The Inverse of a Linear Transformation
How to undo a linear transformation
Exam questions
- \(\mathbf{M=\begin{pmatrix}3 & 4 \\ 2 & -5\end{pmatrix}}\)
\(\det \mathbf{M}=-23\)
- Given that \(\mathbf{A}=\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\), describe the linear transformation given by this matrix.
\(\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\) = a rotation of 90\(^\circ\) anticlockwise.
- Transformation \(\mathbf{A}\) followed by transformation \(\mathbf{B}\) is equivalent to the transformation \(\mathbf{M}\). Find \(\mathbf{B}\).
\(\mathbf{AB}=\mathbf{M}\)
\(\mathbf{IB}=\mathbf{A^{-1}M}\)
\(\mathbf{B}=\mathbf{A^{-1}M}\)
\(\mathbf{A^{-1}}=\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}\)
\(\mathbf{B}=\begin{pmatrix}3 & 4 \\ 2 & -5\end{pmatrix}\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}=\begin{pmatrix}-4 & 3 \\ 5 & 2\end{pmatrix}\)
How many times.... Squashed problem
- Applying the matrix \(\begin{pmatrix}\frac{1}{2} & 0 \\0 & \frac{1}{2}\end{pmatrix}\) co-ordinates halfs the co-ordiantes.
- If we apply this matrix to the point \((2, 6)\) on the quadrilateral shown, we get the point \((1, 3)\) which is not inside the circle of radius 3 at the origin.
- Because this is the co-ordinate with the largest x or y co-ordinate, we can test this to see how many times we should apply the matrix.
- After applying this matrix twice to this point, we get the point \(\left( \frac{1}{2}, \frac{3}{2} \right)\) which is inside the circle at the origin.
- Therefore we have proved that e only need to apply the matrix twice to get the quadrilateral inside the circle.