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?
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
$${}=(x-1)\det\begin{bmatrix}-5 & x-4\\x+2 & 0\end{bmatrix} = -(x-1)(x-4)(x+2). $$