1. Tardigrade
  2. Question
  3. Mathematics
  4. If A =[ beginmatrix0&1&1 1&0&1 1&1&0 endmatrix], then (A2-3I/2)=

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. If $A =\left[\begin{matrix}0&1&1\\ 1&0&1\\ 1&1&0\end{matrix}\right]$, then $\frac{A^{2}-3I}{2}=$

Determinants

Solution:

We have given, $A = \left[\begin{matrix}0&1&1\\ 1&0&1\\ 1&1&0\end{matrix}\right]$
$\therefore \quad$ adj A $=\left[\begin{matrix}-1&1&1\\ 1&-1&1\\ 1&1&-1\end{matrix}\right]\quad$ and $\left|A\right|=-1\left(-1\right)+1\cdot1=2$

Hence, $A^{-1}=\frac{adj\,A}{\left|A\right|}=\frac{1}{2}\left[\begin{matrix}-1&1&1\\ 1&-1&1\\ 1&1&-1\end{matrix}\right] \quad\ldots\left(i\right)$

and $A^{2}=\left[\begin{matrix}0&1&1\\ 1&0&1\\ 1&1&0\end{matrix}\right]\cdot\left[\begin{matrix}0&1&1\\ 1&0&1\\ 1&1&0\end{matrix}\right]=\left[\begin{matrix}2&1&1\\ 1&2&1\\ 1&1&2\end{matrix}\right]\quad\ldots\left(ii\right)$

Now, $\frac{A^{2}-3I}{2}=\frac{1}{2} \left\{\left[\begin{matrix}2&1&1\\ 1&2&1\\ 1&1&2\end{matrix}\right]-\left[\begin{matrix}3&0&0\\ 0&3&0\\ 0&0&3\end{matrix}\right]\right\}$

$=\frac{1}{2}\left[\begin{matrix}-1&1&1\\ 1&-1&1\\ 1&1&-1\end{matrix}\right]=A^{-1}$ $\quad$ [using (i)]