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NCERT Class 12 · Ch 4 Exercise 4.4 · Q12

NCERT Class 12 Maths · Chapter 4 Determinants · Exercise 4.4 Question 12

For A = \(\left[\begin{matrix}3 & 7\\2 & 5\end{matrix}\right]\), B = \(\left[\begin{matrix}6 & 8\\7 & 9\end{matrix}\right]\), verify (AB)⁻¹=B⁻¹A⁻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: \(\left[\begin{matrix}- \frac{61}{2} & \frac{87}{2}\\\frac{47}{2} & - \frac{67}{2}\end{matrix}\right]\)
  1. The determinants are \(1\) and \(-2\), both nonzero.
  2. Compute A⁻¹=\(\left[\begin{matrix}5 & -7\\-2 & 3\end{matrix}\right]\) and B⁻¹=\(\left[\begin{matrix}- \frac{9}{2} & 4\\\frac{7}{2} & -3\end{matrix}\right]\).
  3. Their reversed product B⁻¹A⁻¹=\(\left[\begin{matrix}- \frac{61}{2} & \frac{87}{2}\\\frac{47}{2} & - \frac{67}{2}\end{matrix}\right]\).
  4. Directly AB=\(\left[\begin{matrix}67 & 87\\47 & 61\end{matrix}\right]\) and (AB)⁻¹=\(\left[\begin{matrix}- \frac{61}{2} & \frac{87}{2}\\\frac{47}{2} & - \frac{67}{2}\end{matrix}\right]\), the same matrix.

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