# Thread: matrices problem

1. ## matrices problem

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

2. ## Re: matrices problem

the image of $T$ is a straight line thru $(0,0)$

and

$\displaystyle T(0,1)=(b,1-a)$

3. ## Re: matrices problem

"b)Given that M^2=M and that b and c are non-zero, prove that M is singular."
From M^2= M, M^2- M= M(M-I)= 0. That is sufficient to show that M is singular.

4. ## Re: matrices problem

Originally Posted by Idea
the image of $T$ is a straight line thru $(0,0)$

and

$\displaystyle T(0,1)=(b,1-a)$
sorry but could you explain how you found that the image of $T$ is a straight line thru $(0,0)$

5. ## Re: matrices problem

Originally Posted by bigmansouf
sorry but could you explain how you found that the image of $T$ is a straight line thru $(0,0)$
ImT is a subspace of $R^2$ so it is either trivial, a line thru the origin ( since T(0,0)=(0,0) ),

or the whole plane depending on rank T = 0,1, or 2

T is singular since M is singular, so T is not surjective and $T \neq 0$ therefore

ImT = straight line i.e. dim ImT = 1 and ImT is spanned by the vector $T(0,1)=(b,1-a)$

in a coordinate system that is the line $(1-a)x=b y$