<s>[INST] Find the set of all $\alpha$ such that Matrix A is invertible and calculate the inverse for all $\alpha$ $A=\begin{pmatrix}
        0 & 1 & -1 & 2\\
        2 & -1 & 3 & 0 \\
        \alpha & 0 & 1& 0 \\ 
3 & -1 &4 & 0
        \end{pmatrix}$
I know that a Matrix is invertible only if $det(A)\not=0$
$det(A)=0\begin{vmatrix}
        -1 & 3 & 0 \\
        0 & 1 & 0 \\
        -1 & 4 & 0 \\
        \end{vmatrix}-1\begin{vmatrix}
        2 & 3 & 0 \\
        \alpha & 1 & 0 \\
        3 & 4 & 0 \\
        \end{vmatrix}-1\begin{vmatrix}
        2 & -1 & 0 \\
        \alpha & 0 & 0 \\
        3 & -1 & 0 \\
        \end{vmatrix}-2\begin{vmatrix}
        2 & -1 & 3 \\
        \alpha & 0 & 1 \\
        3 & -1 & 4 \\
        \end{vmatrix}$
$=-[2\begin{vmatrix}
        1 & 0\\
        4 & 0\\
        \end{vmatrix}-3\begin{vmatrix}
        \alpha & 0\\
        3 & 0\\
        \end{vmatrix}]-[2\begin{vmatrix}
        0 & 0\\
        -1 & 0\\
        \end{vmatrix}+\begin{vmatrix}
        \alpha & 0\\
        3 & 0\\
        \end{vmatrix}]-2[2\begin{vmatrix}
        0 & 1\\
        -1 & 4\\
        \end{vmatrix}+\begin{vmatrix}
        \alpha & 1\\
        3 & 4\\
        \end{vmatrix}+3\begin{vmatrix}
        \alpha & 0\\
        3 & -1\\
        \end{vmatrix}]$
$=-2(2+4\alpha-3-3\alpha)=-4-8\alpha+6+6\alpha=-2\alpha+2$
$-2\alpha+2=0 \iff \alpha=1$
So the matrix is invertible for all $\alpha \in \Bbb R| \alpha\not=0$
But how do I calculate the inverse of A for ALL other $\alpha$ without spending the rest of my life on this question? Am I missing something here?
Thanks in advance?
 [/INST] here is another way to do this problem. we will use a slight variation of the  sherman-morrison formula 
$$ (B+ auv^T)^{-1} = B^{-1} - {aB^{-1}uv^TB^{-1}  \over 1 + av^TB^{-1}u }$$ 
where $u, v$ are in $R^4$ and $a$ is a constant. to fit the question, we take
$$u = \pmatrix{0\\0\\1\\0}, v^T=\pmatrix{1,0,0,0} \text{ and } B = \pmatrix{0&1&-1&2\\2&-1&3&0\\0&0&1&0\\3&-1&4&0}, B^{-1} = \pmatrix{0&-1&-1&1\\0&-3&1&2\\0&0&1&0\\1/2&3/2&0&-1}$$  all we from are third column $B^{-1}u=\pmatrix{-1\\1\\1\\0}$ and the first row $v^TB^{-1} = \pmatrix{0&-1&-1&1}$.
we really did not have to compute all elements of $B^{-1}$ 
$$ A^{-1} = B^{-1} - \dfrac{a}{1-a} \pmatrix{-1\\1\\1\\0} \pmatrix{0&-1&-1&1}$$
</s>