Geometry of 2x2 matrices

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Geometry of 2x2 matrices

The geometric effect that a $2\times 2$ matrix has on a plane can be effectively visualised by examining its effect on a unit square, with vertices $(0,0)$, $(1,0)$, $(1,1)$ and $(0,1)$.

In the first instance, we will apply the sleight of hand of considering the vertices of the square to be the endpoints of position vectors and we can then apply our $2\times 2$ matrix to them

Suppose we have a matrix $M$, defined as $$ M = \left(\begin{array}{cc}a & b \\ c & d \end{array}\right) $$ then the following are true

$$ M\left(\begin{array}{c}0 \\ 0 \end{array}\right) = \left(\begin{array}{c}0 \\ 0\end{array}\right) $$ $$ M\left(\begin{array}{c}1 \\ 0 \end{array}\right) = \left(\begin{array}{c}a \\ d\end{array}\right) $$ $$ M\left(\begin{array}{c}0 \\ 1 \end{array}\right) = \left(\begin{array}{c}b \\ c\end{array}\right) $$ $$ M\left(\begin{array}{c}1 \\ 1 \end{array}\right) = \left(\begin{array}{c}a+b \\ c+d\end{array}\right) = M\left(\begin{array}{c}0 \\ 1 \end{array}\right) + M\left(\begin{array}{c}1 \\ 0 \end{array}\right) $$

we can represent this effect geometrically as shown in the image below

matrix tx

The area of the parallelogram is given by $$A_{P} = \det M \times A_{Sq} = (ad - bc)\times 1$$

Note that it is possible for the area to be negative; but a negative result merely indicates that the orientation of the new parallelogram is different from the original square; the unsigned value is the true "area" of the parallelogram

Geometry of 2x2 matrices 1

Let the matrix $M$ be defined as $$ M = \left(\begin{array}{cc} 2 & 1 \\ 0 & 1\end{array}\right) $$ Show the effect of $M$ on the unit square and determine the area of the image

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The image of the unit square is a parallelogram with vertices $$(0,0), (2,0),(3,1), (1,1) $$ The area of this parallelogram is $2$. This can be calculated directly from the geometry or we can use the key formula: $$A_{image} = \det(A)\times A_{original}$$ In this case, we have $$ \begin{align} A_{image} &= \left|\begin{array}{cc} 2 & 1 \\ 0 & 1\end{array}\right| \times 1 \\ &= (2\times 1 - 1\times 0) \times 1 \\ &= 2 \end{align} $$

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