how to find a matrix A given the solution? if we need,for example, to find a nonzero 3x3 matrix A such that we are given a 3x1 vector as a solution to Ax = 0. What is the general procedure we can follow to obtain such Matrix A?
Thank you :)
 A: Let's suppose that your vector $v$ is a column vector. 
One of options is to look at the matrix $B:=vv^T$: it's a $3\times 3$ matrix, and $Av = \|v\|^2v$. Now we can look at the matrix $A:=(\|v\|^2Id-B)$: easy to check that $v$ belongs to its nullspace. We need to check that our $A$ is not zero; indeed, take any nonzero vector $w$ such that $v\bot w$, then $Aw=\|v\|^2w\ne 0$.
Note that $A$ is not uniquely defined, because you have $3$ linear equations on $9$ elements of the matrix.
A: Given that we don't really know anything about A (other than non-zero) we can find values to satisfy the equation, but it will not be unique at all. This question breaks down into a system of three equations, but with 9 unknowns. 
$$
\begin{pmatrix}
  a & b & c \\
  d & e & f \\
  g & h & i
 \end{pmatrix}
\begin{pmatrix}
  x \\
  y \\
  z
 \end{pmatrix}
=
\begin{pmatrix}
  0 \\
  0 \\
  0
 \end{pmatrix}
$$
So it comes down to selecting values that satisfy the three corresponding equations. 
$$
 \left\{
  \begin{array}{l l}
    ax+by+cz=0\\
    dx+ey+fz=0 \\
    gx+hy+iz=0
  \end{array} \right\}.
$$
Again, these values wont be unique, but in this way you can get a matrix A to satisfy your equation.
A: First of all, notice that you can choose the rows of $A$ individually and independently. This simplifies the problem to be solved.
Each row $a$ of $A$ is a solution to $ax=0$ (the product of a $1\times 3$ matrix with the given $3\times1$ matrix, or a dot product if you prefer). This is a system of $1$ homogeneous linear equation in $3$ unknowns, so it will have a $2$-dimensional set of solutions (assuming $x$ is itself not the zero vector), which you can find by Gaussian elimination or even easier techniques. From the solution set you can choose any specific solutions you want for each row of $A$.
Moral: whenever a linear algebra problem can be translated into finding solutions to a system of linear equations, do it and go straight for Gaussian elimination!
