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.
Best Answer
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.