<s>[INST] solving the determinants of matrices So i was given this question
Evaluate by first adding all other rows to the first row
My solution: 
$
det
\left[ {\begin{array}{cc}
x-1 & 2 & 3 \\
2 & -3& x-2 \\
-2 & x & -2
\end{array} } \right]
$ = $
\left[ {\begin{array}{cc}
x-1 & 2 & 3 \\
2 & -3& x-2 \\
-2 & x & -2
\end{array} } \right]
$ = $(x-1)
\left[ {\begin{array}{cc}
1 & 0 & 0 \\
2 & -5& x-4 \\
-2 & x & -2
\end{array} } \right]
$ = $(x-1)
\left[ {\begin{array}{cc}
-5 & x-4\\
x+2 & 0& 
\end{array} } \right]
$=$(x-2)(x-4)(x+2)$ = $(x-2)(x^2-2x-8)$
Is this correct? 
 [/INST] No, this is not correct. You may add any row to any other row without changing the determinant. Using your hint you would get
$$
\det\begin{bmatrix}
x-1 & 2 & 3\\
2   &-3 &x-2 \\
-2  & x & -2
\end{bmatrix} = \det\begin{bmatrix}
x-1 & x-1 & x-1\\
2   &-3 &x-2 \\
-2  & x & -2
\end{bmatrix}
$$
Now we can subtract the first column from the second and third and get
$${} = \det\begin{bmatrix}
x-1 & 0 & 0\\
2   &-5 &x-4 \\
-2  & x+2 & 0
\end{bmatrix}$$
Now we can use the Laplace formula on the first row, because there are a lot of zeros and get
$${}=(x-1)\det\begin{bmatrix}-5 & x-4\\x+2 & 0\end{bmatrix}
= -(x-1)(x-4)(x+2). $$
</s>