why I am not getting the inverse of this matrix:

Question

I want to get the inverse of the matrix:

$$\begin{pmatrix} 1+\lambda & 1 & 1 & 1\\ 1 & 1+\lambda & 1 & 1\\ 1 & 1 & 1+\lambda & 1 \\ 1 & 1 & 1 & 1+\lambda \end{pmatrix}$$

my first step was doing some cofactor expansion:

$$\begin{pmatrix} 1+\lambda & 1 & 1 & 1\\ -\lambda & \lambda & 0 & 0\\ -\lambda & 0 & \lambda & 0 \\ -\lambda & 0 & 0 & \lambda \end{pmatrix}$$

then the cofactor matrix:

$$\begin{pmatrix} \lambda^3 & \lambda^3 & \lambda^3 & \lambda^3\\ -\lambda^2 &\lambda^3+4\lambda^2 & 0 & 0\\ -\lambda^2 & 0 & \lambda^3+4\lambda^2 & 0 \\ -\lambda^2 & 0 & 0 & \lambda^3+4\lambda^2 \end{pmatrix}$$

transposing and putting the determinant: $$\frac{1}{\lambda^3(4+\lambda)}\begin{pmatrix} \lambda^3 & -\lambda^2 & -\lambda^2 & -\lambda^2\\ \lambda^3 &\lambda^3+4\lambda^2 & 0 & 0\\ \lambda^3 & 0 & \lambda^3+4\lambda^2 & 0 \\ \lambda^3 & 0 & 0 & \lambda^3+4\lambda^2 \end{pmatrix}$$

however, according to Wolfram This is wrong I was checking the answer but I could not find my mistake

Answer 1

If $J$ is the matrix with all ones, the linear space of matrices of the form $aI+bJ$ is a commutative algebra. In particular the inverse of $aI+bJ$ has the form $cI+dJ$ with $c=1/a$ and $d=-b/(a^2+nab).$

Answer 2

Let $A$ be the matrix in question, then $A = \lambda I + vv^*$, where $v = \frac{1}{2}\mathbf{1}$ so that $vv^*$ is the $4\times 4$ matrix of all ones. The Sherman-Morrison formula then yields $$ A^{-1} = \frac{1}{\lambda} I - \left(\frac{1}{\lambda^2 + 4\lambda}\right)vv^*. $$

In components, this is $$ A^{-1}= \begin{pmatrix} \frac{1}{\lambda} - \frac{1}{\lambda^2 + 4\lambda} & \frac{1}{\lambda^2 + 4\lambda} & \frac{1}{\lambda^2 + 4\lambda} & \frac{1}{\lambda^2 + 4\lambda} & \\ \frac{1}{\lambda^2 + 4\lambda} & \frac{1}{\lambda} - \frac{1}{\lambda^2 + 4\lambda} & \frac{1}{\lambda^2 + 4\lambda} & \frac{1}{\lambda^2 + 4\lambda} & \\ \frac{1}{\lambda^2 + 4\lambda} & \frac{1}{\lambda^2 + 4\lambda} & \frac{1}{\lambda} - \frac{1}{\lambda^2 + 4\lambda} & \frac{1}{\lambda^2 + 4\lambda} & \\ \frac{1}{\lambda^2 + 4\lambda} & \frac{1}{\lambda^2 + 4\lambda} & \frac{1}{\lambda^2 + 4\lambda} &\frac{1}{\lambda} - \frac{1}{\lambda^2 + 4\lambda} & \\ \end{pmatrix}. $$

Answer 3

Let $A$ be the matrix in question:

$$A = \begin{pmatrix} 1+\lambda & 1 & 1 & 1\\ 1 & 1+\lambda & 1 & 1\\ 1 & 1 & 1+\lambda & 1 \\ 1 & 1 & 1 & 1+\lambda \end{pmatrix}$$

Then the cofactors of the entries are either

$$\left\{ \pm\det\begin{pmatrix} 1+\lambda & 1 & 1\\ 1 & 1+\lambda & 1\\ 1 & 1 & 1+\lambda\\ \end{pmatrix}, \pm\det\begin{pmatrix} 1 & 1 & 1\\ 1 & 1+\lambda & 1\\ 1 & 1 & 1+\lambda\\ \end{pmatrix} \right\}$$

Please check how you got your cofactor matrix with entries with many different absolute values.

I previously suspected that you have obtained the inverse of the matrix after row operations, not the inverse of the original matrix. But such claim is wrong too.

Answer 4

Let $C$ be the $4\times 4$ matrix with every entry equal to $1$. You want the inverse of $C+\lambda I$, where $I$ is the $4\times 4$ identity matrix. Notice that $$ C^2=4C $$ Therefore $C^3=C^2C=4C^2=16C$. Continuing in this fashion gives $C^n=4^{n-1}C$, which leads to an explicit expression for the resolvent of $C$: \begin{align} (\lambda I+C)^{-1}&=\frac{1}{\lambda}(I+\frac{1}{\lambda}C)^{-1} \\ &=\frac{1}{\lambda}(I-\frac{1}{\lambda}C+\frac{1}{\lambda^2}C^2-\frac{1}{\lambda^3}C^3+\cdots) \\ &=\frac{1}{\lambda}(I-\frac{1}{\lambda}C+\frac{1}{\lambda^2}4C-\frac{1}{\lambda^3}4^2C+\cdots) \\ &=\frac{1}{\lambda}I-\frac{1}{\lambda^2}(1-\frac{4}{\lambda}+\frac{4^2}{\lambda^2}+\cdots)C \\ &=\frac{1}{\lambda}I-\frac{1}{\lambda^2}\frac{1}{1+\frac{4}{\lambda}}C \\ &=\frac{1}{\lambda}I-\frac{1}{\lambda(\lambda+4)}C \end{align} You can directly check the validity of this assertion by using the relations derived above: \begin{align} &(\lambda I+C)\left(\frac{1}{\lambda}I-\frac{1}{\lambda(\lambda+4)}C\right) \\ &= I+\frac{1}{\lambda}C-\frac{1}{\lambda+4}C-\frac{1}{\lambda(\lambda+4)}C^2 \\ &= I+\frac{4}{\lambda(\lambda+4)}C-\frac{1}{\lambda(\lambda+4)}C^2 \\ &= I+\frac{1}{\lambda(\lambda+4)}(4C-C^2) \\ &= I. \end{align} Therefore, $$ (\lambda I+C)^{-1}=\frac{1}{\lambda}I-\frac{1}{\lambda(\lambda+4)}C $$