# Math Help - Matrices question

1. ## Matrices question

Can somebody plz try to solve the matrices sum in the attached figure??

2. Originally Posted by snigdha
Can somebody plz try to solve the matrices sum in the attached figure??

This is not sum of matrices but product of matrices. Read carefully the definition of product of matrices.

Tonio

3. Originally Posted by tonio
This is not sum of matrices but product of matrices. Read carefully the definition of product of matrices.

Tonio
well tonio...u misunderstood me...by "sum" i mean to say "problem"... & not that sum which means addition..!!

4. Originally Posted by snigdha
well tonio...u misunderstood me...by "sum" i mean to say "problem"... & not that sum which means addition..!!

Well, is that a weird use of the word sum...!

Tonio

5. After multiplying this, we get,

$\begin{pmatrix}
x^2+y^2 &0\\
0 & -x^2-y^2
\end{pmatrix}
=
\begin{pmatrix}
1 &0\\
0 & -1\\
\end{pmatrix}$

Then, we have,

$x^2+y^2=1$

and all possible pairs are,

$(x,y)=(1,0), (0,1), (-1,0),(0,-1)$

6. Originally Posted by tonio
Well, is that a weird use of the word sum...!

Tonio
Apparently our British friends often use "sum" to mean any kind of mathematics problem.

Doing mathematics problems is "doing sums". If only it were that easy!

7. Originally Posted by kjchauhan
After multiplying this, we get,

$\begin{pmatrix}
x^2+y^2 &0\\
0 & -x^2-y^2
\end{pmatrix}
=
\begin{pmatrix}
1 &0\\
0 & -1\\
\end{pmatrix}$

Then, we have,

$x^2+y^2=1$

and all possible pairs are,

$(x,y)=(1,0), (0,1), (-1,0),(0,-1)$
Assuming that x and y are integers, yes. But that was not given in the problem. For x and y real, x can be any number such that $-1\le x\le 1$ and $y= \pm\sqrt{1- x^2}$

8. Originally Posted by HallsofIvy
Apparently our British friends often use "sum" to mean any kind of mathematics problem.
Even we, the Indians do the same..!!

9. Originally Posted by HallsofIvy
Assuming that x and y are integers, yes. But that was not given in the problem. For x and y real, x can be any number such that $-1\le x\le 1$ and $y= \pm\sqrt{1- x^2}$
Thats true.. Thanks.

10. Originally Posted by snigdha
Even we, the Indians do the same..!!
Yes. Unfortunately, the "Raj" taught you a perverted form of the language!