MatrixProblem?

Azels

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Hi,I had a question in my book and I can't figure out the steps for this, may I please have someone explain the steps? Thanks :D

Given Matrices A=[a b] and B=[e f] use the properties of determinants to prove that if AB=0, then ad=bc.
...........................[c d]............[g h]

The dots aren't of any significance, just so the matrices line up
 
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Hi,I had a question in my book and I can't figure out the steps for this, may I please have someone explain the steps? Thanks :D

Given Matrices A=[a b] and B=[e f] use the properties of determinants to prove that if AB=0, then ad=bc.
...........................[c d]............[g h]

The dots aren't of any significance, just so the matrices line up
One of the properties of determinants is that
det(A B) = det(A) det(B)
If det(B) is not zero, what does that tell you about det(A) and the relationship of ad and bc.

BTW: Notice that the statement as it stands is not true. For example, if A is the identity matrix then AB=B and det(A) = 1.
 
Would that tell me that a=c and b=d?
therefore det AB could equal la al
.........................................lb bl ?
Also that one matrix could either be the 0 matrix or something that would be ad=eh and bc=fg?
 
Would that tell me that a=c and b=d?
therefore det AB could equal la al
.........................................lb bl ?
Also that one matrix could either be the 0 matrix or something that would be ad=eh and bc=fg?
Do you know what the determinate of a 2X2 matrix is? For example, if
A = \(\displaystyle \begin{pmatrix}
a\,\,b\\
c\,\,d
\end{pmatrix}\)
what is det(A)?
 
Thus
det(AB) = det (A) det(B) = (ad-bc) (eh -fg)
So, if det(B) [=(eh-fg)] is not zero, what does that tell you about det(A) [=(ad-bc)]? That is, if
u v = 0
and v is not zero, what is u?
 
that would mean that det(A)det(B) = adeh + -1adfg + -1bceh + bcfg
I'm not sure what v and u are meant to be
 
that would mean that det(A)det(B) = adeh + -1adfg + -1bceh + bcfg
I'm not sure what v and u are meant to be
If one number (u) times another number (v) is zero and one number (v) is not zero, what is the other number (u).
 
ok, I realise what went wrong so

for det(A)x(B)=0, one determinant must be 0
if we assume that B is not equal to zero, then A must equal 0
for A to be equal to zero, when calculating the determinant, it comes to (ad-bc)
and for this to equal 0, ad=bc as when they are subtracted, any other number wouldn't give the product of 0.
so therefore, for det(AxB)=0, where detB is not 0, A must equal 0 and therefore ad=bc?
 
ok, I realise what went wrong so

for det(A)x(B)=0, one determinant must be 0
if we assume that B is not equal to zero, then A must equal 0
for A to be equal to zero, when calculating the determinant, it comes to (ad-bc)
and for this to equal 0, ad=bc as when they are subtracted, any other number wouldn't give the product of 0.
so therefore, for det(AxB)=0, where detB is not 0, A must equal 0 and therefore ad=bc?
Yes
 
yay!! Thank you very much
but I've just realised that I made a terrible mistake when reading the question, it wasn't detAB=0, it was AB=the zero matrix!
using the determinant properties I was meant to prove that when AB=zero matrix, ad=bc. Can this be solved anyway close to the determinant way?
 
yay!! Thank you very much
but I've just realised that I made a terrible mistake when reading the question, it wasn't detAB=0, it was AB=the zero matrix!
using the determinant properties I was meant to prove that when AB=zero matrix, ad=bc. Can this be solved anyway close to the determinant way?
Start working through:
AB = \(\displaystyle \begin{pmatrix}a& b \\ c& d \end{pmatrix}\begin{pmatrix}e& f \\ g& h \end{pmatrix} = \begin{pmatrix}ae+bg& af+bh \\ ce+dg& cf+dh \end{pmatrix}= \begin{pmatrix}0& 0 \\ 0& 0 \end{pmatrix}\)
says
(1) ae + bg = 0
(2) af + bh = 0
(3) ce + dg = 0
(4) cf + dh = 0

Suppose neither e nor f were zero. Write a in terms of b, e, and g from (1) and in terms of b, f, and h from (2). That gives two equations for a which must be equal. What does that say about det(B).

Now suppose that a determinate is zero, i.e.
det(C) =0
where
C = \(\displaystyle \begin{pmatrix}i& j \\ k& l \end{pmatrix}\)
then
det(C) = i l - j k = 0
or
i l = j k
Suppose i and j were not zero, then
l/j = k/i = \(\displaystyle \alpha\)
or
l = \(\displaystyle \alpha\) j
and
k = \(\displaystyle \alpha\) i
Thus if det(B) is zero
g = \(\displaystyle \alpha\) e
and
h = \(\displaystyle \alpha\) f
If either i or j were zero, you would need to work out some similar relationship you could use.

Now take the above along with equations (3) and (4) and see what you can discover about det(A). You will also need to investigate the case of either or both e and f equal to zero.
 
ok, so lets say that n is any real number other than 0 and e and f are not equal to zero
(1) an+bg=0
(2) an+bh=0

Is that what you meant?

therefore det B equals
nh-ng?
I'm not sure what this says about det(B)
 
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