I was struggling with this one, so I used software to compute the reduced row echelon forms (which are unique to each linear system) for each and found them to be equal to each other, thus I need that the first matrix could be reduced to the second. Now, $$\begin{pmatrix} 1&2&-1&-3\\ 3&5&k&-4\\9&(k+13)&6&6\end{pmatrix} -3R_1+R_2\to R_2 \begin{pmatrix} 1&2&-1&-3\\ 0&-1&(3+k)&5\\9&(k+13)&6&9\end{pmatrix}$$ $$9R_1 -R_3 \to R_3 \begin{pmatrix} 1&2&-1&-3\\ 0&-1&(3+k)&5\\0&(5-k)&-15&-36\end{pmatrix} $$ $$-R_2 \to R_2 \begin{pmatrix} 1&2&-1&-3\\ 0&1&(-3-k)&-5\\0&(5-k)&-15&-36\end{pmatrix}$$ $$(5-k)R_2-R_3 \to R_3 \begin{pmatrix} 1&2&-1&-3\\ 0&1&(-3-k)&-5\\0&0&(k^2-2k)&(5k+11)\end{pmatrix}.$$
In short, row reduced echelon form(RREF) of a matrix $A$ is such that
i) Every leading entry is 1
ii) Any nonzero rows are above zero rows
iii) any leading entry is strictly to the right of any leading entries above that row
iv) any other entry in a column containing a leading entry is 0 except for the leading entry.
So it does not have to be put in augmented matrix $[A|b]$ to get a RRE form. You are comparing RRE form of matrix $A$ and $[A|b]$.
To see why the statement is true, suppose that you put the matrix $[A|b]$ into RRE form, so you have a matrix E. If E contains a leading entry in its last column, in terms of system of equations, what does it say? And what is the condition for E to not have any leading entry in last column?
Note: If RRE form of $[A|b]$ does contain a leading entry, then it is different from that of $A$. Also, note that RRE form of $[A|b]$ is m by n+1 whereas that of $A$ is m by n.
Solve:
$x+y=1$
$x+y=2$
Then we have
$\
A =
\left( {\begin{array}{cc}
1 & 1 \\
1 & 1
\end{array} } \right)
$
$\
b =
\left( {\begin{array}{cc}
1 \\
2
\end{array} } \right)
$
and $Ax=b$
If we turn A into RREF, we get
$\
E =
\left( {\begin{array}{cc}
1 & 1 \\
0 & 0
\end{array} } \right)
$
So A has rank 1
and if we put $[A|b]$ into RRE form, we get
$\
E' =
\left( {\begin{array}{cc}
1 & 1 & 0 \\
0 & 0 & 1
\end{array} } \right)
$
So augmented matrix has rank 2. Observe what last row says in terms of equations.
Best Answer
The determinant is a multilinear function of the rows (or columns). Since $(4,6,8) = 2(1,1,1)+2(1,2,3)$, we have $\det C = 2 \det A + 2 \det B$. Hence the answer is $-2$.