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Average Determinant: TIFR GS 2017 Part A Problem 8.

Understand the problem

True or False? Let \(n \geq 2\) be a natural number. Let \(S\) be the set of all \(n \times n\) real matrices whose entries are only \(0,1\) or \(2\). Then the average determinant of a matrix in \(S\) is greater than or equal to 1.
Source of the problem
TIFR GS 2017 Entrance Examination
Topic
Linear Algera
Difficulty Level
Moderate
Suggested Book
Linear Algebra, Fourth Edition, 2015, Paperback, English, Pearson Publisher, by Stephen H. Friedberg, Lawrence E. Spence and Arnold J. Insel, and Linear Algebra, Second Edition, 2015, Paperback, English, Prentice Hall India Learning Private Limited Publisher, by Kenneth Hoffman and Ray Kunze.

Start with hints

Do you really need a hint? Try it first!

The set \(S\) which is the set of all \(n \times n\) real matrices whose entries are only \(0,1\) or \(2\) and whose matrix’s average determinant is asked, can be categorized. Can you find out how?
The set \( S \) can be categorized into three categories: Category 1: the matrices with a positive determinant value Category 2: the matrices with a negative determinant value, and Category 3: the matrices with a zero determinant value. Now you need to find the average value of matrix considering these three categories. How will you do so?
Firstly, you can see that the category 3 whose matrices’ determinant value is always zero doesn’t contribute anything towards the average value. So, you are left with only the first two categories. How can you arrange them to get the average value? Can you find a connection between these two categories?
See if you are given a positive determinant matrix then just interchange any two rows/columns to get a negative determinant matrix or vice versa. That is, there is a one to one correspondence between these two categories of sets. Now, what can be the average value of matrix in \(S\)?
Due to an one to one correspondence between the set of all positive determinant matrix (say, \(X\)) and the set of all negative determinant matrix (say, \(Y\)), the sum of the determinant value of the matrices in \(X\) \(=\) \(-\)the sum of the determinant value of the matrices in \(Y\). Therefore, the total sum becomes zero. Hence the average determinant value of a matrix in \(S\) is zero and not greater than or equal to \(1\). Consequently, the statement is false.

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