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.

• You should ask your teacher what kind of work would be needed to be shown to get full points on this type of question. – Robert Israel Feb 22 '17 at 21:13

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.

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 $\text{rank}(M)=3$.

• Thank you very much, too. I knew the trick with determinant but not the one you have explained here, totally new for me. Only thing I knew was that you calculate the determinant and if it doesn't equal zero, the rank could be full, so in this case 3. – cnmesr Feb 22 '17 at 22:02
• As someone who took linear algebra years ago, I wish I'd known this trick then. – JAB Feb 22 '17 at 22:34
• One should be aware, though, that the mod 2 matrix has rank $\le$ the original matrix, so if it has lower rank (which is a significant possibility), you will have to do more work yet to find the rank of the original matrix. Still, a helpful trick to quickly dispose of many cases, – Paul Sinclair Feb 23 '17 at 17:33
• @PaulSinclair: appreciate that. My remark is just about $$\text{odd determinant}\quad\Longrightarrow\quad\text{full rank}$$ but if the determinant is even, we may still try to consider $M\pmod{3}$, hoping to get a determinant $\not\equiv 0\pmod{3}$ and so on. – Jack D'Aurizio Feb 23 '17 at 17:40
• I know. I just wanted to point this out for noviciates, lest they mistake your method as being more powerful than it is. It is indeed quite handy, and I too wish I had been aware of it in my college days. But like most nifty tricks in mathematics, it doesn't apply in all cases. For a 3x3 matrix, I'm not sure it is worthwhile even checking mod 3 if mod 2 fails. It might be wiser at that point to just do REF. For 4x4 or higher though, checking mod 3 next would be the way to go. – Paul Sinclair Feb 23 '17 at 17:51