Find the set of all $\alpha$ such that Matrix A is invertible and calculate the inverse for all $\alpha$

Question

$A=\begin{pmatrix} 0 & 1 & -1 & 2\\ 2 & -1 & 3 & 0 \\ \alpha & 0 & 1& 0 \\ 3 & -1 &4 & 0 \end{pmatrix}$

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?

Answer 1

The determinant of $A$ is given by $\det(A)=2(1-a)$, and the inverse, for all $a\neq 1$, is given by $$ A^{-1}= \frac{1}{\det(A)}\begin{pmatrix} 0 & -2 & -2 & 2 \cr 0 & 2(4a-3) & 2 & 2(2-3a) \cr 0 & 2a & 2 & -2a \cr 1-a & 3(1-a) & 0 & 2(a-1) \end{pmatrix}. $$ You don't need to spend the rest of your life for its computation. For example, the Gauss elimination with $A$ in parallel with the identity matrix, gives you the result fairly easy. Also, it is useful to get familiar with a computer algebra system "early in life". Almost all people I know do no longer compute inverse matrices of size $4$ by hand.

Answer 2

$A^{-1}=\dfrac{1}{det(A)}co(A)^T$

It is fairly easy (a bit long but...) to calculate the comatrix of $A$ (matrix of cofactors or adjugate matrix)...

Answer 3

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 $B^{-1}$

$$ A^{-1} = B^{-1} - \dfrac{a}{1-a} \pmatrix{-1\\1\\1\\0} \pmatrix{0&-1&-1&1}$$