So basically here I am trying to understand why it is like that?
Suppose Matrix $$
A = \left( \begin{array}{ccc}
a & b & c \\
d & e & f \\
a+d & b+e & c+f \end{array} \right)
$$
Then, $\det (A) = 0$.
Why is that? I have tried with subtraction as well and I get $\det (A) = 0$ as well.
I am seeing a pattern, but I cannot explain in strict mathematical terms.
Best Answer
Here are two properties of determinants which I assume you know:
(1) You can add/subtract two rows and determinant doesn't change.
(2) If you multiply $k^{th}$ row by a constant, the determinant will be multiplied by the same constant.
Suppose that the matrix $A$ has a row, say $k^{th}$ row, equal to sum of two other rows. Use law (1) and subtract those two rows from $k^{th}$ row. You have a matrix with one row equal to zero. Let the determinant of this matrix be equal to $x$. According to (2), when you multiply the $k^{th}$ row by $2$, the determinant of the new matrix equals $2x$, however your matrix is the same. So, $x=2x$, which means $x=0$.
PS. See this, too.