Let \[ A=\begin{pmatrix}1&2\\1&-1\end{pmatrix}. \] Express \(A^{-1}\) as a polynomial in \(A\) with real coefficients.
Solution
1. The matrix \(A\) satisfies\[ A^2=3I. \]
Direct multiplication gives \[ \begin{aligned} A^2 &= \begin{pmatrix} 1&2\\ 1&-1 \end{pmatrix} \begin{pmatrix} 1&2\\ 1&-1 \end{pmatrix}\\ &= \begin{pmatrix} 3&0\\ 0&3 \end{pmatrix} = 3I. \end{aligned} \]
2. The requested polynomial expression is\[ \boxed{A^{-1}=\frac13A}. \]
By step 1, \[ A\left(\frac13A\right) = \frac13A^2 = I. \] Since \(A\) commutes with every polynomial in \(A\), the same computation also gives \[ \left(\frac13A\right)A=I. \] Thus \(A^{-1}=\frac13A\), corresponding to the real polynomial \(p(t)=t/3\).
3. Q.E.D.
Step 2 gives the required polynomial expression.