How to efficiently use a calculator in a linear algebra exam, if allowed We are allowed to use a calculator in our linear algebra exam. Luckily, my calculator can also do matrix calculations.
Let's say there is a task like this:

Calculate the rank of this matrix:
$$M =\begin{pmatrix} 5 & 6 & 7\\  12 &4  &9 \\  1 & 7 & 4
\end{pmatrix}$$

The problem with this matrix is we cannot use the trick with multiples, we cannot see multiples on first glance and thus cannot say whether the vectors rows / columns are linearly in/dependent.
Using Gauss is also very time consuming (especially in case we don't get a zero line and keep trying harder).
Enough said, I took my calculator because we are allowed to use it and it gives me following results:
$$M =\begin{pmatrix} 1 & 0{,}3333 & 0{,}75\\  0 &1  &0{,}75 \\  0 & 0 & 1
\end{pmatrix}$$
I quickly see that $\text{rank(M)} = 3$ since there is no row full of zeroes.
Now my question is, how can I convince the teacher that I calculated it? If the task says "calculate" and I just write down the result, I don't think I will get all the points. What would you do?
And please give me some advice, this is really time consuming in an exam.
 A: You're allowed to use your calculator.  So, if it were me on the test, I'd write something like this:

$$
\pmatrix{5&6&7\\12&4&9\\1&7&4} \overset{REF}{\to} 
\pmatrix{
1 & 0,3333 & 0,75\\  0 &1  &0,75 \\  0 & 0 & 1
}
$$
  because the reduced form of $M$ has no zero rows, $M$ has rank $3$.

REF here stands for row-echelon form.

Note: You should check with your professor whether or not this constitutes a sufficient answer.  It may be the case that your professor wants any matrix-calculations to be done by hand.  See Robert Israel's comment.
If that's the case, then you should do row-reduction by hand.  It only takes 3 row operations, though.
A: There is a very nice trick for showing that such matrix has full rank, it can be performed in a few seconds without any calculator or worrying "moral bending". The entries of $M$ are integers, so the determinant of $M$ is an integer, and $\det M\mod{2} = \det(M\mod{2})$. Since $M\pmod{2}$ has the following structure
$$ \begin{pmatrix} 1 & 0 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 0\end{pmatrix} $$
it is trivial that $\det M$ is an odd integer. In particular, $\det M\neq 0$ and $\text{rank}(M)=3$.
