Cross product problem

**ahmedzoro10** · December 14th 2011, 12:54 PM

Hello,
I'm facing problems while trying to solve this problem

we have u and v and we need to find b that satisfies the following equation:

u x b = v

I've tried to solve it and i got only one coordinates for B and it we should get infinity number of vectors satisfies this equation up there

**skeeter** · December 14th 2011, 02:34 PM

Originally Posted by ahmedzoro10

Hello,
I'm facing problems while trying to solve this problem

we have u and v and we need to find b that satisfies the following equation:

u x b = v

I've tried to solve it and i got only one coordinates for B and it we should get infinity number of vectors satisfies this equation up there

it would be helpful to know what u and v are ...

**ahmedzoro10** · December 14th 2011, 11:51 PM

Originally Posted by skeeter

it would be helpful to know what u and v are ...

u= [-1 3 2] and v= [1 1 -1]

**Deveno** · December 15th 2011, 08:13 AM

if $b = (b_1,b_2,b_3)$ then $u \times b = v$ means that $(3b_3 - 2b_2,2b_1+b_3,-b_2-3b_1) = (1,1,-1)$ .

this is a system of 3 linear equations in 3 unknowns:

$\begin{bmatrix}0&-2&3\\2&0&1\\-3&-1&0\end{bmatrix} \begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix} = \begin{bmatrix}1\\1\\-1\end{bmatrix}$

can you continue?

**Plato** · December 15th 2011, 08:22 AM

Originally Posted by Deveno

if $b = (b_1,b_2,b_3)$ then $u \times b = v$ means that $(3b_3 - 2b_2,2b_1+b_3,-b_2-3b_1) = (1,1,-1)$ .

this is a system of 3 linear equations in 3 unknowns:

${\color{red}\begin{bmatrix}0&-2&3\\2&0&1\\-3&-1&0\end{bmatrix}} \begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix} = \begin{bmatrix}1\\1\\-1\end{bmatrix}$

Be careful:
That matrix is singular.

You might try rref.

**ahmedzoro10** · December 15th 2011, 09:10 AM

Originally Posted by Plato

Be careful:
That matrix is singular.

You might try rref.

I got it . Thanks alot

**ahmedzoro10** · December 15th 2011, 10:18 AM

I got A^(-1) = 0! is that right!
I've used adjoint method getting the inverse of A

**HallsofIvy** · December 15th 2011, 10:41 AM

Originally Posted by ahmedzoro10

I got A^(-1) = 0! is that right!
I've used adjoint method getting the inverse of A

No, that is not right. Plato told you that A does not have an inverse.

(And no invertible matrix has 0 as its inverse.)

**ahmedzoro10** · December 15th 2011, 10:55 AM

Originally Posted by HallsofIvy

No, that is not right. Plato told you that A does not have an inverse.

(And no invertible matrix has 0 as its inverse.)

oh sorry I haven't noticed his post saying "this matrices is singular"

**ahmedzoro10** · December 15th 2011, 11:11 AM

b will equal [ t , 1-3t , 1-2t ] assuming that b1 = t ,right?

**ahmedzoro10** · December 15th 2011, 11:38 AM

one question,please
Why we don't just put
3b3-2b2=1
2b1+b3=1
-3b1-b2=-1
and solve for this equations

**HallsofIvy** · December 15th 2011, 12:55 PM

You certainly can do that. In particular, If you multiply the last equation by 2 and subtract it from the first equation, you get $6b_1+ 3b_3= 3$ . From the second equation, $b_3= 1- 2b_1$ so $6b_1+ 3b_3= 6b_1+3- 6b_1= 3$ which is true for all x. Taking $b_1= t$ gives $b2= 1- 3t$ and $b_3= 1- 2t$ , exactly what you give above.

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