# Matrices

#### Hoempa

Math Team
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
Hey! That's pretty good!

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

#### Hoempa

Math Team
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
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.

#### Hoempa

Math Team
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
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.

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#### Hoempa

Math Team
Ah, good idea while playing a round of cards with my roommates 8)

#### agentredlum

Math Team
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.

#### agentredlum

Math Team
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.

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

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

#### Hoempa

Math Team
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?