Given that M : (x, y) \(\to\) (7x, 3x - y) and N : (x, y) \(\to\) (2x - y; 5x + 3y)
(c) find the coordinates of the point Q whose image is Q(2, 4) under the linear transformation N
(a) Reading the images off column by column:
\[M=\begin{pmatrix}7&0\\3&-1\end{pmatrix},\qquad N=\begin{pmatrix}2&-1\\5&3\end{pmatrix}\]
(b) Image of \(P(2,-3)\) under N then M.
Apply \(N\): \((2x-y,\ 5x+3y)=(2(2)-(-3),\ 5(2)+3(-3))=(7,\ 1)\).
Apply \(M\) to \((7,1)\): \((7x,\ 3x-y)=(7(7),\ 3(7)-1)=(49,\ 20)\).
Image is \((49,\ 20)\).
(c) Point whose image under N is \((2,4)\). Solve \(N(x,y)=(2,4)\):
\[2x-y=2,\qquad 5x+3y=4\]
From the first, \(y=2x-2\). Substituting: \(5x+3(2x-2)=4\Rightarrow11x-6=4\Rightarrow x=\dfrac{10}{11}\).
\[y=2\left(\tfrac{10}{11}\right)-2=-\tfrac{2}{11}\]
The point is \(\left(\dfrac{10}{11},\ -\dfrac{2}{11}\right)\).