The entries in a 3 × 3 determinant are either 1 of –1, then match the following:
Column I | Column II |
(i) Total number of such determinants | [A] 4 |
(ii) The number of determinants whose values are 6 | [B] 3 |
(iii) The maximum value of such | [C] 512 |
(iv) The maximum value of Trace of such determinants | [D] zero |
Text Solution
Verified by Experts(i) [C]; (ii) [D]; (iii) [A]; (iv) [B]
Ans.
(i) [C]
(ii) [D]
(iii) [A]
(iv) [B]
Sol. (i) Since every entry can take 2 values. Total possible determinants = 2 9 = 512
The value six will be attained only when three positive terms are 1 and negative terms = – 1
(ii ) = a 1 b 2 c 3 + a 2 b 3 c 1 + a 3 b 1 c 2 –a 1 b 3 c 2 – a 3 b 2 c 1 – a 2 b 1 c 3
which is impossible since if this is the case the multiplication of all result –1 = 1
(iii) → A, → Β
The maximum value of such a determinant is 4 attained by 
. The matrix trace is obviously three.
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