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Question: Answered & Verified by Expert
$A=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$ then $A^3-4 A^2-6 A$ is equal to :
MathematicsMatricesJEE Main
Options:
  • A 0
  • B $A$
  • C $-A$
  • D I
Solution:
1369 Upvotes Verified Answer
The correct answer is: $-A$
$\because \quad A=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$
$\therefore \quad A^2=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$
$=\left[\begin{array}{lll}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{array}\right]$
$A^3=A^2 \cdot A=\left[\begin{array}{lll}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{array}\right]\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$
$=\left[\begin{array}{lll}41 & 42 & 42 \\ 42 & 41 & 42 \\ 42 & 42 & 41\end{array}\right]$
$\therefore A^3-4 A^2-6 A$
$=\left[\begin{array}{lll}41 & 42 & 42 \\ 42 & 41 & 42 \\ 42 & 42 & 41\end{array}\right]-4\left[\begin{array}{lll}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{array}\right]-6\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$
$=\left[\begin{array}{ccc}41-36-6 & 42-32-12 & 42-32-12 \\ 42-32-12 & 41-36-6 & 42-32-12 \\ 42-32-12 & 42-32-12 & 41-36-6\end{array}\right]$
$=\left[\begin{array}{lll}-1 & -2 & -2 \\ -2 & -1 & -2 \\ -2 & -2 & -1\end{array}\right]$
$=-\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]=-A$

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