# Math Help - 2 quick matrix questions.

1. ## 2 quick matrix questions.

1) let
A= (a b)
.....(c d)
with none of a,b,c,d =0

If A is such that ad-bc=0, show the matrix equation AX+XA=0 has a solution X with X a non-zero 2x2 matrix that relies on 1 parameter only.

Ive written X as w,x,y,z and multiplied out then set the resulting equations in w,x,y,z to 0. It takes a few steps to get a relation between y and z which leads to relation between x,y and z then finally w,x,y and Z, so z=P is enough. My concern is that i dont use ad-bc=0 anywhere. what have i missed ? im guessing if A inverse exists we only get X=0??

2) a transformation im 3d space takes (a,b,c) to (x,y,z) where

(x)=(0 0 1)(a)
(y)=(1 0 0)(b)
(z)=(0 1 0)(c)

this transformation leaves distance between 2 points the same and leaves unaltered the points of the line x=y=z.

Assuming the transformation is a rotation about a line,find the angle of rotation.

the answer is 120 degrees but i cant find a decent way of explaining why. Im guessing the line of rotation is x=y=z and as lengths stay the same when i look at (1,0,0) which goes to (0,1,0) i get 2 equilateral triangles,one either side of the line.

2. [QUOTE=jiboom;444343]1) let
A= (a b)
.....(c d)
with none of a,b,c,d =0

If A is such that ad-bc=0, show the matrix equation AX+XA=0 has a solution X with X a non-zero 2x2 matrix that relies on 1 parameter only.

Ive written X as w,x,y,z and multiplied out then set the resulting equations in w,x,y,z to 0. It takes a few steps to get a relation between y and z which leads to relation between x,y and z then finally w,x,y and Z, so z=P is enough. My concern is that i dont use ad-bc=0 anywhere. what have i missed ? im guessing if A inverse exists we only get X=0??[./quote]
You don't say HOW you solved those equations, eliminating w and x, and that is the crucial point. If you were to put your expression for x, y, and z back into the equations, you would get an equation in z only. What prevents you from solving for z?

2) a transformation im 3d space takes (a,b,c) to (x,y,z) where

(x)=(0 0 1)(a)
(y)=(1 0 0)(b)
(z)=(0 1 0)(c)

this transformation leaves distance between 2 points the same and leaves unaltered the points of the line x=y=z.

Assuming the transformation is a rotation about a line,find the angle of rotation.

the answer is 120 degrees but i cant find a decent way of explaining why. Im guessing the line of rotation is x=y=z and as lengths stay the same when i look at (1,0,0) which goes to (0,1,0) i get 2 equilateral triangles,one either side of the line.
You don't need to guess! Obviously, the only points not changed by a rotation (that does change points- not a rotation by a multiple of $2\pi$) are points on the axis of rotation. Since x= y= z is not changed, it is the axis. Now, applying that transformation to (1, 0, 0) gives (0, 1, 0). Further, the plane perpendicular to x= y= z. containing (1, 0, 0) is (x- 1)+ y+ z= x+ y+ z= 1 and, yes, (0, 1, 0) is in that plane. The line x= y= z crosses that plane when x+ x+ x= 3x= 1 or x= y= z= 1/3. A vector from (1/3, 1/3, 1/3) to (1, 0, 0) is <2/3, -1/3, -1/3> and a vector from (1/3, 1/3, 1/3) to (0, 1, 0) is (-1/3, 2/3, -1/3). The angle between those two vectors can be found from their dot product and is the angle of rotation.

3. with

X=(w x)
....(y z)

i get

2aw+cx+by=0...(1)
bw+(a+d)x+bz=0..(2)
cw+(a+d)y+cz=0..(3)
cx+by+2dz=0...(4)

dz=aw so z=aw/d

(2) and (4) lead to, using z=aw/d

[bw/d]+x=0

[cw/d]+y=0

so letting w=T say solution is

w=T
x=-bT/a
y=-cT/a
z=aT/d

so only need parameter T

I cant see an error, so where does ad-dc=0 come into play here? The first part had the condition a=d=0 which once used gave the need for 2 paramters.

Thank-you for the vector solution. Should have got my hands dirty and the done the math. I was just looking at a pic hoping to spot a nice geometric reason. Can you spot one ?