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Question: Answered & Verified by Expert
If $A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0\end{array}\right]$, then $A^{-1}=$
MathematicsMatricesAP EAMCETAP EAMCET 2023 (17 May Shift 2)
Options:
  • A $\mathrm{A}-2 \mathrm{~A}^2$
  • B $2 \mathrm{~A}-\mathrm{A}^2$
  • C $2 \mathrm{~A}^2+\mathrm{A}$
  • D $2 \mathrm{~A}+\mathrm{A}^2$
Solution:
1865 Upvotes Verified Answer
The correct answer is: $2 \mathrm{~A}-\mathrm{A}^2$
$A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0\end{array}\right]$
$|A-\lambda I|=0 \Rightarrow\left|\begin{array}{ccc}1-\lambda & 0 & 1 \\ 0 & 1-\lambda & 1 \\ 0 & 1 & -\lambda\end{array}\right|=0$
$\begin{aligned} & \Rightarrow(1-\lambda)[-(1-\lambda) \lambda-1]=0 \\ & \Rightarrow \lambda^3-2 \lambda^2+1=0\end{aligned}$
$\because$ Every matrix satispfies its characterstic equation.
$\because A^3-2 A^2+I=0$
Multiplying by $\mathrm{A}^{-1}$ in both sides:
$\begin{aligned} & \mathrm{A}^2-2 \mathrm{~A}+\mathrm{A}^{-1}=0 \\ & \Rightarrow \mathrm{A}^{-1}=2 \mathrm{~A}-\mathrm{A}^2\end{aligned}$

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