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: Use the adjoint property:
For a square matrix of order n, \(\det(\operatorname{adj}M) = (\det M)^{n-1}\). Both matrices are 3 by 3, so \(\det(\operatorname{adj}M) = (\det M)^2\).

Step 2: Matrix A:
A is upper triangular, so its determinant is the product of the diagonal entries: \(\det A = (2k-1)^3\). So \(\det(\operatorname{adj}A) = (2k-1)^6\).

Step 3: Matrix B:
Check \(B^T\): the (1,2) entry is \(2k-1\) and the (2,1) entry is \(1-2k\); the (1,3) entry is 1 and the (3,1) entry is \(-1\); the (2,3) entry is \(k\) and the (3,2) entry is \(-k\); the diagonal is 0. So \(B^T = -B\), a skew-symmetric matrix of odd order, and its determinant is 0. So \(\det(\operatorname{adj}B) = 0\).

Step 4: Solve:
\[ (2k-1)^6 + 0 = 11^6 \Rightarrow 2k-1 = \pm 11 \Rightarrow k = 6 \text{ or } k = -5 \]
Then \(k - 5 = 1\) or \(-10\). Only 1 is in the options.

Step 5: Check the options:
The values 2, 4 and 6 for \(k-5\) would need \(k = 7, 9, 11\), which give \(2k-1 = 13, 17, 21\), none equal to \(\pm11\).

Final Answer:
\(k - 5 = 1\), option (A). \[ \boxed{1} \]