NCERT Class 12 Maths · Chapter 4 Determinants · Miscellaneous Exercise Question 4

✅ Step-by-Step Solution

1. The determinant is \(-5\), so all the inverses used here exist. Compute A⁻¹=\(\left[\begin{matrix}- \frac{14}{5} & \frac{9}{5} & \frac{1}{5}\\\frac{9}{5} & - \frac{4}{5} & - \frac{1}{5}\\\frac{1}{5} & - \frac{1}{5} & \frac{1}{5}\end{matrix}\right]\).
2. The adjugate is \(\left[\begin{matrix}14 & -9 & -1\\-9 & 4 & 1\\-1 & 1 & -1\end{matrix}\right]\); its inverse is \(\left[\begin{matrix}- \frac{1}{5} & - \frac{2}{5} & - \frac{1}{5}\\- \frac{2}{5} & - \frac{3}{5} & - \frac{1}{5}\\- \frac{1}{5} & - \frac{1}{5} & -1\end{matrix}\right]\).
3. Computing the adjugate of A⁻¹ gives the same matrix \(\left[\begin{matrix}- \frac{1}{5} & - \frac{2}{5} & - \frac{1}{5}\\- \frac{2}{5} & - \frac{3}{5} & - \frac{1}{5}\\- \frac{1}{5} & - \frac{1}{5} & -1\end{matrix}\right]\).
4. Finally the inverse of A⁻¹ is \(\left[\begin{matrix}1 & 2 & 1\\2 & 3 & 1\\1 & 1 & 5\end{matrix}\right]\), exactly A.