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The question is:

Prove that If the sum of the elements of each row of a square matrix is k, then the sum of the elements in each row of the inverse matrix is 1/k ?

In the text book the answer is:

Let A be ${m\times m}$, non-singular, with the stated property. Let B be its imverse. Then for $n\leqslant m$, $$ 1 = \sum\limits_{r=1}^m \sigma_{nr} = \sum\limits_{r=1}^m\sum \limits_{s=1}^mb_{ns}a_{sr} = \sum\limits_{s=1}^m\sum \limits_{r=1}^mb_{ns}a_{sr} = k\sum\limits_{s=1}^m b_{ns}$$

(A is singular if K = 0).

I have trouble to understand this proof. Is there another way to prove it?

Oualid
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2 Answers2

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Let $A$ be the invertible square matrix.

The product $A \pmatrix{1\\1\\\vdots\\1} $ gives a column matrix, with elements equal to sum of elements in a row of $A$.

$$\begin{align*} A \pmatrix{1\\1\\\vdots\\1} &= \pmatrix{k\\k\\\vdots\\k} \\ A^{-1}A\pmatrix{1\\1\\\vdots\\1} &= A^{-1} \pmatrix{k\\k\\\vdots\\k}\\ \pmatrix{1\\1\\\vdots\\1} &= A^{-1} \pmatrix{k\\k\\\vdots\\k}\\ \pmatrix{1/k\\1/k\\\vdots\\1/k} &= A^{-1} \pmatrix{1\\1\\\vdots\\1}\\ \end{align*}$$

i.e. each row of $A^{-1}$ sums to $\frac1k$, if $k\ne 0$.

peterwhy
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Let $M$ be the matrix and $u$ be the vector with 1 for all its elements. Then saying that the sum of all the elements in the rows of $M$ is $k$ is equivalent to saying that $$Mu= ku$$ Now multiply that equation by $M^{-1}$ to the left: $$u=kM^{-1}u$$ or $$\frac 1k u=M^{-1}u$$ Which means that the sum of the elements of the rows of $M^{-1}$ all equal $\frac 1k$.

Stefan Lafon
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