Matrices

Hoempa

Math Team
Apr 2010
2,780
361
Overkill:
Let \(\displaystyle A = \[a \ b \\ c \ d \]\)

Then \(\displaystyle A^2 = \[a^2 + bc \ b(a + d) \\ c(a + d) \ bc + d^2 \] = \[4 \ 0 \\ 0 \ 4 \]\)

Which gives
a^2 + bc = 4 [1]
b(a + d) = 0 [2]
c(a + d) = 0 [3]
d^2 + bc = 4 [4]
[1]-[4]: a^2 - d^2 = 0 so \(\displaystyle a = d \vee a = -d\)

a = d gives with [2] and [3]: \(\displaystyle b = c = 0\)
\(\displaystyle a^2 = 4 \rightarrow a = -2 \vee 2\)

So \(\displaystyle A = \[-2 \ 0 \\ 0 \ -2 \] \vee A = \[2 \ 0 \\ 0 \ 2 \]\)
Which might also be found by using \(\displaystyle A^2 = 4I\)

Now we have a = -d
which gives 4-a^2 = bc (and a + d = 0)
So
\(\displaystyle \large A = \begin{bmatrix}
a & b\\
\frac{4 - a^2}{b} & -a
\end{bmatrix}\) (for non-zero b)
For a = -d, b = 0, we find
\(\displaystyle \large A = \begin{bmatrix}
2 & 0\\
c & -2
\end{bmatrix}\) ,\(\displaystyle \large A = \begin{bmatrix}
2 & c\\
0 & -2
\end{bmatrix}\), \(\displaystyle \large A = \begin{bmatrix}
-2 & 0\\
c & 2
\end{bmatrix}\) and \(\displaystyle \large A = \begin{bmatrix}
-2 & c\\
0 & 2
\end{bmatrix}\)
 

agentredlum

Math Team
Jul 2011
3,372
234
North America, 42nd parallel
Hey! That's pretty good!

You found an infinite number of distinct matrices. I think you found them all.

:D
 

Hoempa

Math Team
Apr 2010
2,780
361
Thanks 8)
By the way, for the first problem, \(\displaystyle A^2+B^2=\left( \begin{array}{cc}2&3\\3&2\end{array}\right)\),

I tried to let
\(\displaystyle A = \left( \begin{array}{cc}a&b\\c&d\end{array}\right)\) and \(\displaystyle B = \left( \begin{array}{cc}e&f\\g&h\end{array}\right)\)
and to express e, f, g and h in terms of a, b, c and d without any luck so far.
 

agentredlum

Math Team
Jul 2011
3,372
234
North America, 42nd parallel
I think you read a post that i deleted part of because of me not paying attention enough to your formulas. Disregard me saying there are two more please.

Yes , i also tried to set up 8 variables for the OP but it turned into a mess so i switched focus and re-wrote the problem using the SUM to my advantage , like this ,..

\(\displaystyle A^2 \ = \ \[2 \ 3 \\ 0 \ 0 \]\)

\(\displaystyle B^2 \ = \ \[0 \ 0 \\ 3 \ 2 \]\)

From here it was easy to get an answer to the OP , so , now you know my secret.

:D
 

Hoempa

Math Team
Apr 2010
2,780
361
I came back to see what solutions I had and saw that you had edited your post so I edited the part that said I'd look further. No worries!

Cool secret. I'll try my plan again but you have inspired me to look at

\(\displaystyle A^2 \ = \ \[a \ b \\ c \ d \]\)

\(\displaystyle B^2 \ = \ \[2-a \ 3-b \\ 3-c \ 2-d \]\)

and see what happens. However, I have some homework..
 

agentredlum

Math Team
Jul 2011
3,372
234
North America, 42nd parallel
Hoempa said:
...However, I have some homework..
:lol:

After you're done , take a swig of this , captain.


[attachment=0:1aa31jm4]131_metaxa.jpg[/attachment:1aa31jm4]


There is a 3 star , a 5 star and a 7 star. I like the 5 star the best , smoothest brandy i've ever had.

:D
 

Attachments

Hoempa

Math Team
Apr 2010
2,780
361
Ah, good idea while playing a round of cards with my roommates :p 8)
 

agentredlum

Math Team
Jul 2011
3,372
234
North America, 42nd parallel
Have you ever seen anyone multiply matrices from right to left instead of from left to right?

1) \(\displaystyle \ \ \[0 \ 1 \\ 1 \ 0 \] \ = \ \[1 \ 0 \\ 0 \ 1 \]\) « \(\displaystyle \[1 \ 0 \\ 0 \ 1 \]\)

2) \(\displaystyle \ \ \[0 \ 1 \\ 1 \ 0 \] \ = \ \[0 \ 1 \\ 1 \ 0 \]\) « \(\displaystyle \[0 \ 1 \\ 1 \ 0 \]\)

So that way the answers on the LHS have square roots? I've introduced the « because we have to go strictly from the RIGHT. So , using 1) , to get the entry 1 in the \(\displaystyle \ a_{1,2} \\) position we would do

(first row of matrix on the right of «)×(second column of matrix on the left of «) =

0*0 + 1*1 = 1

We go right to left and top to bottom , contrary to the usual way where we go left to right and top to bottom.

:D
 

agentredlum

Math Team
Jul 2011
3,372
234
North America, 42nd parallel
agentredlum said:
Have you ever seen anyone multiply matrices from right to left instead of from left to right?

1) \(\displaystyle \ \ \[0 \ 1 \\ 1 \ 0 \] \ = \ \[1 \ 0 \\ 0 \ 1 \]\) « \(\displaystyle \[1 \ 0 \\ 0 \ 1 \]\)

2) \(\displaystyle \ \ \[0 \ 1 \\ 1 \ 0 \] \ = \ \[0 \ 1 \\ 1 \ 0 \]\) « \(\displaystyle \[0 \ 1 \\ 1 \ 0 \]\)

So that way the answers on the LHS have square roots? I've introduced the « because we have to go strictly from the RIGHT. So , using 1) , to get the entry 1 in the \(\displaystyle \ a_{1,2} \\) position we would do

(first row of matrix on the right of «)×(second column of matrix on the left of «) =

0*0 + 1*1 = 1

We go right to left and top to bottom , contrary to the usual way where we go left to right and top to bottom.

:D
I have started a discussion about this at the following link.

http://math.stackexchange.com/questions ... redirect=1

My thanx go to anyone who shows an interest in this matter.

:D
 

Hoempa

Math Team
Apr 2010
2,780
361
Hey Agent!

Maybe we should set up a formula for your definition.

The usual matrixmultiplication A*B where A has the same amount of columns as B has rows, gives the product, say C.
A has \(\displaystyle m_A\) rows and \(\displaystyle n_A\) columns, B has \(\displaystyle m_B\) rows and \(\displaystyle n_B\) columns.
Let \(\displaystyle a_{i,j}\) be the element in the i-th row and j-th column, where \(\displaystyle 1 \le i \le m_a\) and \(\displaystyle 1 \le j \le n_a\) so \(\displaystyle A = [a_{i,j}]\) and likewise for B. C will have \(\displaystyle m_A\) rows and \(\displaystyle n_B\) columns, so C consists of elements c_{i,j} where \(\displaystyle 1 \le i \le m_A\) and \(\displaystyle 1 \le j \le n_B\)
Now we must have \(\displaystyle n_A = m_B\).

then \(\displaystyle c_{i,j} = \sum_{k = 1}^{n_A} a_{i,k} \cdot b_{k, j}\)

Now for your product, C = A<<B we seem to have

C with \(\displaystyle m_B\) rows and \(\displaystyle n_A\) columns requiring \(\displaystyle n_B = m_A\) such that
\(\displaystyle c_{i,j} = \sum_{k=1}^{n_B} b_{(i, n_{B}+1-k)} \cdot A_{k, j}\).

Do you get the same result?