If matrix $A=[a_{ij}]_{4 \times 4}$ such that...

Question

If matrix $A=[a_{ij}]_{4 \times 4}$ such that $a_{ij}= \begin{cases} 2&\text{if}\, i=j\\ 0&\text{if}\, i \not= j\\ \end{cases}$ , then $\{\frac{det(adj(adjA))}{7}\}$ is ($\{x\}$ represent fractional part function)

Options

1. $\frac{1}{7}$

2. $\frac{2}{7}$

3. $\frac{3}{7}$

4. None

Answer: $\frac{1}{7}$ (wolframalpha)

We know that determinant $ |adjB|=|B|^{n-1} $ where n is order of matrix. For this case replace $B$ by $adjA$ and n by 4, then the determinant

$ D=|adj(adjA)|=|adjA|^{3}=|A|^{9} $

Also

$A=diag(2,2,2,2)$ hence $|A|=2^4$

Therefore $ D=2^{36} $

Next we have to calculate fractional part of $ \frac{D}{7} $

that is $ \{\frac{2^{36}}{7}\} $

From looking onto the options

One way I can think of is subtracting a constant positive integer smaller than 7 from numerator and proving its divisibility by 7.

$ \frac{2^{36}-k+k}{7}=\frac{2^{36}-k}{7}+\frac{k}{7} $

Then if we prove divisibility of $2^{36}-k$ by 7 then $ \frac{k}{7} $ will be the answer.

But I cannot find a way to do that.

There might be alternative approach to get to final answer.

Answer 1

Hint: $$2^{36} = (2^3)^{12} = (1+7)^{12} = 1 + \sum_{i=1}^{12}\binom{12}{i}7^i = 1+ 7k,$$ where $k$ is an integer.