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NDA 2019 Maths · Mathematics · Determinants · easy

The value of beginvmatrix2&0&1\\1&3&2\\1&1&4endvmatrix is

Single correct option

The value of ∣201132114∣\begin{vmatrix}2&0&1\\1&3&2\\1&1&4\end{vmatrix} is
  1. A. 2222
  2. B. 1414
  3. C. −18-18
  4. D. 1818
Show answer and solution

Answer: (D) 1818

Expand along the first row: 2(3⋅4−2⋅1)−0+1(1⋅1−3⋅1)=2(10)+(−2)=182(3\cdot4-2\cdot1)-0+1(1\cdot1-3\cdot1)=2(10)+(-2)=18. The zero in the first row saves one cofactor. The value 2222 results from missing the negative sign in 1⋅1−3⋅1=−21\cdot1-3\cdot1=-2, and 1414 from using +2+2 with the wrong sign for the last cofactor.
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Original question written in the NDA 2019 Maths pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake