Vector transferring from the left side of the equation to the right

• Feb 7th 2007, 04:25 AM
ditzafar
Vector transferring from the left side of the equation to the right
I have equation from the form a*b=c. where a is a row vector of [1*4] b is a column vector of [4*1] and c is a number. I want to get equation of the forum b=…
How can I do it?
Thanks
Ditza
• Feb 7th 2007, 06:56 AM
ThePerfectHacker
Quote:

Originally Posted by ditzafar
I have equation from the form a*b=c. where a is a row vector of [1*4] b is a column vector of [4*1] and c is a number. I want to get equation of the forum b=…
How can I do it?
Thanks
Ditza

I do not think you can get an equation for b.
Becuase this is a dot product.
But when you have a matrix product,
\$\displaystyle A\bold{x}=\bold{b}\$
Then,
\$\displaystyle \bold{x}=A^{-1}\bold{b}\$
Assuming "A" is an invertible matrix.
• Feb 7th 2007, 07:07 AM
CaptainBlack
Quote:

Originally Posted by ditzafar
I have equation from the form a*b=c. where a is a row vector of [1*4] b is a column vector of [4*1] and c is a number. I want to get equation of the forum b=…
How can I do it?
Thanks
Ditza

Since u.v'=|u||v|cos(theta), where theta is the angle between u and v, there is no way of recovering v from the inner product.

RonL
• Feb 11th 2007, 05:48 AM
Rebesques
Well... I think it can be done if we look at it as matrix multiplications. If the entries of a are nonzero, then the (4*1) vector d with entries the reciprocals of the entries of a, satisfies a*d=4, so if we multiply the equation a*b=c by d and treat everything as matrices (and since the associative law holds in this case), we get d*a*b=c*d or b=(c/4)*d. This is only a specific solution; The dot product - as Captainblack noted - guarantees there are infinitely many.

Edit: In fact, four uknowns (the entries of b) minus one equation (ab=c --- this is only because of the dot product) we get three degrees of freedom for the solution. Will stop blabbering now.