Matrices and Determinants Questions and Answers

Answer: B

The true statement is \(P\) is diagonalizable but \(Q\) is NOT diagonalizable.

A matrix is diagonalizable when each eigenvalue has as many independent eigenvectors as its multiplicity. The repeated eigenvalue is 1, so look at the rank of \(M - I\); the eigenspace dimension is \(3 - \text{rank}\).

\[ P - I = \begin{bmatrix} 3 & -2 & 2 \\ 6 & -4 & 4 \\ 3 & -2 & 2 \end{bmatrix} \]

Every row is a multiple of \((3, -2, 2)\), so the rank is 1 and the eigenspace for 1 has dimension 2. With one more eigenvector for 2, \(P\) has three independent eigenvectors and is diagonalizable.

\[ Q - I = \begin{bmatrix} 2 & -2 & 2 \\ 4 & -5 & 6 \\ 2 & -3 & 4 \end{bmatrix} \]

Subtracting twice row 1 from row 2 and row 1 from row 3 leaves \((0, -1, 2)\) twice, so the rank is 2 and the eigenspace for 1 has dimension only 1. \(Q\) is short of an eigenvector and is not diagonalizable, which rules out the other three combinations.

Answer: C

The eigenvalues are 5 and 4.

Solve the characteristic equation \(\det(A - \lambda I) = 0\):

\[ \begin{aligned} (6 - \lambda)(3 - \lambda) - (1)(-2) &= 0 \\ \lambda^{2} - 9\lambda + 20 &= 0 \\ (\lambda - 5)(\lambda - 4) &= 0 \end{aligned} \]

A quick check: the eigenvalues must add to the trace, 6 + 3 = 9, and multiply to the determinant, 18 + 2 = 20. The pair (3, 6) just reads off the diagonal, which works only for a triangular matrix; (1, −2) and (1, 6) fail both the sum and the product test.