Question

Let \(A = \left[ \begin{array}{ccc}2k-1 & 1 & 1 \\ 0 & 2k-1 & 1 \\ 0 & 0 & 2k-1\end{array} \right]\) and \(B = \left[ \begin{array}{ccc}0 & 2k-1 & 1 \\ 1-2k & 0 & k \\ -1 & -k & 0\end{array} \right]\) where k is a real number. If \(det(\text{adj}A)+det(\text{adj}B) = 11^6\), then the value of \(k-5\) is equal to...

• A. \(1\)
• B. \(2\)
• C. \(4\)
• D. \(6\)

Hint:

Use det(adj M) = (det M)^2 for a 3 by 3 matrix; B is skew-symmetric of odd order so its determinant is 0.

Answer 1

The Correct Option is A

Step 1: Determinant of adj:
Use $|\operatorname{adj}M| = |M|^2$ for a 3 by 3 matrix.

Step 2: A:
The diagonal is $2k-1$ three times and entries below it are 0, so $|A| = (2k-1)^3$ and $|\operatorname{adj}A| = (2k-1)^6$.

Step 3: B:
Every entry satisfies $b_{ij} = -b_{ji}$, so B is skew-symmetric. For odd order, $|B| = |B^T| = |-B| = (-1)^3|B| = -|B|$, which forces $|B| = 0$.

Step 4: Finish:
From $(2k-1)^6 = 11^6$ we get $2k-1 = \pm11$, so $k = 6$ or $-5$. The only matching option is $k - 5 = 1$ (with $k = 6$).

Final Answer:
The answer is 1. \[ \boxed{1} \]