Question

If \(A=(a_{ij})_{4\times 4}\) such that \[ a_{ij}= \begin{cases} 2 & \text{if } i=j\\ 0 & \text{if } i\neq j \end{cases} \] then \[ \left\{\dfrac{\det(\operatorname{adj}(\operatorname{adj}A))}{7}\right\} \] is (where \(\{\ \}\) represents fractional part function)

• A. \(\dfrac{1}{7}\)
• B. \(\dfrac{2}{7}\)
• C. \(\dfrac{3}{7}\)
• D. \(\dfrac{4}{7}\)

Hint:

Important determinant results:
\(\det(\operatorname{adj}A)=(\det A)^{n-1}\)
For identity matrix: \(\det(kI_n)=k^n\)
Fractional part depends only on remainder modulo the denominator

Answer 1

The Correct Option is A

Step 1: The given matrix is \[ A=2I_4 \] Hence, \[ \det(A)=2^4=16 \]
Step 2: For an \(n\times n\) matrix, \[ \det(\operatorname{adj}A)=(\det A)^{\,n-1} \] Here \(n=4\), so \[ \det(\operatorname{adj}A)=16^3=4096 \]
Step 3: Again applying the same property: \[ \det(\operatorname{adj}(\operatorname{adj}A)) =\left(\det(\operatorname{adj}A)\right)^{3} =4096^3 \] Since \(4096=2^{12}\), \[ \det(\operatorname{adj}(\operatorname{adj}A))=2^{36} \]
Step 4: Evaluate the fractional part: \[ \left\{\frac{2^{36}}{7}\right\} \] Using modulo arithmetic: \[ 2^3=8\equiv 1 \pmod{7} \] \[ 2^{36}=(2^3)^{12}\equiv 1^{12}\equiv 1 \pmod{7} \] So, \[ \frac{2^{36}}{7}=\text{integer}+\frac{1}{7} \]
Step 5: Hence, the fractional part is: \[ \boxed{\frac{1}{7}} \]