Question

If $A={\left[a_{ij}\right]}_{4\times 4}$; such that $a_{ij}=\{\begin{array}{c}2wherei=j\\ 0wherei\ne j\end{array}$, then $\left\{\frac{det\left(adj\right(adjA\left)\right)}{7}\right\}$ is (where $\{\cdot \}$ represents fractional part function)

• A. $\frac{1}{7}$
• B. $\frac{2}{7}$
• C. $\frac{3}{7}$
• D. none of these

Answer 1

Answer: (A)

$\frac{1}{7}$

Given $A={\left[a_{ij}\right]}_{4\times 4}$; where $a_{ij}=\{\begin{array}{c}2wherei=j\\ 0wherei\ne j\end{array}$

$\therefore A=\left[\begin{array}{cccc}2& 0& 0& 0\\ 0& 2& 0& 0\\ 0& 0& 2& 0\\ 0& 0& 0& 2\end{array}\right]$

$\left|A\right|=16$

$det\left(adj\right(adjA\left)\right)n=4$

$={\left|A\right|}^{{(n-1)}^2}$

$={\left(16\right)}^{{(4-1)}^2}={\left(16\right)}^6$

Now, $\left\{\frac{det\left(adj\right(adjA\left)\right)}{7}\right\}=\left\{\frac{{\left(16\right)}^6}{7}\right\}$

$=\{{\left(16\right)}^6\times \frac{1}{7}\}$

${\left(16\right)}^6\to$whole number

$\frac{1}{7}\to$Fraction

Fractional part$=\frac{1}{7}$