Why does a matrix have determinant zero if one row is the sum of two other rows? 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. 
 A: 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.
A: Assume that line $k$ is the sum of lines $i$ and $j$ and consider the vector $x$ whose $i$ and $j$ entries are $-1$, entry $k$ is $+2$ and every other entry is $0$. 
Then $A^Tx=0$ (can you show this?) and $x\ne0$ hence $A^T$ is not invertible, which implies that $\det(A^T)=0$. Since $\det(A)=\det(A^T)$ (can you show this?), this shows that $\det(A)=0$.
A: Rows in a matrix represent transforms of horizontal unit vectors.  So if your matrix represents a transform $T$, then:
$$T([1, 0, 0]) = [a, b, c]$$
$$T([0, 1, 0]) = [d, e, f]$$
$$T([0, 0, 1]) = [a+d, b+e, c+f]$$
Notice that those 3 points plus the origin are co-planar (all four points are in the same plane).  This means that transform $T$ maps the points of a cube into some flat 2D shape.
A determinant tells you how much multidimensional volume you gain or lose from a transform.  For example, determinant of $1\over 2$ means the transformed shape has half the volume of the original.  A 2D shape has zero 3D volume, so we expect the determinant of the example matrix to be zero.  And if you work out the algebra from the definitions, it does.
A: You can use this simple fact $det(A)=0$ if and only if $A$ is not invertable.
Since last row is sum of first two rows, If you make ths matrice row reduced then you ill get a matrix has $0$ in one row. And you can easily say that this matrix is not invertable so we are done.
A: One of the rows of $A$ is a linear combination of the others. This means that $A$ is not an invertible matrix and hence its determinant is zero.
A: $$|A| = \left| \begin{array}{ccc}
a & b & c \\
d & e & f \\
a+d & b+e & c+f \end{array} \right|$$
there is this property where you can split determinants (How? refer clause 7). So,
$$|A| = \left| \begin{array}{ccc}
a & b & c \\
d & e & f \\
a & b & c \end{array} \right|+\left| \begin{array}{ccc}
a & b & c \\
d & e & f \\
d & e & f \end{array} \right| ...(1)$$
Next use the property where determinants with same rows have their $determinant=0$ (How? refer Corollary 161 page 69) thus,
$$
|A| =0$$
And yeah I think this $2^{nd} $ link will be very useful for you in your future study.
