3x3 matrix with complex numbers

In summary, the conversation revolved around solving a problem involving imaginary numbers and variables. The participants discussed using different methods to find the determinant and solve for x and y. However, there were some errors in the calculations and further clarification was needed.
  • #1
CJovaras
3
2
Homework Statement
Solve for x and y if

| 1 1+2j 0 |
| 1-j -1 -1 | = 2j
| x 0 yj |


j represents imaginary numbers and x and y represent variables. My entire class is struggling with understanding this.
Relevant Equations
none
The attempt at a solution:

I tried the normal method to find the determinant equal to 2j. I ended up with:
2j = -4yj -2xj -2j -x +y

then I tried to see if I had to factorize with j so I didn't turn the j^2 into -1 and ended up with 2 different options:

1) 0= y(-4j-j^2) -x(2j-1) -2j

2) -2j(2y +x +1) +y -x

After that I get stuck, I've spent at least 8 hours staring at this problem and googling how to find a solution but nothing makes sense to me.
 
  • Like
Likes Pradeep1234
Physics news on Phys.org
  • #2
CJovaras said:
Homework Statement:: Solve for x and y if

| 1 1+2j 0 |
| 1-j -1 -1 | = 2j
| x 0 yj |j represents imaginary numbers and x and y represent variables. My entire class is struggling with understanding this.
Relevant Equations:: none

The attempt at a solution:

I tried the normal method to find the determinant equal to 2j. I ended up with:
2j = -4yj -2xj -2j -x +y
Let's see.
$$
\begin{align*}
\det\begin{pmatrix}1&1+2i&0\\1-i&-1&-1\\x&0&yi\end{pmatrix}&=-iy-(1+2i)((1-i)iy+x)\\
&=-iy-(1+2i)(iy+y+x)\\&=-iy-iy-y-x+2y-2iy-2ix\\&=(-x-y)+i(-2x-4y)
\end{align*}
$$
Looks as if one of us made a mistake. (I developed the determinant along the first row.)

Edit (for future readers): Correction (see discussion below):
the determinant is ##(-x+y)+i(-2x-4y)## since I have forgotten to add the ##+2y## term.
Here is explained how you can type formulas on PF: https://www.physicsforums.com/help/latexhelp/
CJovaras said:
then I tried to see if I had to factorize with j so I didn't turn the j^2 into -1 and ended up with 2 different options:

1) 0= y(-4j-j^2) -x(2j-1) -2j

2) -2j(2y +x +1) +y -x

After that I get stuck, I've spent at least 8 hours staring at this problem and googling how to find a solution but nothing makes sense to me.
 
Last edited:
  • Like
Likes Pradeep1234 and CJovaras
  • #3
fresh_42 said:
Let's see.
$$
\begin{align*}
\det\begin{pmatrix}1&1+2i&0\\1-i&-1&-1\\x&0&yi\end{pmatrix}&=-iy-(1+2i)((1-i)iy+x)\\
&=-iy-(1+2i)(iy+y+x)\\&=-iy-iy-y-x+2y-2iy-2ix\\&=(-x-y)+i(-2x-4y)
\end{align*}
$$
Looks as if one of us made a mistake. (I developed the determinant along the first row.)
You made a mistake too on the sign of the first y :smile:
$$
\ldots =
(-x+y)+i(-2x-4y)
$$
 
  • #4
fresh_42 said:
Here is explained how you can type formulas on PF: https://www.physicsforums.com/help/latexhelp/
@CJovaras -- you can also see the LaTeX that fresh_42 used for his math. Right-click your mouse over the LaTeX and "copy" it to your clipboard, then paste it into a *.txt file to study. :smile:
 
  • Like
Likes scottdave
  • #5
CJovaras said:
2) -2j(2y +x +1) +y -x
Assuming you get the correct equation here, you can then continue by comparing real and imaginary parts. For instance, you will get ##y-x = 0## by comparing the real parts. You can then use that to solve for ##x## using the imaginary parts.
 
  • Like
Likes PeroK
  • #6
DrClaude said:
You made a mistake too on the sign of the first y :smile:
$$
\ldots =
(-x+y)+i(-2x-4y)
$$
I did not. It is ## (-1) \cdot (-i) \cdot (iy) = (-1)^2 i^2 y = -y##
 
  • Like
Likes Pradeep1234
  • #7
fresh_42 said:
I did not.
fresh_42 said:
$$
\begin{align*}
\det\begin{pmatrix}1&1+2i&0\\1-i&-1&-1\\x&0&yi\end{pmatrix}&=-iy-(1+2i)((1-i)iy+x)\\
&=-iy-(1+2i)(iy+y+x)\\&=-iy-iy-y-x+2y-2iy-2ix\\&=(-x-y)+i(-2x-4y)
\end{align*}
$$
You have ##-y## and ##+2y## on the penultimate row:
$$
-iy-iy-y-x+2y-2iy-2ix = (-x+y)+i(-2x-4y)
$$
 
  • #8
fresh_42 said:
And you ignored the minus in front of ##(1+2i)##!
I don't want us to derail this thread, but
(1) the last two lines in post #2 do not form an equality, so one of them is incorrect

(2) when I calculate the determinant (independent of what you wrote) I get (-x+y)+i(-2x-4y)
 
  • Like
Likes scottdave
  • #9
DrClaude said:
I don't want us to derail this thread, but
(1) the last two lines in post #2 do not form an equality, so one of them is incorrect

(2) when I calculate the determinant (independent of what you wrote) I get (-x+y)+i(-2x-4y)
I didn't make a sign error! I lost a complete term (##2y##)! That's why I didn't understand you.
 
  • Like
Likes DrClaude
  • #10
berkeman said:
@CJovaras -- you can also see the LaTeX that fresh_42 used for his math. Right-click your mouse over the LaTeX and "copy" it to your clipboard, then paste it into a *.txt file to study. :smile:
thank you!!!
 
  • #11
CJovaras said:
Homework Statement:: Solve for x and y if

| 1 1+2j 0 |
| 1-j -1 -1 | = 2j
| x 0 yj |j represents imaginary numbers and x and y represent variables. My entire class is struggling with understanding this.
Relevant Equations:: none

The attempt at a solution:

I tried the normal method to find the determinant equal to 2j. I ended up with:
2j = -4yj -2xj -2j -x +y
You're very close here, but you have an extra term. The equation from the determinant simplifies to -4yj -2xj -x +y = 2j. Your equation has an extra term of -2j in it. Calculating the determinant is very tedious and you can easily make a mistake.
Check your determinant again.
CJovaras said:
then I tried to see if I had to factorize with j so I didn't turn the j^2 into -1 and ended up with 2 different options:
It's a mistake to not replace ##j^2## by -1.
CJovaras said:
1) 0= y(-4j-j^2) -x(2j-1) -2j

2) -2j(2y +x +1) +y -x

After that I get stuck, I've spent at least 8 hours staring at this problem and googling how to find a solution but nothing makes sense to me.
As it turns out, x and y have the same value, but I won't say more than that. Once I got my solution, I replaced x and y with the values I found, and calculated the determinant. My result was 2j, which verified that my solution was correct. As mentioned, calculating the determinant was tedious, so I wrote a Python program to do the grunt work for me, using the SymPy library.
 
  • #12
Mark44 said:
You're very close here, but you have an extra term. The equation from the determinant simplifies to -4yj -2xj -x +y = 2j. Your equation has an extra term of -2j in it. Calculating the determinant is very tedious and you can easily make a mistake.
Check your determinant again.It's a mistake to not replace ##j^2## by -1.

As it turns out, x and y have the same value, but I won't say more than that. Once I got my solution, I replaced x and y with the values I found, and calculated the determinant. My result was 2j, which verified that my solution was correct. As mentioned, calculating the determinant was tedious, so I wrote a Python program to do the grunt work for me, using the SymPy library.
Please tell me you're joking?!?! I found an answer where x and y were equal to one another in the first 30 minutes of working on the problem but I had assumed I had made a mistake! Thank you lol.
 
  • Like
Likes scottdave
  • #13
Wolfram can do this for you. Edit: I'm on my phone, hard to switch back-forth between sites.
 
  • #14
WWGD said:
Wolfram can do this for you. Edit: I'm on my phone, hard to switch back-forth between sites.
https://www.symbolab.com/solver/matrix-calculator is better. But I think it is important to know the formula instead of using websites. They should be used after people learned all about determinants, their computation included.
 
  • Like
Likes DaveE
  • #15
Now, a more interesting question to me is the interpretation of a Complex determinant. Can't use volume, area, etc. And linear dependence in Complexes is also different in a Complex vector space ( Complexes over themselves).
 
  • #16
WWGD said:
Now, a more interesting question to me is the interpretation of a Complex determinant. Can't use volume, area, etc. And linear dependence in Complexes is also different in a Complex vector space ( Complexes over themselves).
My thoughts about this are here:
https://www.physicsforums.com/insights/an-overview-of-complex-differentiation-and-integration/

The apparent ambiguity occurs only because we insist that volumes are real. Let them be complex and everything is fine. We have complex Graßmann algebras, too.

If we want real volumes, then forget about complex integration.
 
  • Like
Likes scottdave and WWGD
  • #17
Volumes are Real, but their names have been changed to protect the innocent;).
 
  • Haha
Likes topsquark
  • #18
How about an argument using Complexification, or Extension by Scalars?
 
  • #19
WWGD said:
How about an argument using Complexification, or Extension by Scalars?
I think this discussion now exceeds the limits of this thread tremendously.

Time to close the debate here.
 
  • Like
Likes DaveE

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
3
Views
129
  • Quantum Physics
Replies
3
Views
841
  • Precalculus Mathematics Homework Help
Replies
20
Views
970
  • Precalculus Mathematics Homework Help
Replies
10
Views
966
  • Precalculus Mathematics Homework Help
Replies
19
Views
1K
  • Precalculus Mathematics Homework Help
Replies
14
Views
388
  • Precalculus Mathematics Homework Help
Replies
2
Views
934
Replies
2
Views
3K
  • Quantum Physics
Replies
1
Views
839
  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
Back
Top