Find $\det$ in terms of $k$ Consider the following matrix:
\begin{bmatrix}
        1 & 2 & 3 \\
        2 & k-3 & 4 \\
        3 & 4 & k-4 \\
\end{bmatrix}
I have the following problems:


*

*How to find $\det(B)$ in terms of $k$?

*For what value(s) of $k$ are the column vectors of $B$ linearly dependent?
 A: You can find the determinant by applying row transformations:
\begin{align}
B=\begin{bmatrix}
1 & 2 & 3\\
2 & k-3 & 4\\
3 & 4 & k-4
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 2 & 3\\
0 & k-7 & -2\\
0 & -2 & k-13
\end{bmatrix}
&&\begin{aligned}R_2&\gets R_2-2R_1\\R_3&\gets R_3-3R_1\end{aligned}
\\&\to
\begin{bmatrix}
1 & 2 & 3\\
0 & -2 & k-13 \\
0 & k-7 & -2
\end{bmatrix}
&&R_2\leftrightarrow R_3
\\&\to
\begin{bmatrix}
1 & 2 & 3\\
0 & 1 & (13-k)/2 \\
0 & k-7 & -2
\end{bmatrix}
&&R_2\gets (-1/2)R_2
\\&\to
\begin{bmatrix}
1 & 2 & 3\\
0 & 1 & (13-k)/2 \\
0 & 0 & (k^2-20k+87)/2
\end{bmatrix}
&&R_3\gets R_3+(7-k)R_2
\end{align}
Thus the determinant is
$$
(-1)(-2)\frac{k^2-20k+87}{2}=
k^2-20k+87
$$
The first factor is due to the row swap, the second factor is due to the pivot reduction, the third is the product of the elements in the diagonal.
Now recall that the columns are linearly independent if and only if the determinant is nonzero.
A: here is one way to do this. note that the determinant is quadratic in $k.$
you can fine the coefficient of $k^2$ by finding the determinant of 
$\begin{bmatrix}
        1 & 2 & 3 \\
        0 & 1 & 0 \\
        0 & 0 & 1 \\
\end{bmatrix}$  which is $1.$  we need to find the set for two other vales of $k.$ for $k = 0,$ the det is $87$ and for $k = 1,$ the det is  $68.$
so you have $ \det A = k^2 + (68-1-87)k+87=k^2-20k+87.$
