We consider the matrix $A=I-a\vec{v}\vec{v}^T$, where $I$ is the $n\times n$ identity matrix and $\vec{v}$ is a unit vector $n\times 1$. Find for which value of $a$ the determinant of the matrix $A$ is zero. If the determinant of the matrix $A$ is not zero, find the value of $b$ such that $I+b\vec{v}\vec{v}^T$ is the inverse of $A$. For $a>0$ apply this for the inversion of the matrix $A=\begin{pmatrix}1+a & a & \ldots & a \\ a & 1+a & \ldots & a \\ \vdots & \vdots & \vdots & \vdots \\ a & a & \ldots & 1+a\end{pmatrix}$. 

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Could you give me a hint for the first part, about how to compute $a$ ? To what is $\det (I-a\vec{v}\vec{v}^T)$ equal? 

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As for the second part: 

We want that $A^{-1}=I+b\vec{v}\vec{v}^T$. 

We have the following:\begin{align\*}A\cdot A^{-1}=I &\Rightarrow (I-a\vec{v}\vec{v}^T)\cdot (I+b\vec{v}\vec{v}^T)=I \\ & \Rightarrow I(I+b\vec{v}\vec{v}^T)-a\vec{v}\vec{v}^T(I+b\vec{v}\vec{v}^T)=I \\ & \Rightarrow I+b\vec{v}\vec{v}^T-a\vec{v}\vec{v}^T-ab\vec{v}\vec{v}^T\vec{v}\vec{v}^T=I \\ & \Rightarrow b\vec{v}\vec{v}^T-a\vec{v}\vec{v}^T-ab\vec{v}\vec{v}^T\vec{v}\vec{v}^T=0 \\ & \Rightarrow b\vec{v}\vec{v}^T-ab\vec{v}\vec{v}^T\vec{v}\vec{v}^T=a\vec{v}\vec{v}^T \\ & \Rightarrow b\left (\vec{v}\vec{v}^T-a\vec{v}\vec{v}^T\vec{v}\vec{v}^T\right )=a\vec{v}\vec{v}^T \\ & \Rightarrow b\left (I-a\vec{v}\vec{v}^T\right )\vec{v}\vec{v}^T=a\vec{v}\vec{v}^T\\ & \Rightarrow b\left (I-a\vec{v}\vec{v}^T\right )=a\\ & \Rightarrow b=a\left (I-a\vec{v}\vec{v}^T\right )^{-1}\end{align\*}

Is this correct? 

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As for the last part: 

We have that \begin{align\*}A&=\begin{pmatrix}1+a & a & \ldots & a \\ a & 1+a & \ldots & a \\ \vdots & \vdots & \vdots & \vdots \\ a & a & \ldots & 1+a\end{pmatrix} =\begin{pmatrix}1 & 0 & \ldots & 0 \\ 0 & 1 & \ldots & 0 \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \ldots & 1\end{pmatrix}+\begin{pmatrix}a & a & \ldots & a \\ a & a & \ldots & a \\ \vdots & \vdots & \vdots & \vdots \\ a & a & \ldots & a\end{pmatrix} =I+a\begin{pmatrix}1 & 1 & \ldots & 1 \\ 1 & 1 & \ldots & 1 \\ \vdots & \vdots & \vdots & \vdots \\ 1 & 1 & \ldots & 1\end{pmatrix} \\ & =I+a\begin{pmatrix}1 \\ 1 \\ \vdots \\ 1 \end{pmatrix}\begin{pmatrix}1 & 1 & \ldots & 1\end{pmatrix} =I-a\begin{pmatrix}-1 \\ -1 \\ \vdots \\ -1 \end{pmatrix}\begin{pmatrix}-1 & -1 & \ldots & -1\end{pmatrix}\end{align\*} 

So we have in this case $\vec{v}=\begin{pmatrix}-1 \\ -1 \\ \vdots \\ -1 \end{pmatrix}$. 

Therefore the inverse matrix is $A^{-1}=I+b\vec{v}\vec{v}^T$ with $b=a\left (I-a\vec{v}\vec{v}^T\right )^{-1}$. 

Is this part correct?