1. Tardigrade
  2. Question
  3. Mathematics
  4. A=[1 2 2 2 1 2 2 2 1] then A3-4 A2-6 A is equal to :

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. $A=\begin{bmatrix}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{bmatrix}$ then $A^{3}-4 A^{2}-6 A$ is equal to :

EAMCETEAMCET 2006

Solution:

$\therefore A=\begin{bmatrix}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{bmatrix}$
$\therefore A^{2}=\begin{bmatrix}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{bmatrix}\begin{bmatrix}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{bmatrix}$
$=\begin{bmatrix}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{bmatrix}$
$A^{3}=A^{2} \cdot A=\begin{bmatrix}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{bmatrix}\begin{bmatrix}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{bmatrix}$
$=\begin{bmatrix}41 & 42 & 42 \\ 42 & 41 & 42 \\ 42 & 42 & 41\end{bmatrix}$
$\therefore A^{3}-4 A^{2}-6 A$
$=\begin{bmatrix}41 & 42 & 42 \\ 42 & 41 & 42 \\ 42 & 42 & 41\end{bmatrix}-4\begin{bmatrix}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{bmatrix}-6\begin{bmatrix}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{bmatrix}$
$=\begin{bmatrix}41-36-6 & 42-32-12 & 42-32-12 \\ 42-32-12 & 41-36-6 & 42-32-12 \\ 42-32-12 & 42-32-12 & 41-36-6\end{bmatrix}$
$=\begin{bmatrix}-1 & -2 & -2 \\ -2 & -1 & -2 \\ -2 & -2 & -1\end{bmatrix}$
$=-\begin{bmatrix}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{bmatrix}=-A$