If $A=\left[\begin{array}{ccc}1 & 0 & -2 \ -2 & -1 & 2 \ 3 & 4 & 1\end{array}\right]$, then $A^{-1}=$

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Aicharya · Accepted Answer

Characteristic equation of any square matrix $A$ is given by $\begin{aligned} & |A-\lambda I|=0 \ & \Rightarrow \quad\left[\begin{array}{ccc} 1 & 0 & -2 \ -2 & -1 & 2 \ 3 & 4 & 1 \end{array}\right]-\lambda\left[\begin{array}{lll} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{array}\right]=0 \ & \Rightarrow \quad\left[\begin{array}{ccc} 1 & 0 & -2 \ -2 & -1 & 2 \ 3 & 4 & 1 \end{array}\right]-\left[\begin{array}{ccc} \lambda & 0 & 0 \ 0 & \lambda & 0 \ 0 & 0 & \lambda \end{array}\right]=0 \ & \Rightarrow \quad\left|\begin{array}{ccc} 1-\lambda & 0 & -2 \ -2 & -1-\lambda & 2 \ 3 & 4 & 1-\lambda \end{array}\right|=0 \ & \Rightarrow(1-\lambda)[-(1+\lambda)(1-\lambda)-8]-2[-8+3(\lambda+1)]=0 \ & \Rightarrow \quad(1-\lambda)\left[-\left(1-\lambda^2\right)-8\right]-2(-8+3 \lambda+3)=0 \ & \Rightarrow \quad(1-\lambda)\left(\lambda^2-9\right)-2(-5+3 \lambda)=0 \ & \Rightarrow \quad \lambda^2-9-\lambda^3+9 \lambda-6 \lambda+10=0 \ & \Rightarrow \quad-\lambda^3+\lambda^2+3 \lambda+1=0 \ & \Rightarrow \quad \lambda^3-\lambda^2-3 \lambda-1=0 \end{aligned}$ According to Caley Hamilton, every square matrix satisfies its characteristics equations. $\begin{aligned} & \therefore \quad A^3-A^2-3 A-I=0 \ & \Rightarrow \quad A^{-1}\left(A^3-A^2-3 A-I\right)=0 \ & \Rightarrow \quad A^2-A-3 I-A^{-1}=0 \ & \Rightarrow \quad A^{-1}=A^2-A-3 I \end{aligned}$

If \(A=\left[\begin{array}{ccc}1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1\end{array}\right]\), then \(A^{-1}=\)

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