Question: The matrix M is given by M $\displaystyle = \begin{pmatrix} a & b\\ c & d \end{pmatrix} $, where $\displaystyle a,b,c,d \epsilon \mathbb{R} $.

a)Find $\displaystyle M^{2} $.

b)Given that $\displaystyle M^{2} = M $ and that b and c are non-zero, prove that M is singular.

c)Prove that in this case, the transformation T, is defined by $\displaystyle T:\begin{pmatrix} x\\ y \end{pmatrix} \mapsto M \begin{pmatrix} x\\ y \end{pmatrix} $

maps all points of the plane to points of the line $\displaystyle (1-a) x =by $

my attempt:

$\displaystyle M^2 = \begin{pmatrix} a^2 + bc & ab+bd \\ ac+dc & bc+d^2 \end{pmatrix} $

prove that M is singular

$\displaystyle \begin{pmatrix} a^2 + bc & ab+bd \\ ac+dc & bc+d^2 \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} $

if matrix is singular then $\displaystyle det(M) = 0 $

thus we need to prove that $\displaystyle det(M) = 0 $

$\displaystyle det(M) = ad -bc $

$\displaystyle a^2+bc=a......1 \\ ab+bd = b .....2\\ ac+dc=c ......3 \\ bc+d^2=d .....4 $

from 1

$\displaystyle bc = a- a^2 $

from 2

$\displaystyle ab+ bd =b $

$\displaystyle bd = b - ab $

$\displaystyle bd = b(1-a) $

$\displaystyle d = 1-a $

sub bc and d into $\displaystyle ad-bc $

thus

$\displaystyle a(1-a) - (a - a^2) = 0 $

since $\displaystyle ad -bc =0$ therefore M is singular

I need help with part c

please help