Let $A$ be a singular $2\times 2$ matrix such that $\tr(A)\neq -1$ and let $I$ be the $2\times 2$ identity matrix.
Then prove that the inverse matrix of the matrix $I+A$ is given by the following formula:
\[(I+A)^{-1}=I-\frac{1}{1+\tr(A)}A.\]

Using the formula, calculate the inverse matrix of $\begin{bmatrix}
2 & 1\\
1& 2
\end{bmatrix}$.

We have
\begin{align*}
(I+A)\left(\, I-\frac{1}{1+\tr(A)}A \,\right)&=I-\frac{1}{1+\tr(A)}A+A-\frac{1}{1+\tr(A)}A^2\\[6pt]
&=I-\frac{1}{1+\tr(A)}\left(\, A-(1+\tr(A))A +A^2\,\right)\\[6pt]
&=I-\frac{1}{1+\tr(A)}\left(\, A^2-\tr(A)A \,\right) \tag{*}.
\end{align*}

The Cayley-Hamilton theorem for $2\times 2$ matrices yields that
\[A^2-\tr(A)A+\det(A)I=O.\]
Since $A$ is singular, we have $\det(A)=0$.
Hence it follows that we have
\[A^2-\tr(A)A=O,\]
and we obtain from (*) that
\[(I+A)\left(\, I-\frac{1}{1+\tr(A)}A \,\right)=I.\]
Similarly,
\[\left(\, I-\frac{1}{1+\tr(A)}A \,\right)(I+A)=I.\]

Therefore, we conclude that the inverse matrix of $I+A$ is given by the formula
\[(I+A)^{-1}=I-\frac{1}{1+\tr(A)}A.\]

Find the inverse matrix of $\begin{bmatrix}
2 & 1\\
1& 2
\end{bmatrix}$ using the formula

Now let us find the inverse matrix of $\begin{bmatrix}
2 & 1\\
1& 2
\end{bmatrix}$ using the formula.

We first write
\[\begin{bmatrix}
2 & 1\\
1& 2
\end{bmatrix}=I+A,\]
where
\[A=\begin{bmatrix}
1 & 1\\
1& 1
\end{bmatrix}.\]
Then $A$ is a singular matrix with $\tr(A)=2$.

The formula yields that
\begin{align*}
\begin{bmatrix}
2 & 1\\
1& 2
\end{bmatrix}^{-1}&=(I+A)^{-1}\\[6pt]
&=I-\frac{1}{3}A\\[6pt]
&=\frac{1}{3}\begin{bmatrix}
2 & -1\\
-1& 2
\end{bmatrix}.
\end{align*}

Related Question.

There is a similar formula for inverse matrices of certain $n\times n$ matrices, called Sherman-Woodberry formula.

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Hint.
Find the trace of $A$.
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by definition.
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