I am obliged to find rank of matrix $A$ depending on $ p $.

I know how to do this using Gauss elimination method but I would like to try solve this using minors. I know that the rank of matrix is equal to degree of the biggest non-zero minor.

$A=\left( \begin{array}{ccc} p-1&p-1&1&1\\ 1&p^{2}-1&1&p-1\\ 1&p-1&p-1& 1 \end{array} \right)$

So I am taking this sub-matrix: $A_{1}=\left( \begin{array}{ccc} p-1&1&1\\ 1&1&p-1\\ 1&p-1& 1 \end{array} \right)$

I am counting $det(A_{1})=-p^{2}+5p-6$

$-p^{2}+5p-6=0 \iff p=2 \lor p = 3 $

So now I got two values when the rank of matrix is not equal to 3. No I will test rank for $p=2$ and $p=3$ and so on. Do I have to check only one minor? Am I doing this right?

  • $\begingroup$ No, this is wrong. Some other minor of order $3$ could still turn out to be nonzero for $p = 2$ or $p = 3$, maybe. What your calculation shows is that for all values of $p \ne 2, 3$, the rank is 3. $\endgroup$
    – user121926
    Jan 17, 2014 at 15:39
  • 1
    $\begingroup$ @user121926 the rank of that matrix is never 4 since the rank is the dimension of the column space which is $\le 3$ $\endgroup$
    – Bman72
    Jan 17, 2014 at 15:40
  • $\begingroup$ Sorry, it looked $4 \times 4$ to me! I'll edit my comment. $\endgroup$
    – user121926
    Jan 17, 2014 at 15:41

2 Answers 2


Yes, you are right so far. Since the matrix has $3$ rows, its rank is at most $3$, and you know that for $ p \notin \{2,3\}$ the minor you considered is nonzero, making the rank $3$. Then you look at the cases $p=2$ and $p=3$ (but there you will have to look at other minors).


This is your matrix. You have already checked in a correct way that the rank of the matrix is 3 for $p \neq 2,3$

$$A=\left( \begin{array}{ccc} p-1&p-1&1&1\\ 1&p^{2}-1&1&p-1\\ 1&p-1&p-1& 1 \end{array} \right)$$

Now you have to check what happens for p=2 or 3. For p = 2 you have

$$A=\left( \begin{array}{ccc} 1&1&1&1\\ 1&3&1&1\\ 1&1&1& 1 \end{array} \right)\to \left( \begin{array}{ccc} 1&1&1&1\\ 0&0&0&2\\ 0&0&0& 0 \end{array} \right)$$

I've obtained the second matrix by subtraction each line with first line and by taking the second column vector to the end of the matrix. You now easily see that the rank of your matrix is $2 \space \text{for} \space p=2$. For p = 3 you obtain:

$$A=\left( \begin{array}{ccc} 2&2&1&1\\ 1&8&1&2\\ 1&2&2& 1 \end{array} \right) \to \left( \begin{array}{ccc} 2&2&1&1\\ 0&14&1&3\\ 0&-6&0&- 1 \end{array} \right)$$

where I've subtracted 2 times the second line with the first line and the third line with the second line.

$$\left( \begin{array}{ccc} 2&2&1&1\\ 0&14&1&3\\ 0&-6&0&- 1 \end{array} \right) \to \left( \begin{array}{ccc} 2&2&1&1\\ 0&14&1&3\\ 0&0&6&4 \end{array} \right)$$

Where I've added 14 time the third line with 6 time the second line. You now see that for $p=3$ the rank is $3$


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