Assume that A∈Rn×nA \in \mathbb{R}^{n\times n}A∈Rn×n and A2+4A+6In=OA^2+4A+6I_n=OA2+4A+6In=O. If (A+3In)−1=aA+bIn(A+3I_n)^{-1}=aA+bI_n(A+3In)−1=aA+bIn. What is the value of 2a+b2a+b2a+b?
參考答案與解析
答案 (B) -1。由A2+4A+6I=OA^2+4A+6I=OA2+4A+6I=O得A2=−4A−6IA^2=-4A-6IA2=−4A−6I。設(A+3I)(aA+bI)=I(A+3I)(aA+bI)=I(A+3I)(aA+bI)=I，展開並代入A2=−4A−6IA^2=-4A-6IA2=−4A−6I化簡後得(b−a)A+(3b−6a)I=I(b-a)A+(3b-6a)I=I(b−a)A+(3b−6a)I=I，比較係數：b−a=0b-a=0b−a=0且3b−6a=13b-6a=13b−6a=1，解得a=b=−1/3a=b=-1/3a=b=−1/3，故2a+b=−12a+b=-12a+b=−1。
