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Class 12 Determinants · Long Answer3 Marksdeterminant_evaluation

Class 12 Maths · Determinants · Long Answer #98

Without using determinant properties, evaluate \(\begin{vmatrix}1+x & 1 & 1\\1 & 1+y & 1\\1 & 1 & 1+z\end{vmatrix}\), and state precisely when it is zero.

✅ Step-by-Step Solution

Final Answer: xyz+xy+xz+yz; zero iff xyz+xy+xz+yz=0.
  1. Expand along row 1: D=(1+x)[(1+y)(1+z)−1]−[(1+z)−1]+[1−(1+y)].
  2. The bracketed terms simplify to y+z+yz, z and −y respectively, so D=(1+x)(y+z+yz)−z−y.
  3. Expand and collect: D=xy+xz+yz+xyz. This is not identically zero; for x=y=z=1 it equals 4.
  4. Therefore the determinant equals xyz+xy+xz+yz, and it vanishes exactly when that expression is zero.

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] Expand along row 1: D=(1+x)[(1+y)(1+z)−1]−[(1+z)−1]+[1−(1+y)].
  • [1 mark] The bracketed terms simplify to y+z+yz, z and −y respectively, so D=(1+x)(y+z+yz)−z−y.
  • [1 mark] Expand and collect: D=xy+xz+yz+xyz. This is not identically zero; for x=y=z=1 it equals 4.
  • [0.5 marks] Therefore the determinant equals xyz+xy+xz+yz, and it vanishes exactly when that expression is zero.

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