Bananarama Jul 5 '13 at 19:05