Question

Let $A=[a_{ij}{]}_{3\times 3}$ be such that $a_{ij}=\{\begin{array}{c}3,wheni=j\\ 0,otherwise\end{array}$

Then $\left\{\frac{det\left(adj\right(adjA\left)\right)}{5}\right\}$ equals
[Note:$\left\{k\right\}$ denotes fractional part of k]

• A. $\frac{2}{3}$
• B. $\frac{1}{5}$
• C. $\frac{2}{5}$
• D. $\frac{1}{3}$

Answer 1

The correct option is B $\frac{1}{5}$

$a_{ij}=\{\begin{array}{cc}3,& i=j\\ 0,& otherwise\end{array}$
$A=\left[\begin{array}{ccc}3& 0& 0\\ 0& 3& 0\\ 0& 0& 3\end{array}\right]$
$\left|A\right|=3(9-0)+0+0$
$\left|A\right|=27$
$\left|adj\right(adjA\left)\right|={\left|A\right|}^{{(n-1)}^2}$
$={\left(27\right)}^{2^2}$
$={\left(27\right)}^4$
$\frac{\left|adj\right(adjA\left)\right|}{5}=\frac{{\left(27\right)}^4}{5}$
$\left\{\frac{\left|adj\right(adjA\left)\right|}{5}\right\}=\frac{\left|adj\right(adjA\left)\right|}{5}-\left[\frac{\left|adj\right(adjA\left)\right|}{5}\right]$
$=\frac{{\left(27\right)}^4}{5}-\left[\frac{{\left(27\right)}^4}{5}\right]$
$=\frac{531441}{5}-106288$
$=\frac{531441-531440}{5}$
$=\frac{1}{5}$