For the direction you want to prove, the invertibility of $Q$ is a distraction: suppose that $A$ is not invertible, then $A$ doesn't have full column rank so that there is $x\neq 0$ such that $Ax=0$. Then $(QA)x=Q(Ax)=Q\times 0=0.$
You only need the invertibility of $Q$ if you want to prove the reverse direction: suppose that $A$ is invertible and suppose that $x\neq 0$, then $Ax\neq 0$. Then, because $Q$ is invertible, you have $(QA)x=Q(Ax)\neq 0$. This means that if $A$ is invertible, then $(QA)x=0$ only has the trivial solution.
$A$ is an invertible matrix.
Find $B$ such that $AB = BA = I$ or show that it is impossible.
$A$ is row equivalent to the n x n identity matrix.
Reduce $A$ to reduced row echelon form. See if is identity.
$A$ has n pivot positions
Pretty much the same as above.
The equation $Ax = 0$ has only the trivial solution.
Solve the system.
The columns of $A$ for a linearly independent set.
Write down your columns as vectors in $\Bbb R^n$. See if $\{x_1,x_2,\ldots,x_n\}$ is linearly dependent or independent by your usual means.
The linear transformation $x ↦ Ax$ is one-to-one.
Solve $Av = 0$ as above. Pick one solution $v_0$ and define $y = x + v_0$ for any vector $x$. Note that $x= y$ if and only if $v_0 = 0$.
The equation $Ax = b$ has at least one solution for each $b$ in $ℝ^n$.
If $A$ is invertible, $A^{-1}b$ is the solution. If not, find $b$ such that $Ax = b$ has no solutions (hint: reduce extended matrix $(A|b)$ to row echelon form. If there are zero rows in transformed $A$, choose $b$ such that its transformed component in that row is non-zero).
The columns of $A$ span $ℝ^n$.
Write down columns of $A$ as vectors in $\Bbb R^n$. See if they generate $\Bbb R^n$ or not. Since there are $n$ columns, it is equivalent to check if columns are linearly dependent or not.
The linear transformation $x ↦ Ax$ maps $ℝ^n$ onto $ℝ^n$.
You must check surjectivity of $A$, i.e. for all $b\in ℝ^n$, you must find solution to $Ax = b$ or find $b$ such that there is no solution.
There is an n x n matrix $C$ such that $CA = I$.
Write general matrix $C$, multiply by $A$ and show that coefficients of $C$ can be chosen to get identity or disprove it.
There is an n x n matrix $D$ such that $AD = I$.
As above.
$AT$ is an invertible matrix.
As the first one.
Best Answer
If $\operatorname{rank}(A)=r$ the solutions of $Ax=0$ is $n-r$ dimensional space, so if $Ax=0$ has only trivial solutions it means that $\operatorname{rank}(A)=\operatorname{size}(A)$, so $A$ is invertible.