Matrices and Determinants Questions and Answers

Answer: C

Applying \( A^{2}=A \) to an eigenvector \( x \) with \( Ax=\lambda x \):

\[ A^{2}x = \lambda^{2}x \quad\text{and}\quad Ax = \lambda x \quad\Rightarrow\quad \lambda^{2} = \lambda \]

\[ \lambda(\lambda-1)=0 \quad\Rightarrow\quad \lambda = 0 \text{ or } 1 \]

An involutory matrix, satisfying \( A^{2}=I \), instead admits only \( \lambda = \pm 1 \).

Answer: D

Each relation constrains the eigenvalues in a characteristic way, real for symmetric, of unit modulus for orthogonal, and zero or one for idempotent. Recognising the form is usually quicker than computing the spectrum directly.

Answer: A

Correct answer: 1-Q, 2-R, 3-P

The three cases are exhaustive. Equal ranks mean the system is consistent, and then the common rank against the number of unknowns decides between one solution and infinitely many: equal gives uniqueness, less gives an \( (n-r) \)-parameter family. Unequal ranks - and the augmented rank can only be the larger - mean inconsistency and no solution at all.

Answer: B

Checking the determinant first avoids wasted work on a singular matrix, then the cofactors are found, transposed into the adjoint, and finally divided through. Omitting the transposition is the commonest error in the sequence.

Answer: C

Correct answer: 1 and 2 only are correct

Statements 1 and 2 are standard properties: a row interchange reverses the sign, and adding a multiple of one row to another leaves the value unchanged, which is what makes elimination safe. Statement 3 is wrong - scaling the whole matrix scales every row, so \( |kA| = k^{n}|A| \), not \( k|A| \).

Answer: D

Scaling one row scales the determinant by the same factor, whereas the other three leave the value untouched. It is the distinction that makes elimination safe for determinant evaluation while scaling requires a compensating factor.

Answer: A

Correct answer: its determinant is zero

\( A^{-1} = \mathrm{adj}(A)/|A| \), so a zero determinant makes the inverse undefined; equivalently, the rows are dependent and the map \( A \) collapses space onto a lower dimension, which cannot be undone. A zero trace, skew-symmetry and individual zero entries are all compatible with invertibility - the identity matrix itself is full of zeros.

Answer: B

Correct answer: sum to the trace

The trace equals the sum of the eigenvalues and the determinant equals their product, both following from the characteristic polynomial. Here \( 3+5 = 8 \neq 9 \), so at least one eigenvalue is wrong. It is the quickest check available on an eigenvalue calculation, and it costs nothing to make.

Answer: C

Omitting the transposition leaves every entry in the wrong position, so the result is the transpose of the true inverse. It coincides with the correct answer only when the matrix is symmetric, which is why the slip sometimes escapes notice.

Answer: D

Correct answer: both the assertion and the reason are true and the reason explains the assertion

Both are true and the reason gives the proof. Interchange the two identical rows: the matrix is unchanged, so its determinant is unchanged, but the interchange rule says the determinant has been negated. Hence \( D = -D \), which forces \( D = 0 \). The argument is a complete derivation, not merely a related fact.