Matrices and Determinants 2 Question 24

26. Let P be a matrix of order 3×3 such that all the entries in P are from the set 1,0,1. Then, the maximum possible value of the determinant of P is

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Answer:

Correct Answer: 26. (4)

Solution:

  1. Let Det(P)=|a1b1c1a2b2c2a3b3c3|

=a1(b2c3b3c2)a2(b1c3b3c1)+a3(b1c2b2c1)

Now, maximum value of Det(P)=6

If a1=1,a2=1,a3=1,b2c3=b1c3=b1c2=1

 and b3c2=b3c1=b2c1=1

But it is not possible as

(b2c3)(b3c1)(b1c2)=1 and (b1c3)(b3c2)(b2c1)=1

 i.e., b1b2b3c1c2c3=1 and 1

Similar contradiction occurs when a1=1,a2=1,a3=1,b2c1=b3c1=b1c2=1

 and b3c2=b1c3=b1c2=1

Now, for value to be 5 one of the terms must be zero but that will make 2 terms zero which means answer cannot be 5

Now, |111111111|=4

Hence, maximum value is 4 .



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