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Hi: I'm struggling with this - apparently simple! - matrix problem:

Use elementary row operations to find the value of $k$ so that the rank of the following matrix is $2$:

$$\left(\begin{matrix}3&3&-6&12\\ 3&-6&3&15\\ 1&k&-3&-7\end{matrix}\right)$$

I understand that I have to reduce one of the rows to all zeros, but I can't find a linear combination of any two rows which is equal to a third row! Thanks for you help.

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Try using the Gauss Jordan Method to get the matrix in Reduced Row Echelon Form:

Start with $i=1$, $j=1$

  1. If $a_{ij}=0$ swap the $i$-th row with some other row to guarantee that $a_{ij}\neq 0$. If all the entries in the column are zero, increase $j$ by $1$

  2. Divide the $i$-th row by $a_{ij}$ to make the pivot entry $1$

  3. Eliminate all other entries in the $j$-th colum by subtracting suitable multiples of the $i$-th row from the other rows

  4. Increase $i$ by $1$ and $j$ by $1$. Return to Step $1$

The algorithm stops after we process the last row or the last column of the matrix

Can you continue from here?

Edit: We can use an online calculator to see that the RREF has no one row of zeros so I would tentatively say that there is no value of $k$ which gives a rank of $2$ (I may be wrong here though, it's been a while since I did this stuff). Have you definitely written down/typed the matrix correctly?

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  • $\begingroup$ Thanks for taking the time to look at this. Thanks in particular for showing what the online symbolic calculator is capable of. It seems obvious from the symbolic result (a pivot in each row, whatever the value of k), that there is no solution to this problem. Must have been an error in the exam paper! I've checked it very carefully and that was the question as printed. $\endgroup$
    – Iolo Huw Jones
    Apr 11, 2017 at 12:18
  • $\begingroup$ Also thanks for tidying up my original representation of the matrix in question! $\endgroup$
    – Iolo Huw Jones
    Apr 11, 2017 at 12:23

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