Determinants Questions and Answers

Answer: A

The determinant of (M - N) is 0.

From \(L^{2} = L^{-1}\), multiply by \(L\): \(L^{3} = I\). Then

\[ M = L^{8} = L^{6}L^{2} = (L^{3})^{2}L^{2} = L^{2} = N \]

So \(M - N\) is the zero matrix, and its determinant is 0.

The values 1, 2 and 3 would need M and N to differ. They cannot, because the cube of L is the identity, so every power of L repeats with period 3, and 8 and 2 leave the same remainder modulo 3.

Answer: B

The true statement is \(\det(B) = -\det(A)\).

Compare the rows: B's first row (3 4 1 2) is A's third row, B's third row (1 2 3 4) is A's first row, and the second and fourth rows are the same. So B is A with rows 1 and 3 interchanged, and a single row interchange changes the sign of a determinant.

The other options fail because \(\det(A) \ne 0\): expanding gives \(\det(A) = -160\) and so \(\det(B) = 160\). A non-zero number cannot equal its own negative, so \(\det(A) = \det(B)\) is false, and \(\det(A) = 0\) is false. The determinant is multiplicative, \(\det(AB) = \det(A)\det(B) = -25600\), not the sum of the two, which here would be 0.

Answer: D

The trace is 5.

\(M\) is upper triangular, so its diagonal entries \(1, \alpha, \alpha^{2}, \alpha^{3}, \alpha^{4}\) are its eigenvalues, and \(M^{2}\) is upper triangular with diagonal \(1, \alpha^{2}, \alpha^{4}, \alpha^{6}, \alpha^{8}\).

\[ \begin{aligned} \operatorname{tr} I &= 5 \\ \operatorname{tr} M &= 1 + \alpha + \alpha^{2} + \alpha^{3} + \alpha^{4} = 0 \\ \operatorname{tr} M^{2} &= 1 + \alpha^{2} + \alpha^{4} + \alpha^{6} + \alpha^{8} = 0 \end{aligned} \]

Both sums vanish because \(\alpha\) is a primitive fifth root of unity: the five fifth roots of unity sum to zero, and since 2 is prime to 5, the powers \(\alpha^{2k}\) run through the same five roots. So \(\operatorname{tr}(I + M + M^{2}) = 5 + 0 + 0 = 5\). The values 0 and 3 forget the identity or count only part of it, and -5 has the wrong sign.

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.