Let $A$ be a $10\times 10$ matrix.
and entry on diagonal equal to $1$ and othet entry equal to $-1$.
I need sum of entry of $A^{-1}$.
can you help me?
sum of entry of $A^{-1}$ in matrix $10\times 10$
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linear-algebra
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0Typing your matrix [here](http://www.bluebit.gr/matrix-calculator/calculate.aspx) gives the result as $-\frac {5}{4} $. – 2017-01-31
3 Answers
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Denote by $N$ the matrix consisiting only of $1$'s. We have $A=2I-N$. Then we compute, using $N^2=10N$ that $A(8I-N)=16I$, so that $$ A^{-1}=\frac{1}{16}{(8I-N)}. $$ Hence the matrix $A^{-1}$ has $\frac{7}{16}$ on the diagonal, and any other entry equal to $-\frac{1}{16}$. The sum of all entries of $A^{-1}$ therefore is $-\frac{5}{4}$.
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0I think your $\frac{7}{10}$ is not correct. Otherwise, very elegant. – 2017-01-31
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0Just a typo, corrected! – 2017-01-31
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Hint: Let $x$ denote the column vector whose entries are all $1$s. You're trying to compute $x^TA^{-1}x$. Note that $x$ is an eigenvector of $A$; it's easy to compute $Ax = -8x$.
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Your matrix $A$ has $1$ on the diagonal and $-1$ off the diagonal. Conjecture that $A^{-1}$ has $a$ on the diagonal and $b$ off the diagonal. Calculating $AA^{-1}_{ii}$ for $i=1,\ldots,10$ gives $a-9b=1$. Calculating $AA^{-1}_{ij}$ for $i\neq j$ gives $-a-7b=0$. Solving these two equations and calculating $10a+90b$ gives your answer.