(i) Since \left(\begin{matrix}a&b\\c&d\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right) = \left(\begin{matrix}x\\y\end{matrix}\right), we see that ax+by=x and cx+dy=y. These rearrange to give (a-1)x=-by and (d-1)y=-cx, whence (a-1)(d-1)xy = bcxy, which is exactly the desired equality.

We know that either x=0, y=0, or (a-1)(d-1)-bc=0. If x=0 (i.e. if L_1 is the vertical line along the y-axis) then our original equations tell us that by=0 and dy=y for all y, whence b=0 and d=1, and so (a-1)(d-1)=bc is satisfied. Similarly, if y=0 then cx=0 and ax=x for all x, and so (a-1)(d-1)=bc is satisfied. The last case trivially gives us the desired equality.

(An alternative (and much slicker) method can be given by writing \mathbf{A}\mathbf{x}=\mathbf{x} as \left[\left(\begin{matrix}a&b\\c&d\end{matrix}\right)-\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\right]\left(\begin{matrix}x\\y\end{matrix}\right) = \left(\begin{matrix}0\\0\end{matrix}\right) and arguing that the determinant of \left(\begin{matrix}a-1&b\\c&d-1\end{matrix}\right) must be zero.)

If L_1 does not pass through the origin then it is either the line x=\lambda (for some \lambda\neq0) or of the form y=mx+\lambda (for some \lambda\neq0). In the first case, expanding and rearranging \mathbf{A}\left(\begin{matrix}\lambda\\y\end{matrix}\right)=\left(\begin{matrix}\lambda\\y\end{matrix}\right) tells us that a\lambda+by=\lambda and c\lambda+dy=y, whence a=1, b=0, c=0, and d=1. In the second case, doing the same, but with \left(\begin{matrix}mx+\lambda\\y\end{matrix}\right), we get exactly the same result. That is, in either case, \mathbf{A} is the identity matrix.

(ii) First assume that b\neq0. Then \mathbf{A}\mathbf{x}=\mathbf{x} if and only if \mathbf{x}=\left(\begin{matrix}x\\y\end{matrix}\right) is such that ax+by=x and cx+dy=y. We already know that these equations rearrange to give (a-1)x+by=0 and (d-1)y+cx=0, so it would suffice to show that these two equations in fact describe the same line, since this line would then be exactly the line of invariant points. But if (a-1)x+by=0, then (a-1)(d-1)x+b(d-1)y=0, and so bcx+b(d-1)y=0 (from the aforementioned fact that the determinant of \mathbf{A}-\mathbf{I} is zero), whence (since b\neq0) cx+(d-1)y=0 (and similarly in the other direction).

Now assume that b=0. So either a=1 or d=1. If d=1, then \left(\begin{matrix}a&0\\c&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right) = \left(\begin{matrix}ax\\cx+y\end{matrix}\right) = \left(\begin{matrix}x\\y\end{matrix}\right) tells us that the line of invariant points is given by x=0. If a=1, then \left(\begin{matrix}1&0\\c&d\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right) = \left(\begin{matrix}x\\cx+dy\end{matrix}\right) = \left(\begin{matrix}x\\y\end{matrix}\right), which tells us that the line of invariant points is given by cx+(d-1)y=0.

N.B. for the next question, make sure to be careful with the difference between a line of invariant points, and an invariant line!

(iii) Since L_2 has the form y=mx+k and does not pass through the origin, we know that k\neq0. The invariance of L_2 under \mathbf{A} tells us that \left(\begin{matrix}a&b\\c&d\end{matrix}\right)\left(\begin{matrix}x\\mx+k\end{matrix}\right) = \left(\begin{matrix}x'\\mx'+k\end{matrix}\right) for some x'. That is, ax+b(mx+k)=x' and cx+d(mx+k)=mx'+k for all x. In particular then, for x=0, we see that x'=bk and dk=mx'+k, whence dk=mbk+k, and so (since k\neq0) d=mb+1. Then, for x=1, we see that x''=a+b(m+k) and c+d(m+k)=mx''+k, whence c+d(m+k)=m(a+b(m+k))+k, which simplifies to c=a(m-1). Substituting both of these equations (for c and for d) into m(a-1)(d-1) tells us that m(a-1)(d-1)=mbc.

If m\neq0, then we are done. If m=0, then the invariance property of L_2 tells us that x'=ax+bk and cx+dk=k for all x. In particular, for x=0, dk=k, whence d=1. For x=1, we see that c=0. Thus (a-1)(d-1)=bc is trivially satisfied.

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INCREDIBLE SCENES!! WELL DONE TIM!! :scream: :scream: :scream: :scream: :scream: :scream: :scream: :scream: :scream: :scream: :scream: