# 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 $\begin{bmatrix} 2 & 1\\ 1& 2 \end{bmatrix}$. Add to solve later

## 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. Add to solve later

### 1 Response

1. 07/11/2017

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