Let A = begin{bmatrix} 2k - 1 & 1 & 1 0 & 2k | vedclass

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(A) Step $1$: Calculate $\det(A)$. Since $A$ is an upper triangular matrix, $\det(A) = (2k - 1)^3$.Step $2$: Calculate $\det(	ext{adj } A)$. We know

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$1$

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(A) Step $1$: Calculate $\det(A)$. Since $A$ is an upper triangular matrix, $\det(A) = (2k - 1)^3$.Step $2$: Calculate $\det(	ext{adj } A)$. We know $\det(	ext{adj } A) = (\det A)^{n-1}$. Here $n=3$, so $\det(	ext{adj } A) = ((2k - 1)^3)^2 = (2k - 1)^6$.Step $3$: Calculate $\det(B)$. $B$ is a skew-symmetric matrix of order $3 	imes 3$. The determinant of a skew-symmetric matrix of odd order is $0$. Thus, $\det(B) = 0$.Step $4$: Calculate $\det(	ext{adj } B)$. Since $\det(B) = 0$, $\det(	ext{adj } B) = (\det B)^{3-1} = 0^2 = 0$.Step $5$: Solve the equation $(2k - 1)^6 + 0 = 11^6$. This implies $(2k - 1)^6 = 11^6$, so $2k - 1 = 11$ or $2k - 1 = -11$.Step $6$: If $2k - 1 = 11$, then $2k = 12$, so $k = 6$. Then $k - 5 = 6 - 5 = 1$.Step $7$: If $2k - 1 = -11$, then $2k = -10$, so $k = -5$. Then $k - 5 = -5 - 5 = -10$ (not in options).Therefore, the value is $1$.

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