If A is the 3 × 3 matrix of order 3 × 3, such that det(A) = 1 / 2, tr(A) = 10

Question

If A is the \(3 \times 3\) matrix of order \(3 \times 3\), such that \(\operatorname{det}(A)=\frac{1}{2}, \operatorname{tr}(A)=10\) and B be another matrix of order \(3 \times 3\) and defined as \(B=\operatorname{adj}(\operatorname{adj}(2 A))\), then \(\operatorname{det}(B)+\operatorname{tr}(B)\) is equal to (where tr(A) denotes trace of matrix A)

Answer 1

Answer is : 336

B = adj(adj2A)

\( B=|2 A|^{n-2}\ (2 A),\left[\text{Using} \ \operatorname{adj}(\operatorname{adj} P)=|P|^{n-2} \cdot P\right], \)

for n = 3

\( =|2 A|(2 A)\)

\( =2^{3}|A|(2 A)\)

\( =8 \times \frac{1}{2}(2 A)\) 

= 4(2A)

B = 8A

|B| = |8A|

\( =8^{3}|A|\) 

\( |B|=8^{3} \times \frac{1}{2}=256\) 

B = 8A

[each element is multiplied 8 times]

tr(B) = 8tr(A)

= 80

|B| + tr(B) = 256 + 80

= 336