Matrix square root proof

[BezBog]

Hi
I need some help with a Matrix problem

I have to proof that

2	-2
1/2	0

and it's additive inverse, are the only square roots of

3	-4
1	-1

I have no idea how to even start that...

 

[Deleted member 4993]

Hi
I need some help with a Matrix problem

I have to proof that

2	-2
1/2	0

and it's additive inverse, are the only square roots of

3	-4
1	-1

I have no idea how to even start that...

Please tell us what you know about the additive inverse of a matrix?


Please share your work with us.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...217#post322217

We can help - we only help after you have shown your work - or ask a specific question (e.g. "are these correct?")

 

[stapel]

I have no idea how to even start that...

Maybe you might consider starting by writing down the additive inverse matrix, multiplying each of the matrices by itself, and verifying that each "square" is equal to the given matrix...?

 

[HallsofIvy]

Hi
I need some help with a Matrix problem

I have to proof that

2	-2
1/2	0

Do you mean $\displaystyle \begin{bmatrix}2 & -2 \\ \frac{1}{2} & 0 \end{bmatrix}$?

and it's additive inverse, are the only square roots of

3	-4
1	-1

I have no idea how to even start that...

Do you mean $\displaystyle \begin{bmatrix}3 & -4 \\ 1 & -1\end{bmatrix}$?

To show that "a" is a square root of "b", show that $\displaystyle a^2= b$.
(That's the easy part- showing that they are the only square roots is harder unless you already know that such a matrix cannot have more than two square roots.)

 

[BezBog]

Sorry for not explaning myself... I am new to such forums.

On the topic now... We have not studied roots in matrices, we have done invere matrix, determinant of a matrix, we have also used matrices to solve linear equations, ect... But as it is a bonus question it is prety impossible for us...

However here is what i have done...
The matrices in question are :
$\displaystyle \begin{bmatrix}2 & -2 \\ 1/2 & 0\end{bmatrix}$
$\displaystyle \begin{bmatrix}-2 & 2 \\ -1/2 & 0\end{bmatrix}$
and
$\displaystyle \begin{bmatrix}3 & -4 \\ 1 & -1\end{bmatrix}$

I have to show that the first two are the ONLY sqare roots of the third one.

I said that A^2=B

A=$\displaystyle \begin{bmatrix}a & b \\ c & d\end{bmatrix}$ and B=$\displaystyle \begin{bmatrix}3 & -4 \\ 1 & -1\end{bmatrix}$.

Then I arrived to 4 eq.

a²+bc=3
ac+cd=1
ba+bd=-4
bc+d²=-1

And I am now stuck on thys system, because i have never done someting like that... However i have solved it with WolframAlpha and I found that the matrices in question are the only possible answers to my equasions.

But in my work I have to show my resoning and solve the equations by myself... And now I am realy stuck....

 

[DrPhil]

I said that A^2=B

A=$\displaystyle \begin{bmatrix}a & b \\ c & d\end{bmatrix}$ and B=$\displaystyle \begin{bmatrix}3 & -4 \\ 1 & -1\end{bmatrix}$.

Then I arrived to 4 eq.

(1) a² + bc = 3
(2) ac + cd = -4
(3) ba + bd = 1
(4) bc + d² = -1

Are equations (2) and (3) reversed? I get
(1) a² + bc = 3
(2) ab + bd = -4
(3) ac + cd = 1
(4) bc + d² = -1
Then
(1)-(4) --> a^2 + d^2 = 4
(2) --> b(a + d) = -4
(3) --> c(a + d) = 1
(2)÷(3) --> b/c = -4
(2)×(3) --> bc (a + d)² = bc (4 + 2ad) = -4

Does this lead anywhere?

 

[BezBog]

Are equations (2) and (3) reversed? I get
(1) a² + bc = 3
(2) ab + bd = -4
(3) ac + cd = 1
(4) bc + d² = -1
Then
(1)-(4) --> a^2 + d^2 = 4
(2) --> b(a + d) = -4
(3) --> c(a + d) = 1
(2)÷(3) --> b/c = -4
(2)×(3) --> bc (a + d)² = bc (4 + 2ad) = -4

Does this lead anywhere?

Yes they were reversed, sorry about that
Thanks for your effort, but my level of mathematics is just not enough to solve that.. I tried to do something with those equations and still failed :-(
I need to get that a=+-2 b=+-2 c=+-1/2 and d=+-0
But I think that to get there i must buld a high degree equation with my 4 equations... and I just have no idea how



EDID: I got it thanks verry muach...:idea: