Maths · Matrices, algebra of matrices, type of matrices
Suppose is any non-singular matrix and (\boldsymbol{A}-\mathbf{5} \boldsymbol{I}
Suppose \( A \) is any \( 3 \times 3 \) non-singular matrix and \( (\boldsymbol{A}-\mathbf{3} \boldsymbol{I})(\boldsymbol{A}-\mathbf{5} \boldsymbol{I})=\boldsymbol{O} \) where \( \boldsymbol{I}=\boldsymbol{I}_{3} \) and \( \boldsymbol{O}=\boldsymbol{O}_{3}, \) If \( \boldsymbol{\alpha} \boldsymbol{A}+ \) \( \beta A^{-1}=4 I, \) then \( \alpha+\beta \) is equal to
- A. 8
- B. 7
- C. 13
- D. 12
Step-by-step solution
Given (A-3I)(A-5I)=O, we have A^2 - 8A +15I = O. Multiply by A^{-1} (A is non-singular): A -8I +15A^{-1}=O => A^{-1} = (8I-A)/15. Substitute into αA+βA^{-1}=4I: αA + β(8I-A)/15 =4I => multiply by 15: 15αA +8βI -βA =60I => (15α-β)A = (60-8β)I. Since A is not necessarily a scalar multiple of I, equate coefficients: 15α-β=0 and 60-8β=0. Solving gives β=7.5, α=0.5, so α+β=8. Alternatively, if A=λI with λ=3 or 5, same result. Thus α+β=8.