If \(A=\left[\begin{array}{lll}1&am

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Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Determinants

Question:

If $A=\left[\begin{array}{lll}1&-1&1\\ 2&-1&0\\ 1&0&0\end{array}\right]$ then $A^{-1}$ is

Options:

$A$

$A^2$

$A^3$

$\frac{1}{2}\left(A-2I\right)$

Correct Answer:

$A^2$

Explanation:

$A=\left[\begin{array}{lll}1&-1&1\\ 2&-1&0\\ 1&0&0\end{array}\right]$

so $|A-λI|=\begin{vmatrix}1-λ&-1&1\\2&-1-λ&0\\1&0&-λ\end{vmatrix}$

$=λ((-1+λ)(1+λ)+2)$

$=λ^3-1=0$

$⇒A^3=I$

so $A^{-1}=A^2$