The linear transformation \(\mathrm{T}\) is a shear which transforms a point \(P\) to the point \(P'\) defined by
\(\overrightarrow{PP'}\) makes an acute angle \(\alpha\) (anticlockwise)
with the \(x\)-axis, \(\angle POP'\) is clockwise (i.e. the rotation from \(OP\) to \(OP'\)
clockwise is less than \(\pi),\) \(PP'=k\times PN,\) where \(PN\) is the perpendicular onto the line
\(y=x\tan\alpha,\) where \(k\) is a given non-zero constant.
If \(\mathrm{T}\) is represented in matrix form by $\begin{pmatrix}x'\\
y'
\end{pmatrix}=\mathbf{M}\begin{pmatrix}x\\
y
\end{pmatrix},$ show that
\[
\mathbf{M}=\begin{pmatrix}1-k\sin\alpha\cos\alpha & k\cos^{2}\alpha\\
-k\sin^{2}\alpha & 1+k\sin\alpha\cos\alpha
\end{pmatrix}.
\]
Show that the necessary and sufficient condition for $\begin{pmatrix}p & q\\
r & t
\end{pmatrix}\( to commute with \)\mathbf{M}$ is
\[
t-p=2q\tan\alpha=-2r\cot\alpha.
\]
Show Solution
Solution:
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We can see that \(\mathbf{M}\) sends \(\begin{pmatrix} 1 \\ \tan \alpha \end{pmatrix}\) to itself, and \(\begin{pmatrix} -\tan \alpha \\ 1 \end{pmatrix}\) to \(\begin{pmatrix} -\tan \alpha \\ 1 \end{pmatrix} + k \begin{pmatrix} 1 \\ \tan \alpha \end{pmatrix}\)
Therefore, we have:
\begin{align*}
&& \mathbf{M} \begin{pmatrix} 1 & -\tan \alpha \\ \tan \alpha & 1 \end{pmatrix} &= \begin{pmatrix} 1 & k - \tan \alpha \\ \tan \alpha & 1 + k\tan \alpha \end{pmatrix} \\
&& \sec \alpha \mathbf{M} \begin{pmatrix} \cos \alpha & -\sin\alpha \\ \sin \alpha & \cos \alpha \end{pmatrix} &= \begin{pmatrix} 1 & k - \tan \alpha \\ \tan \alpha & 1 + k\tan \alpha \end{pmatrix} \\
\Rightarrow && \mathbf{M} &= \cos \alpha\begin{pmatrix} 1 & k - \tan \alpha \\ \tan \alpha & 1 + k\tan \alpha \end{pmatrix}\begin{pmatrix} \cos \alpha & \sin\alpha \\ -\sin \alpha & \cos \alpha \end{pmatrix} \\
&&&= \cos\alpha \begin{pmatrix}\cos \alpha -k\sin\alpha + \frac{\sin^2 \alpha}{\cos \alpha} & \sin \alpha + k \cos \alpha - \sin \alpha \\ \sin \alpha - \sin \alpha - k\frac{\sin^2 \alpha}{\cos \alpha} & \frac{\sin^2 \alpha}{\cos \alpha} + \cos\alpha + k \sin \alpha \end{pmatrix} \\
&&&= \begin{pmatrix}1-k\sin\alpha\cos\alpha & k\cos^{2}\alpha\\
-k\sin^{2}\alpha & 1+k\sin\alpha\cos\alpha
\end{pmatrix}
\end{align*}
Suppose $\begin{pmatrix}p & q\\
r & t
\end{pmatrix} \mathbf{M} = \mathbf{M} \begin{pmatrix}p & q\\
r & t
\end{pmatrix}$ then,
\begin{align*}
&& \begin{pmatrix}p & q\\
r & t
\end{pmatrix} \mathbf{M} &= \mathbf{M} \begin{pmatrix}p & q\\
r & t
\end{pmatrix} \\
\Leftrightarrow && \small \begin{pmatrix} p(1-k\sin\alpha\cos\alpha) + q(-k\sin^{2}\alpha) & pk\cos^{2}\alpha + q(1+k\sin\alpha\cos\alpha)\\
r(1-k\sin\alpha\cos\alpha) + t(-k\sin^{2}\alpha) & rk\cos^{2}\alpha + t(1+k\sin\alpha\cos\alpha)\end{pmatrix} &= \\
&& \qquad \small \begin{pmatrix} p(1-k\sin\alpha\cos\alpha) + rk\cos^{2}\alpha & q(1-k\sin\alpha\cos\alpha) + tk\cos^{2}\alpha \\
-pk\sin^{2}\alpha + r(1+k\sin\alpha\cos\alpha) & -qk\sin^{2}\alpha+t (1+k\sin\alpha\cos\alpha)
\end{pmatrix} \\
\Leftrightarrow && \begin{cases} p(1-k\sin\alpha\cos\alpha) + q(-k\sin^{2}\alpha) &= p(1-k\sin\alpha\cos\alpha) + rk\cos^{2}\alpha \\
pk\cos^{2}\alpha + q(1+k\sin\alpha\cos\alpha) &=q(1-k\sin\alpha\cos\alpha) + tk\cos^{2}\alpha \\
r(1-k\sin\alpha\cos\alpha) + t(-k\sin^{2}\alpha) &=-pk\sin^{2}\alpha + r(1+k\sin\alpha\cos\alpha) \\
rk\cos^{2}\alpha + t(1+k\sin\alpha\cos\alpha) &= -qk\sin^{2}\alpha+t (1+k\sin\alpha\cos\alpha)
\end{cases} \\
\Leftrightarrow && \begin{cases} -q\tan^{2}\alpha &= r \\
p\cos^{2}\alpha + q\sin\alpha\cos\alpha &=-q\sin\alpha\cos\alpha + t\cos^{2}\alpha \\
-r\sin\alpha\cos\alpha + -t\sin^{2}\alpha &=-p\sin^{2}\alpha + r\sin\alpha\cos\alpha \\
r &= -q\tan^{2}\alpha
\end{cases} \\
\Leftrightarrow && \begin{cases} -q\tan^{2}\alpha &= r \\
2q\sin\alpha\cos\alpha &=(t-p)\cos^{2}\alpha \\
(p-t)\sin^{2}\alpha &=2r\sin\alpha\cos\alpha
\end{cases} \\
\Leftrightarrow && \begin{cases} -q\tan^{2}\alpha &= r \\
2q\tan \alpha &=(t-p)
\end{cases} \\
\end{align*}
as required
