Proving Singular (2x2) Matrix Can Be Written in 2 Forms

[vilhelm]

I have to proove that a singular (2 x 2)-matrix can be written as

a b
ta tb

or

ta tb
a b

My attempt is not a real proof, and as I'm very inexperienced with writing proofs, maybe someone could write it, so that I will understand it in the future.

Attempt.

Let B =
a b
c d

From the definition, det(B)= ad-bc.

For a singular matrix, det(B) = 0.

Hence ad-bc=0 <=> ad=bc <=> a/c=b/d.

We have that one row is a multiple of the other.

If A=
a b
ta tb
then we have a/ta=b/tb <=> 1/t=1/t, and that's true for all real t > 0.

And if
A =
ta tb
a b
Then we have ta/a=tb/b <=> t=t,
which is true for all t,a,c.

 

[HallsofIvy]

The final portion of your proof is backwards.

Your are saying that if $A= \begin{bmatrix}a & b \\at & bt\end{bmatrix}$ then |A|= 0.

What you want to prove is the other way: if |A|= 0 then $A= \begin{bmatrix}a & b \\at & bt\end{bmatrix}$.

You are correct that if ad- bc= 0, then a/c= b/d. Let t= a/c= b/d.

 

[dikmikkel]

How about assuming the matrix is on the form:
$\left(\begin{array}{cc}<br /> a & b \\<br /> ta & tb\end{array}\right)$
Then the determinant is:
$atb - tab = 0,\forall t,a,b\in\mathbb{R}$
q.e.d. Can you work the other one out then?