How to find k given determinant? So I've got this matrix here, and need to solve for $k$
$$\text{det}\;\begin{pmatrix}
3 & 2 & -1 & 4 \\
2 & k & 6  & 5 \\
-3& 2 & 1  & 0 \\
6 & 4 & 2  & 3 \\
\end{pmatrix}=33$$
Doing some row operations $(R3+R1) \to R3\text{ and}\; (R4-2R1)\to R4$), I end up with
$$\text{det}\;\begin{pmatrix}
3 & 2 & -1& 4 \\
2 & k & 6 & 5 \\
0 & 4 & 0 & 4 \\
0 & 0 & 4 & -5\\
\end{pmatrix}=33$$
I expand along the first column and somehow my $k$ value is a decimal. Am I doing this correctly? I've tried making this into an upper and lower diagonal matrix and it just gets messy.
 A: How about continuing your work! You've already got
$$
\text{det}\;\begin{pmatrix}
3 & 2 & -1& 4 \\
2 & k & 6 & 5 \\
0 & 4 & 0 & 4 \\
0 & 0 & 4 & -5\\
\end{pmatrix}=33.
$$
Subtract column 2 (yes, we do column reduction now!) from column 4 gives
$$
\text{det}\;\begin{pmatrix}
3 & 2 & -1& 2 \\
2 & k & 6 & 5-k \\
0 & 4 & 0 & 0 \\
0 & 0 & 4 & -5\\
\end{pmatrix}=33.
$$
Now, only one nonzero element remains in the third row. Expand along the third row, we only need to evaluate one $3\times3$ determinant. Hooray~~~!!!
$$
33=
\text{det}\;\begin{pmatrix}
3 & 2 & -1& 2 \\
2 & k & 6 & 5-k \\
0 & 4 & 0 & 0 \\
0 & 0 & 4 & -5\\
\end{pmatrix}=-4
\ \text{det}\;\begin{pmatrix}
3 & -1& 2 \\
2 & 6 & 5-k \\
0 & 4 & -5\\
\end{pmatrix}=-48(k-12).
$$
This gives $k=\frac{33}{-48}+12=-\frac{11}{16}+12=\frac{181}{16}$.
A: Your row reductions seem fine, thus far. You can row reduce some more: 
try to get an upper triangular matrix; short of that, it will simplify the calculation of the determinant.
$(a)$ You can factor out $4$ from the third row. 
$$\text{ det}
\begin{pmatrix}
3 & 2 & -1& 4 \\
2 & k & 6 & 5 \\
0 & 4 & 0 & 4 \\
0 & 0 & 4 & -5\\
\end{pmatrix}
=33\iff 4 \text{ det} \begin{pmatrix}
3 & 2 & -1& 4 \\
2 & k & 6 & 5 \\
0 & 1 & 0 & 1 \\
0 & 0 & 4 & -5\\
\end{pmatrix}
=33$$
$$
\text{Add}\;-2(R_3) \text{ to}\;R_1 \implies 4 \text{ det} \begin{pmatrix}
3 & 0 & -1& 2 \\
2 & k & 6 & 5 \\
0 & 1 & 0 & 1 \\
0 & 0 & 4 & -5\\
\end{pmatrix}
= 33$$
$$
\text{Add }\;R_4 \text{ to}\; R_2\implies 4\text{ det} \begin{pmatrix}
3&0&-1&2\\
2&k&10&0\\
0&1&0&1\\
0&0&4&-5\\
\end{pmatrix}
= 33$$
At this point, I'd suggest simply expanding along the column containing $k$; with the additional row reduction, that may simplify the process. Don't worry if you end up with with an equation in which $k$ evaluates to a fraction (decimal)! 
$$
\text {det} \begin{pmatrix}
3 & 0 & -1& 2 \\
2 & k & 10 & 0 \\
0 & 1 & 0 & 1 \\
0 & 0 & 4 & -5\\
\end{pmatrix}
= \frac{33}{4}\tag{1}$$
$$k\text{ det}\; \begin{pmatrix}
3 &-1&2\\
0&0&1\\
0&4&-5\\
\end{pmatrix}
- \text{ det}\; \begin{pmatrix}
3&-1&2\\
2&10&0\\
0&4&-5\\
\end{pmatrix}\tag{2}$$
$$= k\text{ det}\; \begin{pmatrix}
3 &-1&2\\
0&0&1\\
0&4&-5\\
\end{pmatrix}
- \left(3\text{ det}\; \begin{pmatrix}
10&0\\
4&-5\\
\end{pmatrix} 
-2\text{ det}\; \begin{pmatrix}
-1&2\\
4&-5\\
\end{pmatrix}\right)= \frac{33}{4}
\tag{3}$$
$$-12k - [3(-50) - 2(-3)] = \frac{33}{4}$$ 
$$-12k +144 = \frac{33}{4}$$
$$k=\frac{181}{16}$$
Note:In $(1)$, I simply divided both sides of the equation by $4$.
In $(2)$, I expanded along the second column, the column containing $k$. Note the sign of each resulting determinants. 
In $(3)$, I expanded along the first column of the second matrix in $2$; again, we need to keep. 
The rest is simplification. 
