# 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.

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What exactly are you trying to do, calculate the determinant or the value of $k$? If it is the latter, you are missing an equality. –  Pragabhava Nov 16 '12 at 2:47
If what you are trying to do is find the value(s) of $k$ that make the determinant, say, 17, then you don't need to do any row reduction --- you can just compute the determinant by expanding along the row (or column) containing $k$. And if $k$ turns out to be a decimal --- so what? If you're big enough to do determinants, you're big enough to do decimals. –  Gerry Myerson Nov 16 '12 at 2:53

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.

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You deserve a lot of credit for TeXing out those matrices. –  Isaac Solomon Nov 16 '12 at 4:29
$k=\cfrac{181}{16}$ according to the first matrice in the question. –  user31280 Nov 16 '12 at 15:22
@F'OlaYinka That's what I (just) saw and corrected, and it follows from the equation immediately above. Thanks for bringing my attention to my goof! –  amWhy Nov 16 '12 at 15:37

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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