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Class 12 Determinants · Long Answer3 Marksdeterminant_evaluation

Class 12 Maths · Determinants · Long Answer #97

Without using determinant properties, evaluate \(\begin{vmatrix}0 & 2 & -3\\-2 & 0 & 4\\3 & -4 & 0\end{vmatrix}\) and prove it is zero.

✅ Step-by-Step Solution

Final Answer: 0
  1. Expand along row 1 using signs +,−,+: D=0−2det([[-2,4],[3,0]])−3det([[-2,0],[3,−4]]).
  2. The first 2×2 minor is (−2)·0−4·3=−12; the second is (−2)(−4)−0·3=8.
  3. Thus D=−2(−12)−3(8)=24−24=0.
  4. The determinant is zero, proved by direct expansion only.

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] Expand along row 1 using signs +,−,+: D=0−2det([[-2,4],[3,0]])−3det([[-2,0],[3,−4]]).
  • [1 mark] The first 2×2 minor is (−2)·0−4·3=−12; the second is (−2)(−4)−0·3=8.
  • [1 mark] Thus D=−2(−12)−3(8)=24−24=0.
  • [0.5 marks] The determinant is zero, proved by direct expansion only.

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