- #1
vilhelm
- 37
- 0
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.
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.