Question

Let $A=[a_{ij}{]}_{4\times 4}$ be a matrix such that $a_{ij}=\{\begin{array}{cc}2,& \text{if}i=j\\ 0,& \text{if}i\ne j\end{array}.$
Then the value of $\left\{\frac{det(adj\left(adjA\right))}{7}\right\}$ is
$\left(\right\{.\}$ represents the fractional part function $)$

• A. $\frac{1}{7}$
• B. $\frac{2}{7}$
• C. $\frac{3}{7}$
• D. $\frac{6}{7}$

Answer 1

The correct option is A $\frac{1}{7}$

$A$ is a scalar matrix, therefore $|A|=2^4$
We know that for a $n\times n$ matrix $A,$
$|adjA|=|A{|}^{n-1}$
$\Rightarrow |adj\left(adjA\right)|=|adjA{|}^{n-1}=|A{|}^{(n-1{)}^2}$
$\Rightarrow |adj\left(adjA\right)|=|A{|}^{(4-1{)}^2}=(2^4{)}^9=2^{36}$
$\therefore \left\{\frac{det(adj\left(adjA\right))}{7}\right\}=\left\{\frac{2^{36}}{7}\right\}$
$=\left\{\frac{(7+1{)}^{12}}{7}\right\}$
$=\left\{\frac{{}^{12}{C}_0{7}^{12}+{}^{12}{C}_1{7}^{11}+{}^{12}{C}_2{7}^{10}+\cdots +{}^{12}{C}_{11}{7}^1+{}^{12}{C}_{12}{7}^0}{7}\right\}$
$=\left\{\frac{7k+1}{7}\right\};k\in N$
$=\frac{1}{7}$