Discussion
Consider the matrix
\[ A = \begin{pmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}. \]
The value of \(\det(A^{-1})\) is:
Answer: B
\(\det(A^{-1})\) is 1.
For any invertible matrix \(\det(A^{-1}) = 1/\det(A)\), so it is enough to find \(\det A\). Expanding along the first row:
\[ \begin{aligned} \det A &= 0(0 - 0) - 1(0 - 1) + 1(0 - 0) \\ &= 1 \end{aligned} \]
So \(\det(A^{-1}) = 1/1 = 1\). This fits the structure of \(A\): it is the cyclic permutation matrix, whose determinant is \(+1\), with its second row added to its first, an operation that leaves the determinant unchanged.
0 would mean \(A\) had no inverse, which contradicts the question, and 4 and 1/4 do not arise from this matrix.
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