I have 2 vectors that are scalar multiples of each other but their components have an unknown 'z'. ie
v1=[a, bz+c, e-zd ]
v2=[e, fz+g, h+z]
Only z is a variable/unknown, the rest are just numeric coefficients. How do i solve this using matlab??
I tried using linsolve but it doesnt seem to support symbolic arguments (z).
On
Your system of equations can be written as
$$
V =
\left ( \begin{array}{ccc}
a & bz+c & e-zd \\
e & fz+g & h+z
\end{array} \right )
$$
where $V$ is a matrix V = [v1 ; v2].
The first column of $V$ as well as of the right-hand side are irrelevant to finding $z$ (though the entries should of course match). Using $V'$ for Vp = V(:, 2 : 3) we get
$$
V' =
\left ( \begin{array}{ccc}
bz+c & e-zd \\
fz+g & h+z
\end{array} \right )
=
\left ( \begin{array}{ccc}
b & -d \\
f & 1
\end{array} \right )
~ z +
\left ( \begin{array}{ccc}
c & e \\
g & h
\end{array} \right ).
$$
This matrix equation represents four different equations for $z$ which may or may not have the same solution. If not, there is no general solution.
The four separate solutions can be found in Matlab using element-wise division:
(Vp - [c e ; g h]) ./ [b -d ; f 1]
A solution exists if the four elements of the value of this expression are all the same.
On
You can easily do this entirely symbolically in Matlab if your wish. Not with linsolve, but with solve:
syms a b c d e f g h z k
v1 = [a b*z e-z*d];
v2 = [e f*z+g h+z];
s = solve(v1==k*v2)
It's possible since you seem to be using an older version of Matlab, that you may need to to call solve using the old string format:
s = solve('a=k*e','b*z=k*(f*z+g)','e-z*d=k*(h+z)')
In both case you'll see that solve returns three outputs as there were three equations.
$k$ and $z$ can be found fairly quickly by hand. However, since you are interested in formulating it using Matlab, let us consider that instead. The equations to be solved are
\begin{align} a &= k e \\ b z + c &= k (f z + g) \\ e - z d &= k (h + z), \end{align}
where $k$ is some unknown constant of proportionality. Note: Since $k$ and $z$ are both unknowns, this is no longer a system of linear equations; instead it is a system of nonlinear equations. To formulate this in Matlab, we can use
fsolve. First let us reformulate the problem as follows. First\begin{align} a - k e &= 0 \\ b z + c - k (f z + g) &= 0 \\ e - z d - k (h + z) &= 0. \end{align}
The system of nonlinear equations can now be formulated as solving the equation $\textbf{F}(\textbf{x}) = \textbf{0}$, where
\begin{equation} \textbf{F}(\textbf{x}) = \begin{bmatrix} a - k e \\ b z + c - k (f z + g) \\ e - z d - k (h + z) \end{bmatrix} \text{ and } \textbf{x} = \begin{bmatrix} k \\ z \end{bmatrix}. \end{equation}
For example, to use
fsolveto solve the above function, you could typewhere the initial guess for the solver is
x0 = [0; 0], the solution (if it exists) isx,fvalshould be approximately equal to0, andexitflagdescribes iffsolvewas able to find a solution (see thefsolvehelp page for more info). If there is no solution, then you might need to do some nonlinear optimization, but hopefully there will be a solution.