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If $A = \begin{bmatrix} 2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2 \end{bmatrix}$ and $P = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & 2 \\ 0 & 1 & 5 \end{bmatrix}$, then $\left|P^{-1}AP - 2I\right| =$}

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For any scalar $k$ and similarity transformation, $\left|P^{-1}AP - kI\right| = \left|A - kI\right|$. You never need to calculate $P^{-1}$ or perform full matrix multiplication in such problems!

Updated On: Jul 9, 2026

$55$
$69$
$96$
$38$

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The Correct Option is B

Solution and Explanation

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If $A = \begin{bmatrix} 2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2 \end{bmatrix}$ and $P = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & 2 \\ 0 & 1 & 5 \end{bmatrix}$, then $\left|P^{-1}AP - 2I\right| =$}

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The Correct Option is B

Solution and Explanation

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