# The Formula for the Inverse Matrix of $I+A$ for a $2\times 2$ Singular Matrix $A$

## Problem 505

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}$.

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Contents

## Proof.

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**.

See the post ↴

Sherman-Woodbery Formula for the Inverse Matrix

for the statement of the Sherman-Woodberry formula and its proof.

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