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$A$ = $\begin{vmatrix} 1 & 2 & 3 &4\\ 5 & 6 & 7 &8\\ 6 & 8 & 10 &12\\ 151&262&373&484 \end{vmatrix}$

After some row operation $(R_2=R_2-5R_1),(R_3=R_3-6R_1),(R_4=R_4-151R_1)$ and simplifying we have

$A$ = $\begin{vmatrix} 1 & 2 & 3 &4\\ 0 & -1 & -2&-3\\ 0 & -1 &-2 &-3\\ 0&-1&-2&-3 \end{vmatrix}$ and I get rank $2$ from here

I always convert given form into row echelon form to find rank. But Is there any quick way to find rank? Row echelon form is sometimes time-consuming.

Daman
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  • The third row is blatantly the sum of the previous two, so the rank is at most three. The same phenomenon occurs for a $3\times 3$ minor and there is an invertible $2\times 2$ minor on the main diagonal, so the rank is $2$. In some cases this trick might be helpful in reducing the computation time – Jack D'Aurizio Jan 17 '19 at 08:22
  • @JackD'Aurizio I am not familiar with word mod you used in your answer can you elaborate that answer in simple language? – Daman Jan 17 '19 at 08:25
  • If you are dealing with a matrix with integer entries, the parity of the determinant is also the parity of the determinant of the same matrix, with every entry being reduced to $0$ or $1$ according to its parity. The key observation is that any odd determinant differs from zero, leading to a matrix with full rank. – Jack D'Aurizio Jan 17 '19 at 08:27
  • Of course if you get an even determinant you need further steps to distinguish it (or equate it) to zero. – Jack D'Aurizio Jan 17 '19 at 08:29
  • This is more or less the same idea used for factoring polynomials over $\mathbb{Q}$: the factorization of a polynomial over a finite field always is a simple problem, so the factorization over $\mathbb{Q}$ can be obtained by "glueing back" the factorizations over a suitable number of finite fields. – Jack D'Aurizio Jan 17 '19 at 08:32
  • @JackD'Aurizio Thanks for your explanation but I couldn't understand it. It's because I lack knowledge. I'll be back after gaining some experience with linear algebra and will talk to you. – Daman Jan 17 '19 at 08:47

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