I need to determine the rank of this matrix. I have tried it, but I get lost with the row operations. Any tips and trick how to do it?

Question

anonymous · Answer

I need to determine the rank of the matrix for every  $$(a,b) \epsilon \mathbb{R} ^{2}$$

rational · Answer

first step :  divide the 3rd row by (a+1)

rational · Answer

second step :   R2 - R1

loser66 · Answer

$\left[\begin{matrix}1&(a-1)&2&(a+2)&(a+b)\1&2a&0&a&2a+b\0&-(a+1)&(2a+2)&0&0\0&(2a+2)&(4a-4)&(a^2+a-8)&(4a+ab+b)\end{matrix}\right]$



$R_3*2 +R_4-->R_4$ $\left[\begin{matrix}1&(a-1)&2&(a+2)&(a+b)\1&2a&0&a&2a+b\0&-(a+1)&(2a+2)&0&0\0&0&((4a)^2-16)&(a^2+a-8)&(4a+ab+b)\end{matrix}\right]$

$R_2/(a+1)$--> $\left[\begin{matrix}1&(a-1)&2&(a+2)&(a+b)\1&2a&0&a&2a+b\0&-1&2&0&0\0&0&((4a)^2-16)&(a^2+a-8)&(4a+ab+b)\end{matrix}\right]$


$R_1*(-1)+R_2)$-->  $\left[\begin{matrix}1&(a-1)&2&(a+2)&(a+b)\0&(1+a)&-2&-2&a\0&-1&2&0&0\0&0&((4a)^2-16)&(a^2+a-8)&(4a+ab+b)\end{matrix}\right]$

loser66 · Answer

$R_3+R_2-->R_2$--> $\left[\begin{matrix}1&(a-1)&2&(a+2)&(a+b)\0&a&0&-2&a\0&-1&2&0&0\0&0&((4a)^2-16)&(a^2+a-8)&(4a+ab+b)\end{matrix}\right]$ 

$(R_2 +R_3)(-1)+R_1-->R_1$ $\left[\begin{matrix}1&0&0&(a+4)&b\0&a&0&-2&a\0&-1&2&0&0\0&0&((4a)^2-16)&(a^2+a-8)&(4a+ab+b)\end{matrix}\right]$

loser66 · Answer

That's all I can do for you. They are linearly independent, so that rank A =4

anonymous · Answer

Thanks @Loser66 - I'll try to go through the steps now!