Q57 / 160 · AP EAMCET 2006 · Mathematics

$A=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$ then $A^3-4 A^2-6 A$ is equal to :

  • A
    0
  • B
    $A$
  • Correct answer
    C
    $-A$
  • D
    I

Solution

$\because \quad A=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$ $\therefore \quad A^2=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$ $=\left[\begin{array}{lll}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{array}\right]$ $A^3=A^2 \cdot A=\left[\begin{array}{lll}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{array}\right]\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$ $=\left[\begin{array}{lll}41 & 42 & 42 \\ 42 & 41 & 42 \\ 42 & 42 & 41\end{array}\right]$ $\therefore A^3-4 A^2-6 A$ $=\left[\begin{array}{lll}41 & 42 & 42 \\ 42 & 41 & 42 \\ 42 & 42 & 41\end{array}\right]-4\left[\begin{array}{lll}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{array}\right]-6\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$ $=\left[\begin{array}{ccc}41-36-6 & 42-32-12 & 42-32-12 \\ 42-32-12 & 41-36-6 & 42-32-12 \\ 42-32-12 & 42-32-12 & 41-36-6\end{array}\right]$ $=\left[\begin{array}{lll}-1 & -2 & -2 \\ -2 & -1 & -2 \\ -2 & -2 & -1\end{array}\right]$ $=-\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]=-A$

Hence, the correct answer is option: 'C'.

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