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I would like to solve a system of linear equations, all of which equal to zero. Is there a way to do this using linear algebra without decomposing into separate equations.

The equation is simple, F is a known matrix, lets say a 3x3

x is a vector of unknown variables, for example, x,y,z

0 is a zero vector, in this example 0,0,0

The equation is F*x = 0

What is the best approach to solving this?

If I did inv(F)*F*x = inv(F)*0 the answer I get is x = 0, y = 0, z = 0

Whats the best approach to solving this problem using matrices? Like adding an identity matrix to both sides?

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  • $\begingroup$ Would you be able to build a matrix $F$ for which $det(F)\neq0$ that admit any other solution that the trivial $(0,0,0)$ solution ? $\endgroup$ – MGirard Oct 10 '17 at 18:43
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Consider the equation $A\textbf{x} = \textbf{0}$ with $A$ an invertible matrix. Then clearly you can multiply the inverse matrix to both sides and get

$$ A^{-1}A\textbf{x} = A^{-1}\textbf{0} \implies \textbf{x} = \textbf{0} $$

Thus, the zero vector is the unique solution to the equation $A\textbf{x} = \textbf{0}$. So, if your matrix is invertible, then you have already solved the problem.

Assuming that the matrix $A$ is not invertible, then the above manipulation does not work, because $A^{-1}$ doesn't exist. To solve the equation in this case you can write down the augmented matrix $\left [ A \ | \ \textbf{0} \right ]$ and perform Gaussian elimination to find all solutions.

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The solutions to the equation $F x = 0$ constitute the null space or kernel of the linear operator corresponding to the matrix $F$. A standard way of finding it is Gaussian elimination.

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  • $\begingroup$ Interesting, there is no way to find a solution using linear algebra? $\endgroup$ – Mich Oct 10 '17 at 20:07
  • $\begingroup$ Gaussian elimination is linear algebra. $\endgroup$ – Robert Israel Oct 11 '17 at 0:34

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