Class 12 Matrices · Long Answer3 Marksmatrix_polynomials_and_inverse
Class 12 Maths · Matrices · Long Answer #68
For A=\(\begin{bmatrix}2 & 3\\-1 & 2\end{bmatrix}\), prove A²−4A+7I=O. Use the result to calculate A⁵.
✅ Step-by-Step Solution
Final Answer: \(\begin{bmatrix}-118 & -93\\31 & -118\end{bmatrix}\)
- Compute A²=[[1,12],[−4,1]]. This equals 4A−7I, proving the stated identity.
- Multiply successively and reduce powers: A³=4A²−7A=9A−28I, then A⁴=9A²−28A=8A−63I.
- Therefore A⁵=8A²−63A=8(4A−7I)−63A=−31A−56I.
- Substitution gives A⁵=\(\begin{bmatrix}-118 & -93\\31 & -118\end{bmatrix}\).
📋 CBSE Stepwise Marking Scheme
- [0.5 marks] Compute A²=[[1,12],[−4,1]]. This equals 4A−7I, proving the stated identity.
- [1 mark] Multiply successively and reduce powers: A³=4A²−7A=9A−28I, then A⁴=9A²−28A=8A−63I.
- [1 mark] Therefore A⁵=8A²−63A=8(4A−7I)−63A=−31A−56I.
- [0.5 marks] Substitution gives A⁵=\(\begin{bmatrix}-118 & -93\\31 & -118\end{bmatrix}\).
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