Give all values of h for which the matrix A fails to be invertible Can someone please help me out with this please.  I know the answer is h = 8 and I know the determinant is 21h - 168 and I even know the steps to find those answers.  For some reason this is giving me fits and I must be making silly mathematical mistakes that I keep missing.  First I have been simply using cofactor expansion and cannot come up with the correct answer. Then I used row reduction where I had a seven and two zeros in the first column.
I came up with 21h - 28 at one point and 119h + 28 at one point.
Thanks to anyone taking a look at this.  I know this is a pretty simple problem but I'm not getting something right.  I am studying for a test and fill like I am understanding everything pretty well, but this one problem is making me crazy!
First I use two row reductions.  R2 becomes R1(-2)+R2 and R3 becomes R1(-1)+R3
So I get {{7,-5,3},{0,3,-5},{0,-3,h-3}}


\begin{align*}
\operatorname{det}\begin{bmatrix} 7 & -5 & 3 \\ 14 & -7 & 1 \\ 7 & -8 & h \end{bmatrix} 
&= 3\operatorname{det}\begin{bmatrix} 14 & -7 \\ 7 & -8 \end{bmatrix} - 1\operatorname{det}\begin{bmatrix} 7 & -5 \\ 7 & -8 \end{bmatrix} + h\operatorname{det}\begin{bmatrix} 7 & -5 \\ 14 & -7\end{bmatrix} \\
&= 3(14(-8) + 7(7)) - (7(-8) + 7(5)) + h(7(-7)+14(5))\end{align*}
From here, I suggest factoring $7$ out from every term. If you agree with what I have so far, then your problem is arithmetic, not linear algebra.

An easier way is to row-reduce to find $h$ such that the homogeneous system corresponding to the matrix has nontrivial kernel:
$$  \begin{bmatrix} 7 & -5 & 3 \\ 14 & -7 & 1 \\ 7 & -8 & h \end{bmatrix} \longrightarrow \begin{bmatrix} 7 & -5 & 3 \\ 0 & 3 & -5 \\ 0 & -3 & h-3 \end{bmatrix}\longrightarrow \begin{bmatrix} 7 & -5 & 3 \\ 0 & 3 & -5\\0 & 0 & h-8 \end{bmatrix}$$
The determinant is zero exactly when $h=8.$
