# Area of figure, resulting from unit square transformation

1. May 29, 2015

### Myr73

1- Let the linear operator on R^2 have the following matrix:

A = 1 0
-1 3

What is the area of the figure that results from applying this transformation to the unit square?

2- I am abit confused here, I thought that the matrix for the unit square would be,
0 0
1 0
0 1
1 1....
But apparently that is not it.

And after finding the unit square matrix, I blv I would find the images of it by computing A^Tx

I am not sure though

2. May 29, 2015

### DEvens

A square would be four points. You seem to have collected four points that make up one particular unit square, and arranged them as a 2x4 array. That is kind of odd.

What does it mean to be a linear operator on R^2? What does the operator A operate on? Hint: R^2 means there are two real numbers involved. But there seem to be 8 real numbers involved in your square. So how is that going to work?

3. May 29, 2015

### Myr73

oh. ok, so are they asking to find the Ax? where x would be the unit square.. so like
[10. ] [ 1 0]
[ 1 3 ] [ 01 ] ...

Last edited: May 29, 2015
4. May 30, 2015

anyone?

5. Jun 1, 2015

### HallsofIvy

That matrix is NOT the unit square. The unit square is a geometric figure, not a matrix, that has vertices at (0, 0), (1, 0), (1, 1), and (0, 1).
The bottom side, from (0, 0) to (1, 0), the vector $\begin{bmatrix}1 \\ 0 \end{bmatrix}$, is mapped into
$$\begin{bmatrix}1 & 0 \\ -1 & 3\end{bmatrix}\begin{bmatrix}1 \\ 0 \end{bmatrix}= \begin{bmatrix}1 \\ -1\end{bmatrix}$$

The left side, from (0, 0) to (0, 1), the vector $\begin{bmatrix}0 \\ 1 \end{bmatrix}$, is mapped into
$$\begin{bmatrix}1 & 0 \\ -1 & 3 \end{bmatrix}\begin{bmatrix}0 \\ 1 \end{bmatrix}= \begin{bmatrix}0 \\ 3\end{bmatrix}$$

The other two sides get mapped into equivalent vectors so this is a parallelogram with vertices at (0, 0), (1, -1), (1, 2), and (0, 3).
It should be easy to find the area of that parallelogram.

(And, it is worth noting that this matrix has determinant 3.)