I have just began studying the basic properties of determinants of square matrices and am trying to apply these basic properties in problems and trying all possible methods for each sum to try to simplify the determinant to make expanding easier and make my understanding better. I encountered the below question in my book which is listed as a solved example with the solution provided. However, at the end of the solution a note is provided which suggests the reader to try to solve the same question using the steps provided in the note.
I have understood the solution of the solved example but I am completely unable to work out the solution using the method described in the note. I have added the solution given in the book below and I followed it with the method described in the note.
$$ \text{Show that} \begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \\ \end{vmatrix} = abc(1 + 1/a + 1/b + 1/c) = abc + bc + ac + ab $$
Solution
Taking out factors a,b,c common from $R_1, R_2$ and $R_3$, we get
$$ L.H.S = abc \begin{vmatrix} \frac 1a+ 1 & \frac1a & \frac1a \\ \frac1b & \frac1b + 1 & \frac 1b\\ \frac1c & \frac1c& \frac1c + 1\\ \end{vmatrix} $$
Applying $R_1 \rightarrow R_1 + R_2 + R_3,$ we have
$$ \Delta = abc \begin{vmatrix} 1+ \frac 1a+ \frac1b + \frac1c & 1+ \frac 1a+ \frac1b + \frac1c& 1+ \frac 1a+ \frac1b + \frac1c\\ \frac1b & \frac1b + 1 & \frac 1b\\ \frac1c & \frac1c& \frac1c + 1\\ \end{vmatrix} $$
$$ = abc(1+\frac1a + \frac1b + \frac1c) \begin{vmatrix} 1 & 1 & 1 \\ \frac1b & \frac1b + 1 & \frac 1b\\ \frac1c & \frac1c& \frac1c + 1\\ \end{vmatrix} $$ Now applying
$C_2 \rightarrow C_2 - C_1,$
$ C_3 \rightarrow C_3 - C_1,$ we get
$$ \Delta = abc(1+\frac1a + \frac1b + \frac1c) \begin{vmatrix} 1 & 0 & 0\\ \frac1b & 1 & 0\\ \frac1c & 0& 1\\ \end{vmatrix}$$
$$ =abc(1+\frac1a + \frac1b + \frac1c)[1(1-0)]$$ $$ =abc(1+\frac1a + \frac1b + \frac1c) = abc + bc + ca + ab = R.H.S$$
The note given in the book
NOTE : Alternately try applying $C_1 \rightarrow C_1 - C_2 \text{ and }C_3 \rightarrow C_3 - C_2,\text{ then apply }C_1 \rightarrow C_1 - aC_3$
The note is given after the solution and hence is most probably referring to be applied straight to the initial question. However, I tried it for the initial question and also in a few intermediate steps but wherever I applied the steps given in the note, I wasn’t left with something that actually made the determinant easier to expand. Especially the last step : $\text{ “then apply }C_1 \rightarrow C_1 - aC_3”$ is something that I couldn’t understand as that $a$ was making the determinant more complicated. I am very eager to know why this NOTE is written as I couldn’t solve the sum using the given steps. I would like to know the solution so that I can understand how to approach determinant expansions in different methods. I hope someone can help me and tell me
How do I solve the sum using the alternative method provided in the NOTE?