# Writing $B^{-1}$ in terms of $I, B, B^2$. [closed]

Give an expression for an inverse of a matrix, i.e., $$B^{-1}$$, using only $$I, B, B^2$$.

This seems like it shouldn't be too difficult but I am unable to find anything close.

## closed as unclear what you're asking by Dietrich Burde, Yanior Weg, Lord Shark the Unknown, Lee David Chung Lin, max_zornMay 7 at 4:36

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• Do you mean the inverse of $B^{-1}$? That's clearly $B$. – Jerry Chang May 5 at 13:25
• No, I am asking for $B^{-1}=...$, not $(B^{-1})^{-1}=...$. Sorry for the confusion. – Montes May 5 at 13:27
• See this question. If $B^3=0$ we only need $I,B,B^2$. But your question is not clear. – Dietrich Burde May 5 at 13:32
• If the matrix is $3\times3$, then, given the Cayley-Hamilton theorem, we have that $B$ satisfy its own characteristic equation $B^3 + a_1 B^2 + a_2 B + a_3 I = 0$, so that $B^3$ can be expressed in terms of $B^2, B, I$. – enzotib May 5 at 13:41

If you want to express $$M^{-1}$$ by a linear combination of $$I_n$$, $$M$$ and $$M^2$$, this work if $$n\leq 3$$ using Cayley–Hamilton theorem. But its not possible if $$n\geq 4$$. Take $$M=\left(\begin{array}{cccc} 1&1&0&0\\ 0&1&1&0\\ 0&0&1&1\\ 0&0&0&1\\ \end{array}\right)$$
Then $$M^2=\left(\begin{array}{cccc} 1&2&1&0\\ 0&1&2&1\\ 0&0&1&2\\ 0&0&0&1\\ \end{array}\right)$$
So $$Vect(I_4,M,M^2)\subset\{A\in M_4(\Bbb{R})~|~A_{1,4}=0\}$$.
But $$M^{-1}=\left(\begin{array}{cccc} 1&-1&1&-1\\ 0&1&-1&1\\ 0&0&1&-1\\ 0&0&0&1\\ \end{array}\right)$$
So $$M^{-1}$$ is not a linear combination of $$I_4,M$$ and $$M^2$$.