Class 12 Matrices · Long Answer3 Marksmatrix_polynomials_and_inverse
Class 12 Maths · Matrices · Long Answer #67
For A=\(\begin{bmatrix}2 & -1\\-1 & 2\end{bmatrix}\), show A²=4A−3I and hence find A⁻¹.
✅ Step-by-Step Solution
Final Answer: \(\begin{bmatrix}2/3 & 1/3\\1/3 & 2/3\end{bmatrix}\)
- Multiplication gives A²=[[5,−4],[−4,5]].
- Also 4A−3I=[[8,−4],[−4,8]]−[[3,0],[0,3]]=[[5,−4],[−4,5]], proving the polynomial identity.
- Rearrange to A(4I−A)=(4I−A)A=3I. Hence (4I−A)/3 is a two-sided inverse, without first assuming invertibility.
- Therefore A⁻¹=(4I−A)/3=\(\begin{bmatrix}2/3 & 1/3\\1/3 & 2/3\end{bmatrix}\).
📋 CBSE Stepwise Marking Scheme
- [0.5 marks] Multiplication gives A²=[[5,−4],[−4,5]].
- [1 mark] Also 4A−3I=[[8,−4],[−4,8]]−[[3,0],[0,3]]=[[5,−4],[−4,5]], proving the polynomial identity.
- [1 mark] Rearrange to A(4I−A)=(4I−A)A=3I. Hence (4I−A)/3 is a two-sided inverse, without first assuming invertibility.
- [0.5 marks] Therefore A⁻¹=(4I−A)/3=\(\begin{bmatrix}2/3 & 1/3\\1/3 & 2/3\end{bmatrix}\).
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