For what value of $t$ is that ranks of the following matrix $ A$ equal $3$ ?

Question

For what value of $t$ is that ranks of the following matrix $ A$ equal $3$ ?

$A=\begin{vmatrix}t & 1 & 1 & 1\\ 1& t & 1 & 1\\ 1 & 1 & t & 1\\ 1& 1 & 1 & t\end{vmatrix}.$

My answer : i take t = 1, then the rank of A = 1 so it t=1 is not possible, as i take take t=0 then Rank of A = 4 so it is not possible...

pliz help me,,,,is there any tricks to calculate Matrix $Ranks A = 3$

any hints /solution will be aprreciated.

thanks u

Answer 1

If the rank is to be less than $4$, then the determinant must be $0$. The determinant can only be $0$ for a very limited number of values of $t$, and you've already found one of them ($t=1$), so $t-1$ divides the determinant. In fact, since $t=1$ gives a rank lower than $3$, $t-1$ divides the determinant several times.

The rest are not very hard to find by direct calculation and polynomial division. Check those and you're done.

Answer 2

If you know some spectral graph theory you can see that $t=-3$ gives the Laplacian of $K_4$ and by Kirchofs Theorem the rank is correct.