\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
Since \(L_1\) is a line of invariant points, for each point \((x, y) \in L_1\), we have \[ \begin {pmatrix} a & b \\ c & d \end {pmatrix} \begin {pmatrix} x \\ y \end {pmatrix} = \begin {pmatrix} x \\ y \end {pmatrix}, \] and hence \[ ax + by = x, cx + dy = y. \]
Hence, \[ (1 - a) x = by, (1 - d)y = cx, \] and hence \[ (1 - a) x (1 - d)y = bycx, \] which simplifies to \[ [(a - 1)(d - 1) - bc] xy = 0. \]
If the line \(L_1\) is the line \(x = 0\), then \(by = 0\) for all \(y\) and \(dy = y\) for all \(y\), giving \(d = 1\) and \(b = 0\). Hence, \((a - 1) (d - 1) - bc = 0\).
Similarly, if the line \(L_1\) is the line \(y = 0\), then \(ax = x\) for all \(x\) and \(cx = 0\) for all \(y\), giving \(a = 1\) and \(c = 0\). Hence, \((a - 1) (d - 1) - bc = 0\).
Otherwise, there must be a point \((x, y) \in L_1\) such that \(xy \neq 0\), which means \((a - 1) (d - 1) - bc = 0\).
Hence, in all cases, we must have \((a - 1) (d - 1) = bc\) as desired.
If \(L_1\) does not pass through the origin, then \(y = mx + k\) for some \(k \neq 0\), or \(x = k\) for some \(k \neq 0\).
In the first case, we have \[ ax + b(mx + k) = x, \] and hence \[ (a + bm - 1)x + bk = 0 \] for all \(x\), meaning \(a + bm - 1 = 0\) and \(bk = 0\).
Similarly, \[ cx + d(mx + k) = mx + k, \] and hence \[ (c + dm - m)x + (d - 1)k = 0 \] for all \(x\), meaning \(c + dm - m = 0\) and \((d - 1)k = 0\).
Since \(k \neq 0\), \(bk = 0\) and \((d - 1)k = 0\) implies \(b = 0\) and \(d = 1\) respectively. Putting those back into the first corresponding equations, this solves to \(a = 1\) and \(c = 0\), which means \[ \vect {A} = \begin {pmatrix} 1 & \\ & 1 \end {pmatrix} = \vect {I}_2. \]
In the second case where \(x = k\) for some \(k \neq 0\), we have \[ ak + by = k, \] and hence \[ by + (a - 1)k = 0 \] for all \(y\), meaning \(b = 0\) and \((a - 1)k = 0\).
Similarly, \[ ck + dy = y, \] and hence \[ (d - 1)y + ck = 0 \] for all \(y\), meaning \(d - 1 = 0\) and \(ck = 0\).
Since \(k \neq 0\), \((a - 1)k = 0\) and \(ck = 0\) implies \(a = 1\) and \(c = 0\) respectively. Hence, \[ \vect {A} = \begin {pmatrix} 1 & 0 \\ 0 & 1 \end {pmatrix} = \vect {I}_2. \]
Therefore, \(L_1\) not passing through the origin must imply that \(\vect {A}\) is precisely the \(2\) by \(2\) identity matrix.
If \((x, y)\) is an invariant point, we have \[ (a - 1) x + by = 0, cx + (d - 1)y = 0. \]
If \(b = 0\), then \((a - 1)(d - 1) = bc = 0\), and hence \(a = 1\) or \(d = 1\).
In the case where \(a = 1\), the first equation is trivially true, and the second equation simplifies to \[ cx + (d - 1)y = 0, \] and hence the line \(L: cx + (d - 1)y = 0\) is a line of invariant points.
In the case where \(d = 1\), the original equation simplifies to \[ (a - 1)x = 0, cx = 0, \] and hence the line \(L: x = 0\) is a line of invariant points.
If \(b \neq 0\), we want to show that all points on the line \(L: (a - 1)x + by = 0\) satisfy the second equation. We multiply \((d - 1)\) on both sides of the equation, and hence \[ (a - 1)(d - 1)x + b (d - 1)y = 0, \] which is \[ bcx + b(d - 1)y = 0. \]
Since \(b \neq 0\), we divide \(b\) on both sides, giving \[ cx + (d - 1)y = 0, \] which is precisely the second equation. Hence, \(L: (a - 1)x + by = 0\) is a line of invariant points under this case.
We have \(L_2: y = mx + k\), \(k \neq 0\), we therefore have \[ \begin {pmatrix} a & b \\ c & d \end {pmatrix} \begin {pmatrix} x \\ mx + k \end {pmatrix} = \begin {pmatrix} X \\ mX + k \end {pmatrix}, \] and hence \[ ax + b(mx + k) = X, cx + d(mx + k) = mX + k. \]
Putting the first equation into the second one gives us \[ cx + d(mx + k) = m(ax + b(mx + k)) + k, \] which simplifies to \[ (c + dm - am - bm^2)x + (dk - mbk - k) = 0, \] which is \[ (bm^2 + (a - d)m - c)x + (mb - d + 1)k = 0. \]
Since this is true for all \(x\) and \(k \neq 0\), we must have \[ bm^2 + (a - d)m - c = 0, bm - d + 1 = 0. \]
If \(b = 0\), then \[ (a - d)m = c, d - 1 = 0, \] and hence \(d = 1\), \((a - 1)m = c\), and \[ (a - 1)(d - 1) = 0, bc = 0, \] which gives \[ (a - 1)(d - 1) = bc. \]
If \(b \neq 0\), the second of those equations solve to \[ m = \frac {d - 1}{b}, \] and putting this back into the first equation, we have \[ b \cdot \frac {(d - 1)^2}{b^2} + \frac {(a - d)(d - 1)}{b} - c = 0, \] and multiplying both sides by \(b\) gives \[ (d - 1)^2 + (a - d)(d - 1) = bc, \] and hence \[ (a - 1)(d - 1) = bc. \]
Therefore, in both cases, we have \((a - 1)(d - 1) = bc\), as desired.