<s>[INST] 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.
 [/INST] 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 $k=\frac{33}{-48}+12=-\frac{11}{16}+12=\frac{181}{16}$.
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