4
$\begingroup$

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.

  • 2
    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.2012-11-16
  • 2
    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.2012-11-16

2 Answers 2