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Invertible matrix with sum of each row 1 (TIFR 2015 problem 1)


Let \(A\) be an invertible \(10 \times 10\) matrix with real entries such that the sum of each row is 1. Then

A. The sum of the entries of each row of the inverse of A is 1

B. The sum of the entries of each column of the inverse of A is 1

C. The trace of the inverse of A is non-zero.

D. None of the above.


The sum of each row of \(A\) is 1, means that the sum of the columns of A is the vector \((1,1,…,1)^T \) .

Note that i-th column of \(A\) is given by \(Ae_i \). Therefore, \(\sum_{i=1}^{10} Ae_i = (1,1,…,1)^T \).

Since left multiplication by \(A\) is a linear transformation, the left-hand side of the last expression can be written as \(A(\sum_{i=1}^{10}e_i)\).

Now, \(\sum_{i=1}^{10}e_i = (1,1,…,1)^T \).

Hence we get \(A(1,1…,1)^T = (1,1,…,1)^T \).

Another way of saying the last expression is that the vector \( (1,1…,1)^T \) is fixed by A.

Since A is invertible, applying \(A^{-1}\) on both sides of the last expression we get \((1,…,1)^T = A^{-1}(1,1,…,1)^T \).

By the linearity argument as above, this gives \( (1,1…,1)^T = \sum_{i=1}^{10} A^{-1} (e_i) \). And the i-th term in right-hand side expression is the i-th column of \(A^{-1}\). Therefore, sum of columns of \(A^{-1}\) is the vector \((1,1,…,1)^T \). This is same as saying that sum of entries of each row of \(A^{-1}\) is 1.


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