Can the given transformation possible for given determinant?

Question

In forth step $(x-1)(x-2)$ is obtained by applying transformation R_{$1 \frac{1}{(x+1)}$} and R_{$2 \frac{1}{(x+2)}$}.

But we get value of $x = -1$ or $ x = -2$ so $\frac{1}{(x+1)}$ and $\frac{1}{(x+2)}$ will be undefined because value of x can be -1 and -2.

So my question is can we apply R_1i and R_2i?

Answer 1

In Step four, $(x+1) , (x+2)$ is obtained not by applying transformation $R_{1\frac{1}{x+1}}$ and $R_{2\frac{1}{x+2}}$. Since determinant function is linear in each row, we have \begin{align}det \begin{bmatrix} c.a_{1,1} & c.a_{1,2} & c.a_{1,3}\\ a_{2,1} & a_{2,2} & a_{2,3}\\ a_{3,1} & a_{3,2} & a_{3,3} \end{bmatrix} = c .det \begin{bmatrix} a_{1,1} & a_{1,2} & a_{1,3}\\ a_{2,1} & a_{2,2} & a_{2,3}\\ a_{3,1} & a_{3,2} & a_{3,3} \end{bmatrix} \end{align} Above identity is true for every row not just first row. Hence we have the step four using the linearity of determinant in first and third row.

Answer 2

Note that if $(x+1)=0$, the left hand side in the fourth step is zero. So the transformation still holds.