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?
Best Answer
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 $k\ne 0$.