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If \(A = \begin{bmatrix} 1 & 3 & 2 \\\ 2 & 0 & -1 \\\ 1 & 2 & 3 \end{bmatrix}\), then which one of the following is correct?

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  1. A3 - 3A2 - 4A + 11I = 0
  2. A3 - 4A2 - 3A + 11I = 0
  3. A3 + 4A2 - 3A + 11I = 0
  4. A3 - 3A2 + 4A + 11I = 0

Answer (Detailed Solution Below)

Option 2 : A3 - 4A2 - 3A + 11I = 0
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Detailed Solution

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∵ \(|A - λ I| = \begin{bmatrix} 1- λ & 3 & 2 \\\ 2 & -λ & -1 \\\ 1 & 2 & 3-λ \end{bmatrix}=0\)

(1 - λ) (-3λ + λ2 + 2) - 3(6 - 2λ + 1) + 2(4 + λ) = 0

(1 - λ) (λ2 - 3λ + 2) - 3(7 - 2λ) + 2(4 + λ) = 0

λ3 - 4λ2 - 3λ + 11 = 0

By C-H theorem replace λ by A

A3 - 4A2 - 3A + 11I = 0

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