Another complex system of linear equations

1. Jan 17, 2005

3.14159265358979

I guess i didn't get the last one completely because i've been having a hard time with this one. Solve

3x + iy + (2+i)z = 3i

-ix + y + z = 1

x + y + (2+i)z = i

i've tried dividing through by the leading coefficent, and re-arranging it...but can't seem to get the right answer! Thanks again for any help you can provide.

2. Jan 17, 2005

cronxeh

$$\left( \begin{array}{xyzk} 3 & i & 2+i & 3i\\ -i & 1 & 1 & 1\\ 1 & 1 & 2+i & 1 \end{array} \right)$$$$\ = \left( \begin{array}{xyzk} 3 & i & 2+i & 3i\\ 1 & i & i & i\\ 1 & 1 & 2+i & 1 \end{array} \right)$$
Multiply Row 2 by i, remember you can multiply by any constant - and i is just $$\sqrt{-1}$$. i*i = -1, i*i*i = -i, i*i*i*i = 1

$$\ = \left( \begin{array}{xyzk} 2 & 0 & 2 & 2i\\ 1 & i & i & i\\ 1 & 1 & 2+i & 1 \end{array} \right)$$$$\ = \left( \begin{array}{xyzk} 2 & 0 & 2 & 2i\\ 0 & i-1 & -2 & i-1\\ 1 & 1 & 2+i & 1 \end{array} \right)$$$$\ = \left( \begin{array}{xyzk} 1 & 0 & 1 & i\\ 0 & i-1 & -2 & i-1\\ 1 & 1 & 2+i & 1 \end{array} \right)$$
R1 = R1 - R2 and R2 = R2 - R3 and R1 = 1/2 R1

$$\ = \left( \begin{array}{xyzk} 1 & 0 & 1 & i\\ -1 & i-1 & -3 & -1\\ 1 & 1 & 2+i & 1 \end{array} \right)$$$$\ = \left( \begin{array}{xyzk} 1 & 0 & 1 & i\\ 0 & i & i-1 & 0\\ 1 & 1 & 2+i & 1 \end{array} \right)$$$$\ = \left( \begin{array}{xyzk} 1 & 0 & 1 & i\\ 0 & -1 & -1-i & 0\\ 1 & 1 & 2+i & 1 \end{array} \right)$$
R2 = R2 - R1 and R2 = R2 + R3 and R2 = i*R2

$$\ = \left( \begin{array}{xyzk} 1 & 0 & 1 & i\\ 0 & -1 & -1-i & 0\\ 0 & 1 & 1+i & 1-i \end{array} \right)$$$$\ = \left( \begin{array}{xyzk} 1 & 0 & 1 & i\\ 0 & 0 & 0 & 1-i\\ 0 & 1 & 1+i & 1-i \end{array} \right)$$
R3 = R3 - R1 and R2 = R2 + R3

$$\ = \left( \begin{array}{xyzk} 1 & 0 & 1 & i\\ 0 & 0 & 0 & 1-i\\ 0 & 1 & 1+i & 1-i \end{array} \right)$$
You swap Row2 and Row3 and you get:

$$\ = \left( \begin{array}{xyzk} 1 & 0 & 1 & i\\ 0 & 1 & 1+i & 1-i\\ 0 & 0 & 0 & 1-i \end{array} \right)$$$$\ = \left( \begin{array}{xyzk} 1 & 0 & 1 & i\\ 0 & 1 & 1+i & 0\\ 0 & 0 & 0 & 0 \end{array} \right)$$
R2 = R2 - R3 and you can eliminate Col 4 of R3

And viola! You end up with:
x + z = i
y + (1+i)z = 0

3. Jan 17, 2005

Hurkyl

Staff Emeritus
Arithmetic errors tend to be common on these types of problems (not just by students!!). The added complexity (haha!) of complex arithmetic only makes the problem worse.

I've found that the only reliable method (other than a symbolic calculator!) is to do it, check the answer, and if it's not right, try again.

4. Jan 17, 2005

3.14159265358979

wow! thanks for the detailed solution! I need to practice these complex systems. Thanks!

5. Jan 17, 2005

3.14159265358979

here's what i got...

$$\left(\begin{array}{xyzk}3 & i & 2+i & 3i\\-i & 1 & 1 & 1\\1 & 1 & 2+i & 1\end{array}\right)$$

R2 = R2 * i
$$\left(\begin{array}{xyzk}3 & i & 2+i & 3i\\1 & i & i & i\\1 & 1 & 2+i & 1\end{array}\right)$$

R1 = R1 - R2
$$\left(\begin{array}{xyzk}2 & 0 & 2 & 2i\\1 & i & i & i\\1 & 1 & 2+i & 1\end{array}\right)$$

R2 = R2 - R3 and R1 = R1 / (2)
$$\left(\begin{array}{xyzk}1 & 0 & 1 & i\\0 & i-1 & -2 & i-1\\1 & 1 & 2+i & 1\end{array}\right)$$

R2 = R2 - R1
$$\left(\begin{array}{xyzk}1 & 0 & 1 & i\\-1 & i-1 & -3 & -1\\1 & 1 & 2+i & 1\end{array}\right)$$

R2 = R2 + R1
$$\left(\begin{array}{xyzk}1 & 0 & 1 & i\\0 & i-1 & -2 & -1+i\\1 & 1 & 2+i & 1\end{array}\right)$$

R3 = R3 - R1
$$\left(\begin{array}{xyzk}1 & 0 & 1 & i\\0 & i-1 & -2 & -1+i\\0 & 1 & 1+i & 1-i\end{array}\right)$$

R3 = R3 - [(-i-1)/2]R2
$$\left(\begin{array}{xyzk}1 & 0 & 1 & i\\0 & i-1 & -2 & -1+i\\0 & 0 & 0 & -i\end{array}\right)$$

Doesn't that mean there is no solution?

6. Jan 17, 2005

3.14159265358979

ah, the question was written down wrong in the matrix....it's suppose to be

$$\left(\begin{array}{xyzk}3 & i & 2+i & 3i\\-i & 1 & 1 & 1\\1 & 1 & 2+i & i\end{array}\right)$$

(notice the last element is an 'i' not a 1) Thanks for everyones reply...i'm going to work with this, and see where it gets me...

7. Jan 17, 2005

cronxeh

oh wow I'm sorry my mistake. Give me a minute to correct this

8. Jan 17, 2005

cronxeh

$$\left(\begin{array}{xyzk} 3 & i & 2+i & 3i\\ -i & 1 & 1 & 1\\ 1 & 1 & 2+i & i \end{array}\right)$$$$\left(\begin{array}{xyzk} 3 & i & 2+i & 3i\\ 1 & i & i & i\\ 1 & 1 & 2+i & i \end{array}\right)$$$$\left(\begin{array}{xyzk} 2 & 0 & 2 & 2i\\ 1 & i & i & i\\ 1 & 1 & 2+i & i \end{array}\right)$$$$\left(\begin{array}{xyzk} 1 & 0 & 1 & i\\ 0 & i-1 & -2 & 0\\ 1 & 1 & 2+i & i \end{array}\right)$$$$\left(\begin{array}{xyzk} 1 & 0 & 1 & i\\ 0 & i-1 & -2 & 0\\ 0 & 1 & 1+i & 0 \end{array}\right)$$

$$\left(\begin{array}{xyzk} 1 & 0 & 1 & i\\ 0 & i & -1+i & 0\\ 0 & 1 & 1+i & 0 \end{array}\right)$$$$\left(\begin{array}{xyzk} 1 & 0 & 1 & i\\ 0 & -1 & -i-1 & 0\\ 0 & 1 & 1+i & 0 \end{array}\right)$$$$\left(\begin{array}{xyzk} 1 & 0 & 1 & i\\ 0 & 0 & 0 & 0\\ 0 & 1 & 1+i & 0 \end{array}\right)$$$$\left(\begin{array}{xyzk} 1 & 0 & 1 & i\\ 0 & 1 & 1+i & 0\\ 0 & 0 & 0 & 0 \end{array}\right)$$

x + z = i -> x = i - z
y + (1+i)z = 0 -> y = -z - iz

Check (from original equation):
3x + iy + (2+i)z = 3i -> 3(i-z) + i(-z -iz) + (2+i)z = 3i -> 3i - 3z -iz +1z + 2z +iz = 3i -> 3i = 3i

-ix + y + z = 1 -> -i(i-z) + (-z-iz) + z = 1 -> 1+iz -z -iz + z = 1 -> 1 = 1

x + y + (2+i)z = i -> i-z -z -iz +(2+i)z = i -> i-2z -iz + 2z +iz = i -> i=i

9. Jan 17, 2005

3.14159265358979

thank you cronxeh, you've been very helpful!

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