LA2.2.7

$A=\begin{bmatrix} 1 & 1 & -1 & 0 \\ 4 & 4 & -3 & 1 \\ b & 2 & 2 & 2 \\ 9 & 9 & b & 3\end{bmatrix}\,$

Applying $R_2-4R_1,R_3-2R_1,R_4-9R_1\,$ to the above matrix,we get

$\begin{bmatrix} 1 & 1 & -1 & 0 \\ 0 & 0 & 1 & 1 \\ b-2 & 0 & 4 & 2 \\ 0 & 0 & b+9 & 3\end{bmatrix}\,$

Applying $R_3-4R_2,R_4-3R_2\,$ to the above,we get

$\begin{bmatrix} 1 & 1 & -1 & 0 \\ 0 & 0 & 1 & 1 \\ b-2 & 0 & 0 & -2 \\ 0 & 0 & b+6 & 0\end{bmatrix}\,$

Now $R_4 - R_3=\begin{bmatrix} 1 & 1 & -1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & b+6 & 0 \\ b-2 & 0 & 0 & -2\end{bmatrix}\,$

If $b=2,|A|=1\cdot 0\cdot 8\cdot(-2)=0\,$ ,Therefore,the rank of A=3.

If $b=-6\,$ number of non-zero rows is 3,rank of A=3.