Solving for an matrix containing complex numbers
$$Ax = 0$$
$$A x= begin{bmatrix}-6 -2i & 5\0& 0end{bmatrix} begin{bmatrix}x_1\x_2end{bmatrix} = begin{bmatrix}0\0end{bmatrix}$$
Simple question, but how do I solve for $x_1$ and $x_2$?
One potential answer is:
$$begin{bmatrix}x_1\x_2end{bmatrix} = begin{bmatrix}-1+2i\-2+2iend{bmatrix}$$
I'm not sure how to get a result like this with complex numbers though. Can I get a detailed explanation on how to do this?
linear-algebra complex-numbers
edited Nov 30 '18 at 3:40
Robert Howard
1,9161822
asked Nov 30 '18 at 2:57
pylabpylab
103
add a comment |
$$Ax = 0$$
$$A x= begin{bmatrix}-6 -2i & 5\0& 0end{bmatrix} begin{bmatrix}x_1\x_2end{bmatrix} = begin{bmatrix}0\0end{bmatrix}$$
Simple question, but how do I solve for $x_1$ and $x_2$?
One potential answer is:
$$begin{bmatrix}x_1\x_2end{bmatrix} = begin{bmatrix}-1+2i\-2+2iend{bmatrix}$$
I'm not sure how to get a result like this with complex numbers though. Can I get a detailed explanation on how to do this?
linear-algebra complex-numbers
edited Nov 30 '18 at 3:40
Robert Howard
1,9161822
asked Nov 30 '18 at 2:57
pylabpylab
103
add a comment |
$$Ax = 0$$
$$A x= begin{bmatrix}-6 -2i & 5\0& 0end{bmatrix} begin{bmatrix}x_1\x_2end{bmatrix} = begin{bmatrix}0\0end{bmatrix}$$
Simple question, but how do I solve for $x_1$ and $x_2$?
One potential answer is:
$$begin{bmatrix}x_1\x_2end{bmatrix} = begin{bmatrix}-1+2i\-2+2iend{bmatrix}$$
I'm not sure how to get a result like this with complex numbers though. Can I get a detailed explanation on how to do this?
linear-algebra complex-numbers
edited Nov 30 '18 at 3:40
Robert Howard
1,9161822
asked Nov 30 '18 at 2:57
pylabpylab
103
$$Ax = 0$$
$$A x= begin{bmatrix}-6 -2i & 5\0& 0end{bmatrix} begin{bmatrix}x_1\x_2end{bmatrix} = begin{bmatrix}0\0end{bmatrix}$$
Simple question, but how do I solve for $x_1$ and $x_2$?
One potential answer is:
$$begin{bmatrix}x_1\x_2end{bmatrix} = begin{bmatrix}-1+2i\-2+2iend{bmatrix}$$
I'm not sure how to get a result like this with complex numbers though. Can I get a detailed explanation on how to do this?
linear-algebra complex-numbers
linear-algebra complex-numbers
edited Nov 30 '18 at 3:40
Robert Howard
1,9161822
asked Nov 30 '18 at 2:57
pylabpylab
103
edited Nov 30 '18 at 3:40
Robert Howard
1,9161822
asked Nov 30 '18 at 2:57
pylabpylab
103
edited Nov 30 '18 at 3:40
Robert Howard
1,9161822
edited Nov 30 '18 at 3:40
Robert Howard
1,9161822
edited Nov 30 '18 at 3:40
Robert Howard
1,9161822
1,9161822
asked Nov 30 '18 at 2:57
pylabpylab
103
asked Nov 30 '18 at 2:57
pylabpylab
103
asked Nov 30 '18 at 2:57
pylabpylab
103
103
add a comment |
add a comment |
1 Answer
1
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oldest
votes
Let $x_1=-6+2i$
$$(-6-2i)(-6+2i)+5x_2=0$$
$$36-(2i)^2+5x_2=0$$
$$40+5x_2=0$$
$$x_2=-8$$
answered Nov 30 '18 at 3:02
Siong Thye GohSiong Thye Goh
100k1465117
How would it be possible to get $x1$ and $x2$ in the form $a+bi$?
– pylab
Nov 30 '18 at 3:24
what do you mean? isn't $x_1=-6+2i$ and $x_2=-10 + 0i$ in that form?
– Siong Thye Goh
Nov 30 '18 at 3:53
If I substitute in the values $-6 + 2i$ and $-10$ into $x1$ and $x2$, I get the matrix $<-10, 0>$, but it should be $<0, 0>$
– pylab
Nov 30 '18 at 4:07
Hi, I made a mistake earlier.
– Siong Thye Goh
Nov 30 '18 at 4:24
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let $x_1=-6+2i$
$$(-6-2i)(-6+2i)+5x_2=0$$
$$36-(2i)^2+5x_2=0$$
$$40+5x_2=0$$
$$x_2=-8$$
answered Nov 30 '18 at 3:02
Siong Thye GohSiong Thye Goh
100k1465117
How would it be possible to get $x1$ and $x2$ in the form $a+bi$?
– pylab
Nov 30 '18 at 3:24
what do you mean? isn't $x_1=-6+2i$ and $x_2=-10 + 0i$ in that form?
– Siong Thye Goh
Nov 30 '18 at 3:53
If I substitute in the values $-6 + 2i$ and $-10$ into $x1$ and $x2$, I get the matrix $<-10, 0>$, but it should be $<0, 0>$
– pylab
Nov 30 '18 at 4:07
Hi, I made a mistake earlier.
– Siong Thye Goh
Nov 30 '18 at 4:24
add a comment |
Let $x_1=-6+2i$
$$(-6-2i)(-6+2i)+5x_2=0$$
$$36-(2i)^2+5x_2=0$$
$$40+5x_2=0$$
$$x_2=-8$$
answered Nov 30 '18 at 3:02
Siong Thye GohSiong Thye Goh
100k1465117
How would it be possible to get $x1$ and $x2$ in the form $a+bi$?
– pylab
Nov 30 '18 at 3:24
what do you mean? isn't $x_1=-6+2i$ and $x_2=-10 + 0i$ in that form?
– Siong Thye Goh
Nov 30 '18 at 3:53
If I substitute in the values $-6 + 2i$ and $-10$ into $x1$ and $x2$, I get the matrix $<-10, 0>$, but it should be $<0, 0>$
– pylab
Nov 30 '18 at 4:07
Hi, I made a mistake earlier.
– Siong Thye Goh
Nov 30 '18 at 4:24
add a comment |
Let $x_1=-6+2i$
$$(-6-2i)(-6+2i)+5x_2=0$$
$$36-(2i)^2+5x_2=0$$
$$40+5x_2=0$$
$$x_2=-8$$
answered Nov 30 '18 at 3:02
Siong Thye GohSiong Thye Goh
100k1465117
Let $x_1=-6+2i$
$$(-6-2i)(-6+2i)+5x_2=0$$
$$36-(2i)^2+5x_2=0$$
$$40+5x_2=0$$
$$x_2=-8$$
answered Nov 30 '18 at 3:02
Siong Thye GohSiong Thye Goh
100k1465117
edited Nov 30 '18 at 4:24
answered Nov 30 '18 at 3:02
Siong Thye GohSiong Thye Goh
100k1465117
answered Nov 30 '18 at 3:02
Siong Thye GohSiong Thye Goh
100k1465117
answered Nov 30 '18 at 3:02
Siong Thye GohSiong Thye Goh
100k1465117
100k1465117
How would it be possible to get $x1$ and $x2$ in the form $a+bi$?
– pylab
Nov 30 '18 at 3:24
what do you mean? isn't $x_1=-6+2i$ and $x_2=-10 + 0i$ in that form?
– Siong Thye Goh
Nov 30 '18 at 3:53
If I substitute in the values $-6 + 2i$ and $-10$ into $x1$ and $x2$, I get the matrix $<-10, 0>$, but it should be $<0, 0>$
– pylab
Nov 30 '18 at 4:07
Hi, I made a mistake earlier.
– Siong Thye Goh
Nov 30 '18 at 4:24
add a comment |
How would it be possible to get $x1$ and $x2$ in the form $a+bi$?
– pylab
Nov 30 '18 at 3:24
what do you mean? isn't $x_1=-6+2i$ and $x_2=-10 + 0i$ in that form?
– Siong Thye Goh
Nov 30 '18 at 3:53
If I substitute in the values $-6 + 2i$ and $-10$ into $x1$ and $x2$, I get the matrix $<-10, 0>$, but it should be $<0, 0>$
– pylab
Nov 30 '18 at 4:07
Hi, I made a mistake earlier.
– Siong Thye Goh
Nov 30 '18 at 4:24
How would it be possible to get $x1$ and $x2$ in the form $a+bi$?
– pylab
Nov 30 '18 at 3:24
How would it be possible to get $x1$ and $x2$ in the form $a+bi$?
– pylab
Nov 30 '18 at 3:24
what do you mean? isn't $x_1=-6+2i$ and $x_2=-10 + 0i$ in that form?
– Siong Thye Goh
Nov 30 '18 at 3:53
what do you mean? isn't $x_1=-6+2i$ and $x_2=-10 + 0i$ in that form?
– Siong Thye Goh
Nov 30 '18 at 3:53
If I substitute in the values $-6 + 2i$ and $-10$ into $x1$ and $x2$, I get the matrix $<-10, 0>$, but it should be $<0, 0>$
– pylab
Nov 30 '18 at 4:07
If I substitute in the values $-6 + 2i$ and $-10$ into $x1$ and $x2$, I get the matrix $<-10, 0>$, but it should be $<0, 0>$
– pylab
Nov 30 '18 at 4:07
Hi, I made a mistake earlier.
– Siong Thye Goh
Nov 30 '18 at 4:24
Hi, I made a mistake earlier.
– Siong Thye Goh
Nov 30 '18 at 4:24
add a comment |
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