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NCERT Class 12 · Ch 4 Miscellaneous Exercise · Q3

NCERT Class 12 Maths · Chapter 4 Determinants · Miscellaneous Exercise Question 3

Given A⁻¹=\(\left[\begin{matrix}3 & -1 & 1\\-15 & 6 & -5\\5 & -2 & 2\end{matrix}\right]\) and B=\(\left[\begin{matrix}1 & 2 & -2\\-1 & 3 & 0\\0 & -2 & 1\end{matrix}\right]\), find (AB)⁻¹.

💡 Concept Hints

  1. Identify whether this needs determinant expansion, an inverse, or the absolute determinant formula for area.
  2. For a determinant use signed cofactors. For a system first check invertibility; for an area take half the absolute determinant.

✅ Step-by-Step Solution

Final Answer: \(\left[\begin{matrix}9 & -3 & 5\\-2 & 1 & 0\\1 & 0 & 2\end{matrix}\right]\)
  1. Use (AB)⁻¹=B⁻¹A⁻¹; the given matrix is already A⁻¹.
  2. The determinant of B is \(1\), and B⁻¹=\(\left[\begin{matrix}3 & 2 & 6\\1 & 1 & 2\\2 & 2 & 5\end{matrix}\right]\).
  3. Multiply in the reversed order to get B⁻¹A⁻¹=\(\left[\begin{matrix}9 & -3 & 5\\-2 & 1 & 0\\1 & 0 & 2\end{matrix}\right]\).

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