# Thread: Linear transformation, volume, problem

1. ## Linear transformation, volume, problem

True or false :
Let $\displaystyle T:\mathbb{R}^3 \to \mathbb{R}^3$ be a linear transformation defined by $\displaystyle T(1,0,0)=(1,1,0)$, $\displaystyle T(0,1,0)=(-2,1,1)$, $\displaystyle T(0,0,1)=(0,3,-2)$.
If $\displaystyle C$ is the cube formed by the vertices $\displaystyle (0,0,0)$, $\displaystyle (1,0,0)$, $\displaystyle (0,1,0)$, $\displaystyle (0,0,1)$, then $\displaystyle T(C)=3^2$.
My attempt : By intuition its false because $\displaystyle 3^2$ seems more like an area, but it could be a volume.
Ok, so I have formed $\displaystyle [T]_c^{c}$ in order to find out that $\displaystyle T(x,y,z)=(x-2y,x+y+3z,y-2z)$. Then I found the vertices passed by the transformation to be $\displaystyle (0,0,0)$, $\displaystyle (1,1,0)$, $\displaystyle (-2,1,1)$ and $\displaystyle (0,3,-2)$. Now it remains to calculate the volume of the solid formed by these vertices. I'm not really good when it comes to draw 3 dimensional solids. Especially this "prism" looking volume. I've no idea how to find out its volume.

Oh.... let me see, I remember I've read in a calculus III book that I must calculate a jacobian matrix and check out the determinant, or something like that. If it's 1 then the volume of the cube is the same as the volume of this prism. And if it's 2 then the volume of the prism is half or twice the volume of the cube, etc. I don't know which jacobian to find though.

2. Write the transformation as a matrix. That's very easy since you have been told what it does to the basis vectors. It is
1 -2 0
1 1 3
0 1 -2
(or the transpose if you want to write vectors using rows, so that you calculate the action by doing v*M rather than M*v)
The determinant of the matrix is the volume of your transformed cube.

3. Originally Posted by arbolis
$\displaystyle T(C)=3^2$.
Can you explain how?
$\displaystyle T(C)=3^2$

4. You can do it directly. The three edges of C from (0,0,0) are given by the vectors <1, 0, 0>, <0, 1, 0>, and <0, 0, 1>. Applying T to each of those, we get, as we are told, <1, 1, 0>, <-2, 1, 1>, and <0, 3, -2>. Now use the fact that the volume of such a "parallelopiped" is given by the absolute value of the determinant
$\displaystyle \left|\begin{array}{ccc}1 & 1 & 0 \\ -2 & 1 & 1 \\ 0 & 3 & -2\end{array}\right|$$\displaystyle = \left|\begin{array}{cc}1 & 1 \\ 3 & -2\end{array}\right|- \left|\begin{array}{cc}-2 & 0 \\ 0 & -2\end{array}\right|$= (-2-3)- 4= -9
so the absolute value is $\displaystyle 9= 3^2$. The fact that a number happens to be a perfect square doesn't mean it must be an area!