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 $\begin{bmatrix}
2 & 1\\
1& 2
\end{bmatrix}$.

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.

The post The Formula for the Inverse Matrix of $I+A$ for a \times 2$ Singular Matrix $A$ first appeared on Problems in Mathematics.

Let $\mathbf{u}$ and $\mathbf{v}$ be vectors in $\R^n$, and let $I$ be the $n \times n$ identity matrix. Suppose that the inner product of $\mathbf{u}$ and $\mathbf{v}$ satisfies
\[\mathbf{v}^{\trans}\mathbf{u}\neq -1.\] Define the matrix
\[A=I+\mathbf{u}\mathbf{v}^{\trans}.\]

Prove that $A$ is invertible and the inverse matrix is given by the formula
\[A^{-1}=I-a\mathbf{u}\mathbf{v}^{\trans},\] where
\[a=\frac{1}{1+\mathbf{v}^{\trans}\mathbf{u}}.\] This formula is called the Sherman-Woodberry formula.

Proof.

Let us put
\[B=I-a\mathbf{u}\mathbf{v}^{\trans},\] the matrix given by the Sherman-Woodberry formula.
We compute $AB$ and $BA$ and show that they are equal to the identity matrix $I$.

Let us first compute the matrix product $AB$. We have
\begin{align*}
AB&=(I+\mathbf{u}\mathbf{v}^{\trans})(I-a\mathbf{u}\mathbf{v}^{\trans})\\
&=I-a\mathbf{u}\mathbf{v}^{\trans}+\mathbf{u}\mathbf{v}^{\trans}-a\mathbf{u}\mathbf{v}^{\trans}\mathbf{u}\mathbf{v}^{\trans}\\
&=I+(1-a)\mathbf{u}\mathbf{v}^{\trans}-a\mathbf{u}\mathbf{v}^{\trans}\mathbf{u}\mathbf{v}^{\trans} \tag{*}
\end{align*}
By using the defining formula for $a=\frac{1}{1+\mathbf{v}^{\trans}\mathbf{u}}$, we have
\[a(1+\mathbf{v}^{\trans}\mathbf{u})=1,\] and thus
\[1-a=a\mathbf{v}^{\trans}\mathbf{u}. \tag{**}\]

Note that the third term in (*) $-a\mathbf{u}\mathbf{v}^{\trans}\mathbf{u}\mathbf{v}^{\trans}$ contains $\mathbf{v}^{\trans}\mathbf{u}$ in the middle, and $\mathbf{v}^{\trans}\mathbf{u}$ is just a number. Thus we can factor out this number and get
\begin{align*}
-a\mathbf{u}\mathbf{v}^{\trans}\mathbf{u}\mathbf{v}^{\trans}&=-a\mathbf{u}(\mathbf{v}^{\trans}\mathbf{u})\mathbf{v}^{\trans}\\
&=-a(\mathbf{v}^{\trans}\mathbf{u})\mathbf{u}\mathbf{v}^{\trans} \tag{***}
\end{align*}
Inserting (**) and (***) into (*), it follows that we have
\begin{align*}
AB&=I+(a\mathbf{v}^{\trans}\mathbf{u})\mathbf{u}\mathbf{v}^{\trans}-a(\mathbf{v}^{\trans}\mathbf{u})\mathbf{u}\mathbf{v}^{\trans}\\
&=I.
\end{align*}
Thus we have proved $AB=I$.

Now we compute $BA$. We have
\begin{align*}
BA&=(I-a\mathbf{u}\mathbf{v}^{\trans})(I+\mathbf{u}\mathbf{v}^{\trans})\\
&=I+\mathbf{u}\mathbf{v}^{\trans}-a\mathbf{u}\mathbf{v}^{\trans}-a\mathbf{u}\mathbf{v}^{\trans}\mathbf{u}\mathbf{v}^{\trans}\\
&=I+(1-a)\mathbf{u}\mathbf{v}^{\trans}-a\mathbf{u}\mathbf{v}^{\trans}\mathbf{u}\mathbf{v}^{\trans}
\end{align*}
and this is exactly the expression (*), hence $BA=AB=I$.

Therefore, we conclude that the matrix $A$ is invertible and the inverse matrix is $B$. Hence
\[A^{-1}=I-a\mathbf{u}\mathbf{v}^{\trans}\] and we have proved the Sherman-Woodberry formula.

Comment.

The invertible matrix theorem says that once we have $AB=I$, then we have automatically $BA=I$ and the inverse matrix of $A$ is $B$, that is, $A^{-1}=B$.
So in the above proof, after proving $AB=I$, you may conclude that $A$ is invertible and $A^{-1}=B$.

Related Question.

See the post ↴
The Formula for the Inverse Matrix of $I+A$ for a $2\times 2$ Singular Matrix $A$
for a proof of this problem.

The post Sherman-Woodbery Formula for the Inverse Matrix first appeared on Problems in Mathematics.