# Area of figure, resulting from unit square transformation

• Myr73
In summary, the linear operator on R^2 with the matrix A = [1 0; -1 3] maps a unit square to a parallelogram with vertices at (0, 0), (1, -1), (1, 2), and (0, 3). The area of this parallelogram can be easily calculated. It is also worth noting that the determinant of this matrix is 3.
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

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?

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:
anyone?

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.)

## 1. What is the formula for finding the area of a figure resulting from unit square transformation?

The formula for finding the area of a figure resulting from unit square transformation is to first count the number of unit squares that make up the figure, and then multiply that number by the area of one unit square, which is equal to 1 square unit.

## 2. How do you determine the number of unit squares in a figure?

To determine the number of unit squares in a figure, you can count the number of unit squares that completely cover the figure. If the figure is irregularly shaped, you may need to break it down into smaller, regular shapes and then count the unit squares in each shape before adding them together.

## 3. Can the area of a figure resulting from unit square transformation be a fraction?

No, the area of a figure resulting from unit square transformation is always a whole number. Since the area of one unit square is equal to 1 square unit, the total area of the figure will be a multiple of 1, resulting in a whole number.

## 4. How does the concept of unit square transformation apply to real life scenarios?

The concept of unit square transformation is applicable in many real life scenarios, such as when determining the number of tiles needed to cover a floor, or the number of pixels in a digital image. It is also useful in calculating the area of land, where the unit square can represent a specific unit of measurement, such as square meters or square feet.

## 5. Is the area of a figure resulting from unit square transformation affected by the orientation of the unit squares?

No, the orientation of the unit squares within a figure does not affect its total area. As long as the unit squares are not overlapping or leaving gaps, the total area will remain the same regardless of their orientation.

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