For AI agents: use /llms.txt for the Nakafa content index.
A 2 × 2 2 \times 2 2 × 2
A point in the Cartesian plane, often symbolized by the ordered pair ( x , y ) (x,y) ( x , y )
( x y ) \begin{pmatrix} x \\ y \end{pmatrix} ( x y ) If a point P ( x , y ) P(x,y) P ( x , y ) M = ( a b c d ) M = \begin{pmatrix} a & b \\ c & d \end{pmatrix} M = ( a c b d ) P ′ ( x ′ , y ′ ) P'(x',y') P ′ ( x ′ , y ′ )
Thus, x ′ = a x + b y x' = ax + by x ′ = a x + b y y ′ = c x + d y y' = cx + dy y ′ = c x + d y
If ( x y ) \begin{pmatrix} x \\ y \end{pmatrix} ( x y ) ( 0 − 1 1 0 ) ( x y ) \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} ( 0 1 − 1 0 ) ( x y )
It can be observed that the point ( x , y ) (x,y) ( x , y ) ( 0 − 1 1 0 ) \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} ( 0 1 − 1 0 ) ( − y , x ) (-y,x) ( − y , x ) 90 ∘ 90^\circ 9 0 ∘
Find the image of △ A B C \triangle ABC △ A B C A ( 1 , 1 ) A(1,1) A ( 1 , 1 ) B ( 4 , 1 ) B(4,1) B ( 4 , 1 ) C ( 4 , 2 ) C(4,2) C ( 4 , 2 ) ( − 1 0 0 − 1 ) \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} ( − 1 0 0 − 1 )
First, we can write the coordinates of the points as columns of a matrix, i.e., ( 1 4 4 1 1 2 ) \begin{pmatrix} 1 & 4 & 4 \\ 1 & 1 & 2 \end{pmatrix} ( 1 1 4 1 4 2 )
Next, multiply this matrix from the left by ( − 1 0 0 − 1 ) \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} ( − 1 0 0 − 1 )
The result of the transformation is a new triangle △ A ′ B ′ C ′ \triangle A'B'C' △ A ′ B ′ C ′ A ′ ( − 1 , − 1 ) A'(-1,-1) A ′ ( − 1 , − 1 ) B ′ ( − 4 , − 1 ) B'(-4,-1) B ′ ( − 4 , − 1 ) C ′ ( − 4 , − 2 ) C'(-4,-2) C ′ ( − 4 , − 2 )
The matrix ( − 1 0 0 − 1 ) \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} ( − 1 0 0 − 1 ) 180 ∘ 180^\circ 18 0 ∘
Find the product of ( 0 1 − 1 0 ) ( x y ) \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} ( 0 − 1 1 0 ) ( x y )
A transformation is associated with the matrix ( 1 2 0 1 ) \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} ( 1 0 2 1 ) A ( 2 , 0 ) A(2,0) A ( 2 , 0 )
The point ( x , y ) (x,y) ( x , y ) ( y , − x ) (y,-x) ( y , − x )
This is a − 90 ∘ -90^\circ − 9 0 ∘ 270 ∘ 270^\circ 27 0 ∘
Transformation matrix M = ( 1 2 0 1 ) M = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} M = ( 1 0 2 1 )
Vertices: A ( 2 , 0 ) A(2,0) A ( 2 , 0 )
Point matrix: ( 2 2 0 0 1 1 ) \begin{pmatrix} 2 & 2 & 0 \\ 0 & 1 & 1 \end{pmatrix} ( 2 0 2 1 0 1 )
Image vertices: A ′ ( 2 , 0 ) A'(2,0) A ′ ( 2 , 0 ) B ′ ( 4 , 1 ) B'(4,1) B ′ ( 4 , 1 ) C ′ ( 2 , 1 ) C'(2,1) C ′ ( 2 , 1 )
(This transformation is known as a shear)
