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Describe the geometric transformations represented by the following matrices, applied to the unit square.

Mathematics

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8 steps

Here are the descriptions of the transformations for each matrix, using the unit square with vertices O(0,0)O(0,0), A(1,0)A(1,0), B(0,1)B(0,1), and C(1,1)C(1,1).

Part a· 2 steps

For the matrix (0110)\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}

  1. Apply the transformation to the vertices of the unit square.

    O′=(0110)(00)=(00)O' = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} A′=(0110)(10)=(01)A' = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} B′=(0110)(01)=(10)B' = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} C′=(0110)(11)=(11)C' = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}

  2. Describe the transformation.

    The original vertices are (0,0),(1,0),(0,1),(1,1)(0,0), (1,0), (0,1), (1,1). The transformed vertices are (0,0),(0,1),(1,0),(1,1)(0,0), (0,1), (1,0), (1,1). This transformation swaps the x and y coordinates.
    The transformation is a reflection about the line y=xy=x.

Part b· 2 steps

For the matrix (0−110)\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}

  1. Apply the transformation to the vertices of the unit square.

    O′=(0−110)(00)=(00)O' = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} A′=(0−110)(10)=(01)A' = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} B′=(0−110)(01)=(−10)B' = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 0 \end{pmatrix} C′=(0−110)(11)=(−11)C' = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \end{pmatrix}

  2. Describe the transformation.

    The original vertices are (0,0),(1,0),(0,1),(1,1)(0,0), (1,0), (0,1), (1,1). The transformed vertices are (0,0),(0,1),(−1,0),(−1,1)(0,0), (0,1), (-1,0), (-1,1). This corresponds to rotating the x-axis to the y-axis and the y-axis to the negative x-axis.
    The transformation is a rotation of 90∘90^\circ counter-clockwise about the origin (0,0)(0,0).

Part c· 2 steps

For the matrix (2002)\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}

  1. Apply the transformation to the vertices of the unit square.

    O′=(2002)(00)=(00)O' = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} A′=(2002)(10)=(20)A' = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 0 \end{pmatrix} B′=(2002)(01)=(02)B' = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 2 \end{pmatrix} C′=(2002)(11)=(22)C' = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \end{pmatrix}

  2. Describe the transformation.

    The original vertices are $(0,0), (1,0), (0,1), (1,1)$. The transformed vertices are $(0,0), (2,0), (0,2), (2,2)$. All coordinates are multiplied by 2.
    The transformation is an enlargement (or dilation) with a scale factor of 2, centered at the origin $(0,0)$.

Part d· 2 steps

For the matrix (0−1−10)\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}

  1. Apply the transformation to the vertices of the unit square.

    O′=(0−1−10)(00)=(00)O' = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} A′=(0−1−10)(10)=(0−1)A' = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ -1 \end{pmatrix} B′=(0−1−10)(01)=(−10)B' = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 0 \end{pmatrix} C′=(0−1−10)(11)=(−1−1)C' = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ -1 \end{pmatrix}

  2. Describe the transformation.

    The original vertices are (0,0),(1,0),(0,1),(1,1)(0,0), (1,0), (0,1), (1,1). The transformed vertices are (0,0),(0,−1),(−1,0),(−1,−1)(0,0), (0,-1), (-1,0), (-1,-1). This transformation maps (x,y)(x,y) to (−y,−x)(-y,-x).
    The transformation is a reflection about the line y=−xy=-x.

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