Matrices and Determinants

The zero matrix and the identity both satisfy \(A^{3} = A\), and they rule out the first three options: \(A = 0\) is not invertible (A), \(A = I\) has determinant 1 (B), and \(A = 0\) has trace 0 (C).

For (D), multiplying by a matrix never increases rank, so \(\operatorname{rank}(A^{2}) \le \operatorname{rank}(A)\). Also \(A = A^{3} = A \cdot A^{2}\), so \(\operatorname{rank}(A) \le \operatorname{rank}(A^{2})\). The two ranks are therefore always equal.

Another view: \(A\) is diagonalisable with eigenvalues in \(\{-1, 0, 1\}\), and squaring keeps exactly the same eigenvalues non-zero.

Answer: D

Put the vectors in the columns of \(X = [x_1\ x_2\ \cdots\ x_n]\). Then \(A_{ij} = x_i^{\top} x_j\) says \(A = X^{\top} X\), a Gram matrix.

For any non-zero \(z\), \(z^{\top} A z = (Xz)^{\top}(Xz) = \|Xz\|^{2}\). The columns of \(X\) are linearly independent, so \(Xz \ne 0\) and \(z^{\top} A z \gt 0\). Hence \(A\) is positive definite, which rules out (D).

A positive definite matrix has only positive eigenvalues, so \(\det A = (\det X)^{2} \gt 0\) and \(A\) is invertible. Being symmetric, its singular values are these same positive eigenvalues, so 0 is not one of them. Options B and C fail.

Answer: A

Factor first: \(\boldsymbol{M}^{2} + 12\boldsymbol{M} = \boldsymbol{M}(\boldsymbol{M} + 12I)\), so its determinant is \(\det(\boldsymbol{M}) \cdot \det(\boldsymbol{M} + 12I)\).

Expanding along the first row,

\[\det(\boldsymbol{M}) = 1(6 - 9) - 2(18 - 12) + 3(9 - 4) = -3 - 12 + 15 = 0.\]

This is no accident: row 3 is the sum of rows 1 and 2, so \(\boldsymbol{M}\) is singular. A zero factor makes the whole product zero, so \(\det(\boldsymbol{M}^{2} + 12\boldsymbol{M}) = 0\) without computing \(\boldsymbol{M}^{2}\) at all.

Answer: 0