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TIFR 2015 Problem 1 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India’s premier institution for advanced research in Mathematics. The Institute runs a graduate programe leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
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Problem:

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.

Discussion:

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.

Helpdesk

• What is this topic: Linear Algebra
• What are some of the associated concept: Linear Transformation,Invertible Matrix
• Book Suggestions: Introduction to Linear Algebra by Gilbert Strang