Different form of determinant, does it make mine wrong? 
Calculate the determinant of the following $(n+1) \times (n+1)$ matrix:
$$A = \pmatrix{1 & 1 & 1 & 1 &\cdots & 1 \\ 1 & a_1 & 0 & 0 &\cdots & 0 \\ 1  & 0 & a_2 & 0 &\cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\  & & & & \cdots &  \\ 1 &0 & 0 & 0 & \cdots & a_n }$$

So I first did the row operations where I divided all rows $i > 1$ by $\frac{1}{a_1}$ to get
$$\pmatrix{1 & 1 & 1 & 1 &\cdots & 1 \\ \frac{1}{a_1} & 1 & 0 & 0 &\cdots & 0 \\ \frac{1}{a_2}  & 0 & 1 & 0 &\cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\  & & & & \cdots &  \\ \frac{1}{a_n} &0 & 0 & 0 & \cdots & 1 }$$
and then did the row operations $R_1 - R_{i}$ and got
$$\pmatrix{1 - \sum_{i = 1}^na_1 & 0 & 0 & 0 &\cdots & 0 \\ \frac{1}{a_1} & 1 & 0 & 0 &\cdots & 0 \\ \frac{1}{a_2}  & 0 & 1 & 0 &\cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\  & & & & \cdots &  \\ \frac{1}{a_n} &0 & 0 & 0 & \cdots & 1 }$$
which gives me a lower triangular matrix and so my determinant is just the product of all the elements in the main diagonal which is
$$\left( 1 - \sum_{i = 1}^n \frac{1}{a_i} \right) \cdot (1)^n = 1 - \sum_{i = 1}^n \frac{1}{a_i}$$
but in the answers it says it should be
$$a_1 \cdots a_n \sum_{i = 1}^n \prod_{i \neq j}a_j $$
which seems to be my matrix without having done the $\frac{R_i}{a_i}$ row operations. Does this make my determinant wrong?
 A: Your determinant is not wrong in the sense you can still use it test for singularity, as long as you never scale by zero. It does not however possess other nice, elegant and important properties of the standard determinant.
Taking the standard as a reference, you computation breaks one rule:
Multiplying the row (or column) of a matrix by a scalar $s$ changes its determinant also by a factor of $s$. This is a direct result of the definition of determinant as given in linear algebra. Alternatively, if you consider determinants as defined by permutation expansion, it is easy to verify that this rule emerges as a property.
I can think of three intuitive ways to justify this rule:


*

*Take the identity matrix, and multiply one row by $s=0$. Clearly it is now singular, so its determinant must have changed (from one) to zero.

*If you like the geometric signed volume explanation of determinants, it is then also easy to see that if you multiply a vector by $s$, the volume will change accordingly.

*Take a $2 \times 2$ matrix as a prototype for determinants of any dimension, with entries $a,b,c,d$, and work out when it is that you can always solve it as a linear equation.
$ax_1+bx_2=c_1$
$cx_1+dx_2=c_2$
gives
$ax_1+bx_2=c_1$
$0+(d-bc/a)x_2=c_2$
Which is always solvable given that $(d-bc/a) \neq 0$ and so $ad-bc \neq 0$. If you then agree that $ad-bc$ is the determinant, then obviously, scaling a row will scale the determinant. 


What you calculated is the standard determinant multiplied by all the row scalings of your first step. To 'fix' it, you have to multiply it by their inverses.  
Note: One comment says that doing the row operation alters your matrix and hence your determinant. Keep in mind that this holds for the scaling row operation, but not when linearly combining two rows. In that case, your matrix also 'changes' but the determinant does not.
