TS EAMCET · Maths · Matrices
\(A=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]\) then \(A^3-4 A^2-6 A\) is equal to :
- A 0
- B \(A\)
- C \(-A\)
- D I
Answer & Solution
Correct Answer
(C) \(-A\)
Step-by-step Solution
Detailed explanation
\(\because \quad A=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]\) \(\therefore \quad A^2=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]\)…
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