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Problem Statement:

Find all real $x$ so that the matrix is invertible. $$\begin{bmatrix} 2+x & 2 & 2 & 2 \\ 2 & 2+x & 2 & 2 \\ 2 & 2 & 2+x & 2 \\ 2 & 2 & 2 & 2+x \end{bmatrix}$$

Assuming that $A^{-1}$ exists find the sum of all the elements of $A^{-1}$ without explicitly computing $A^{-1}$.

For the first part I calculated the determinant (came out to be $x^4+8x^3$) and put it to $0$ to find that the matrix is not-invertible if $x=-8$ or $x=0$. But I've not idea how to approach the second part of the problem.

  • Did you try with smaller 1x1, 2x2 matrices first? – Somos Mar 19 '18 at 20:48
  • Hint: If the sum of all elements in each row of $A$ is constant, say $c\neq 0$, then $c$ is an eigenvalue of $A$ and $1/c$ is an eigenvalue of $A^{-1}$ (in fact, the sum of all elements in each row of $A^{-1}$ is $1/c$, a constant). – Prasun Biswas Mar 19 '18 at 20:50

1 Answers1

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As suggested here, the sum of all elements in a matrix $M$ is given by $\langle Me, e\rangle$ where $e = \pmatrix{1 \\ 1 \\ 1 \\ 1}$.

In that case, if $v$ is a vector such that $Av = e$, then we have $$\langle A^{-1}e, e\rangle = \langle v, e\rangle$$

Therefore, we have to solve the system

$$\pmatrix{2+x & 2 & 2 &2 \\ 2 & 2+x & 2 & 2 \\ 2 & 2 & 2+x &2 \\ 2 & 2 & 2 & 2+x} \pmatrix{v_1 \\ v_2 \\ v_3 \\ v_4} = \pmatrix{1 \\ 1 \\ 1 \\ 1}$$

Since $A$ is invertible, the solution is unique and it is given by $$v = \frac{1}{x+8}\pmatrix{1 \\ 1 \\ 1 \\ 1}$$

Therefore the sum is equal to

$$\langle v, e\rangle = \frac1{x+8}\|e\|^2 = \frac{4}{x+8}$$

Note that solving the linear system above is in general faster than inverting the matrix, especially in this case since your $A$ is symmetric.

mechanodroid
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