Problem
How many solutions equation $Ax=0$ has when $A$ is defined as:
$$ A = \begin{bmatrix} 2 & 3 & 5 \\ 1 & 4 & 0 \\ -4 & 2 & 1 \end{bmatrix} $$
Attemtp to solve
One way would be to solve this problem with row reduction and see how many solutions equation has. Another way would be to compute determinant since, it tells if linear transformation squishes all possible vectors to a line.
$$ \det(A)=95 $$
It looks like this linear transformation doesn't squish everything to single line. Ax=0 will only have trivial solution of $x=\begin{bmatrix} 0 & 0 & 0 \end{bmatrix}^T$. When determinant is $0$ equation $Ax=0$ has infinite amount of solutions. Linear transformation squishes a whole line of vectors to $0$ meaning $Ax=0$ has full line of solutions or infinite amount.
I would like to know if this intuition is correct or am I missing something or perhaps just simply wrong ?
$\endgroup$ 42 Answers
$\begingroup$You are correct. A square matrix $A$ whose determinant is nonzero is invertible, and then the equation $Ax = y$ always has a unique solution.
$\endgroup$ 1 $\begingroup$$Ax=0$ may not only have a line of solutions, it may have a plane, or, in the case of the zero transformation, the whole three dimensional space.
In general, the null space will have dimension $3-\operatorname{rank}A$, by the Rank-nullity theorem.
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