Class 12 Determinants · Long Answer3 Marksmatrix_inverse

Class 12 Maths · Determinants · Long Answer #96

For A=\(\begin{bmatrix}3 & 7\\2 & 5\end{bmatrix}\) and B=\(\begin{bmatrix}6 & 8\\7 & 9\end{bmatrix}\), verify (AB)⁻¹=B⁻¹A⁻¹.

✅ Step-by-Step Solution

Final Answer: \(\begin{bmatrix}-61/2 & 87/2\\47/2 & -67/2\end{bmatrix}\)

Compute AB=\(\begin{bmatrix}67 & 87\\47 & 61\end{bmatrix}\), with determinant -2≠0.
Invert the product directly: (AB)⁻¹=\(\begin{bmatrix}-61/2 & 87/2\\47/2 & -67/2\end{bmatrix}\).
Independently, A⁻¹=\(\begin{bmatrix}5 & -7\\-2 & 3\end{bmatrix}\) and B⁻¹=\(\begin{bmatrix}-9/2 & 4\\7/2 & -3\end{bmatrix}\). Their reversed product is \(\begin{bmatrix}-61/2 & 87/2\\47/2 & -67/2\end{bmatrix}\).
The two results agree, verifying (AB)⁻¹=B⁻¹A⁻¹.

[0.5 marks] Compute AB=\(\begin{bmatrix}67 & 87\\47 & 61\end{bmatrix}\), with determinant -2≠0.
[1 mark] Invert the product directly: (AB)⁻¹=\(\begin{bmatrix}-61/2 & 87/2\\47/2 & -67/2\end{bmatrix}\).
[1 mark] Independently, A⁻¹=\(\begin{bmatrix}5 & -7\\-2 & 3\end{bmatrix}\) and B⁻¹=\(\begin{bmatrix}-9/2 & 4\\7/2 & -3\end{bmatrix}\). Their reversed product is \(\begin{bmatrix}-61/2 & 87/2\\47/2 & -67/2\end{bmatrix}\).
[0.5 marks] The two results agree, verifying (AB)⁻¹=B⁻¹A⁻¹.

Practise interactive questions, flashcards, visual mind maps, and timed board tests for Determinants.