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Summing an inverse

Linear algebraHard

Let $A$ be the $3 \times 3$ matrix with $2$ in every diagonal entry and $1$ in every other entry. What is the sum of all $9$ entries of $A^{- 1}$ ? Give the answer as a fraction in lowest terms.

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Hint

Look at what $A$ does to the vector $(1, 1, 1)$ .

Solution

Each row of $A$ sums to $2 + 1 + 1 = 4$ , so $A u = 4 u$ for $u = (1, 1, 1)$ .
Applying $A^{- 1}$ to both sides gives $A^{- 1} u = \frac{u}{4}$ , so every row of $A^{- 1}$ sums to $\frac{1}{4}$ .
The sum of all entries is the sum of the $3$ row sums, $3 \times \frac{1}{4} = \frac{3}{4}$ .

Answer: $\frac{3}{4}$

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