Class 12 Determinants · Short Answer2 Marksmatrix_inverse
Class 12 Maths · Determinants · Short Answer #51
Find (AB)⁻¹ if A=\(\begin{bmatrix}1 & 0\\-4 & 2\end{bmatrix}\) and B⁻¹=\(\begin{bmatrix}3 & 1\\5 & 2\end{bmatrix}\).
✅ Step-by-Step Solution
Final Answer: \(\begin{bmatrix}5 & 1/2\\9 & 1\end{bmatrix}\)
- det(A)=2 and A⁻¹=(1/2)[[2,0],[4,1]]=[[1,0],[2,1/2]].
- Reverse the factors: (AB)⁻¹=B⁻¹A⁻¹=[[3,1],[5,2]][[1,0],[2,1/2]]=[[5,1/2],[9,1]].
- Therefore (AB)⁻¹=\(\begin{bmatrix}5 & 1/2\\9 & 1\end{bmatrix}\).
📋 CBSE Stepwise Marking Scheme
- [0.5 marks] det(A)=2 and A⁻¹=(1/2)[[2,0],[4,1]]=[[1,0],[2,1/2]].
- [1 mark] Reverse the factors: (AB)⁻¹=B⁻¹A⁻¹=[[3,1],[5,2]][[1,0],[2,1/2]]=[[5,1/2],[9,1]].
- [0.5 marks] Therefore (AB)⁻¹=\(\begin{bmatrix}5 & 1/2\\9 & 1\end{bmatrix}\).
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