Question

If element of matrix $A={\left[a_{ij}\right]}_{(3\times 3)}$ where

$A=\left\{\begin{array}{ll}{(-1)}^{j-i}& i<j\\ 2& i=j\\ {(-1)}^{i+j}& i>j\end{array}\right.$

, then the value of $|3adj(2A^{-1}\left)\right|$is:

• A. $72$
• B. $36$
• C. $108$
• D. $48$

Answer 1

The correct option is C

$108$

Finding the value of :

It is given that the elements of the matrix are:

$A=\left\{\begin{array}{ll}{(-1)}^{j-i}& i<j\\ 2& i=j\\ {(-1)}^{i+j}& i>j\end{array}\right.$

$A=\left|\begin{array}{ccc}2& -1& 1\\ -1& 2& -1\\ 1& -1& 2\end{array}\right|$

$\begin{array}{rcl}\left|A\right|& =& 2(4-1)+1(-2+1)+1(1-2)\\ & =& 2\left(3\right)+1(-1)+1(-1)\\ & =& 4\\ |3adj(2A^{-1}\left)\right|& =& 3^3\left|adj\right(2A^{-1}\left)\right|\\ & =& 3^3\times |2A^{-1}{|}^2\\ & =& 3^3\times 2^6\times |A^{-1}{|}^2\\ & =& 3^3\times 2^5\times 1/|A{|}^2\\ & =& 27\times 64\times \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$16$}\right.\\ & =& 27\times 4\\ & =& 108\end{array}$

Hence, option (C) is the answer.