Show that $B$ is invertible if $B=A^2-2A+2I$ and $A^3=2I$

If $A$ is $40\times 40$ matrix such that $A^3=2I$ show that $B$ is invertible where $B=A^2-2A+2I$.

I tried to evaluate $B(A-I)$ , $B(A+I)$ , $B(A-2I)$ ... but I couldn't find anything.

Solution 1:

You can find the inverse directly by solving $$\left(\alpha A^2+\beta A +\gamma I\right)B=I.$$ If you do, you will find that $\alpha = 1/10$, $\beta = 3/10$, and $\gamma = 4/10$. Hence, $$B^{-1} = \tfrac{1}{10}A^2+\tfrac{3}{10}A+\tfrac{4}{10}I.$$ This can be easily checked by multiplying out and using the relation $A^{3}=2I$.

Solution 2:

Here's something I call the "miracle method" for this type of problem. Suspend your disbelief for a moment and suppose $A$ and $B$ were scalars, not matrices. Then, we would simply be looking for $$ \frac{1}{B} = \frac{1}{A^2 - 2A - 2} = \frac{1}{2}+\frac{A}{2}+\frac{A^2}{4}-\frac{A^4}{8}-\frac{A^5}{8} + \cdots$$ where in the power series expansion, the coefficient of $A^n$ is $$ c_n = \frac{1+i}{2^{n+2}} \left((1-i)^n-i (1+i)^n\right). $$ But we know that $A^3 = 2$, so this becomes $$ \frac{1}{2}+\frac{A}{2}+\frac{A^2}{4}-\frac{A^4}{8}-\frac{A^5}{8} + \cdots = \frac{1}{2}+\frac{A}{2}+\frac{A^2}{4}-\frac{A}{4}-\frac{A^2}{4} + \cdots $$ and by summing the resulting coefficients on $1$, $A$, and $A^2$, we find that $$ \frac{1}{B} = \frac{2}{5} + \frac{3}{10}A + \frac{1}{10}A^2. $$ Now, what we've just done should be total nonsense if $A$ and $B$ are really matrices, not scalars. But try setting $B^{-1} = \frac{2}{5}I + \frac{3}{10}A + \frac{1}{10}A^2$, and you'll find that, miraculously, this answer works!

Solution 3:

The eigenvalues of $A$ are the cubic roots of $2$. Then $$B=(A-(1+i)I)(A-(1-i)I)$$ is a regular matrix.

Solution 4:

You have $$B=A^3+A^2-2A=A(A-I)(A+2I)\ .$$ $A$ is invertible, and since $$A^3-I=I\qquad\Rightarrow\qquad(A-I)(A^2+A+I)=I$$ and $$A^3+8I=10I\qquad\Rightarrow\qquad(A+2I)(A^2-2A+4I)=10I$$ also $(A-I)$ and $(A+2I)$ are invertible.

Now, since $B$ is the product of invertible matrices, it is invertible.