Matrices and Determinants Questions and Answers

Answer: A

Distinct eigenvalues do guarantee independence and so make the matrix diagonalisable, but they are sufficient rather than necessary. A repeated eigenvalue may still supply enough independent eigenvectors, as the identity matrix shows.

Answer: B

For a two by two matrix the pair of relations determines the roots completely, and for larger matrices they still provide two independent arithmetic checks. Both hold whatever the symmetry or the multiplicity of the eigenvalues.

Answer: C

Similar matrices represent the same map in different bases, so the characteristic polynomial and with it the eigenvalues, trace and determinant all survive. The entries, the triangular form and the basis itself all change.

Answer: B

A homogeneous system has a non-trivial solution only when the determinant of its coefficient matrix is zero:

Expanding along the first row: \((k^2 + 4k - 9) - 2(3k - 3) - (12 - 2k - 14) = k^2 - 1\). So \(k^2 = 1\) and \(k = \pm 1\). The options -1 and 1 each give only one of the two roots, and k = 0 gives determinant -1, so the system then has only the trivial solution.

Answer: A

The rank of the matrix is 2. First find the determinant: \[ \begin{aligned} &1(3\times 2 - 4\times 2) \\ &- 5(2\times 2 - 4\times(-1)) \\ &+ 6(2\times 2 - 3\times(-1)) \end{aligned} \] \[ = 1(-2) - 5(8) + 6(7) = -2 - 40 + 42 = 0 \] So the rank is less than 3. A 2 by 2 minor is non-zero: \[ \begin{vmatrix} 1 & 5 \\ 2 & 3 \end{vmatrix} = 3 - 10 = -7 \neq 0 \] The rank is therefore 2. The values 4, 6 and 9 are larger than the number of rows and cannot be the rank of a 3 by 3 matrix.

Answer: C

The constraint holds for both the identity matrix and the null matrix.

A matrix with \(A^{2} = A\) is called idempotent. For the identity, \(I^{2} = I \cdot I = I\); for the null matrix, \(O^{2} = O\). So both satisfy the constraint, and the options saying 'the identity matrix only' or 'the null matrix only' each leave out a matrix that works.

'No square matrix' is plainly false. Other idempotent matrices also exist (any projection matrix, such as \(\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\)), but among the four options the only one that is true is the one naming both the identity and the null matrix.

Answer: C

The system has no solution when k = 4.

Two equations in three unknowns either have infinitely many solutions or none. They have none only when the left-hand sides are proportional while the right-hand sides are not. Proportional coefficients need

\[ \frac{2}{1} = \frac{k}{2} = \frac{8}{k} \]

which gives \(k = 4\). Then the first equation doubled reads \(2x + 4y + 8z = 2\), while the second says \(2x + 4y + 8z = 3\), a contradiction, so there is no solution.

For k = 0, 2 or 8 the two coefficient rows are not proportional, the rank is 2, and the system has infinitely many solutions.

Answer: A

Another eigenvalue is 18.33.

The determinant of A is zero:

\[ \begin{aligned} |A| &= 2(10 - 150) - 7(25 - 25) + 10(30 - 2) \\ &= -280 - 0 + 280 = 0 \end{aligned} \]

so one eigenvalue is 0. The eigenvalues add up to the trace, \(2 + 2 + 5 = 9\). With 0 and -9.33, the third is \(9 - 0 + 9.33 = 18.33\). As a check, the sum of the principal 2 x 2 minors, \(-31 + 0 - 140 = -171\), equals \(18.33 \times (-9.33) \approx -171\).

A real matrix has complex eigenvalues only in conjugate pairs, so a lone \(18.33 \pm 9.33i\) cannot sit beside the real eigenvalues; and -18.33 would make the trace wrong.

Answer: B

The eigenvalues of \(A^{2}\) are \(\lambda_{i}^{2}\).

If \(Ax = \lambda_{i}x\), apply \(A\) again:

\[ \begin{aligned} A^{2}x &= A(\lambda_{i}x) \\ &= \lambda_{i}(Ax) = \lambda_{i}^{2}x \end{aligned} \]

So each eigenvector of \(A\) is an eigenvector of \(A^{2}\), with the eigenvalue squared. In general the eigenvalues of \(A^{k}\) are \(\lambda_{i}^{k}\). \(\lambda_{i}^{4}\) would belong to \(A^{4}\), and \(\lambda_{i}^{1/2}\) and \(\lambda_{i}^{1/4}\) to square and fourth roots of \(A\).

Answer: B

The statement that is NOT TRUE is det(A+B) = det(A) + det(B).

The determinant is multiplicative, not additive. A quick counterexample: take A = B = I, the 2 × 2 identity. Then det(A + B) = det(2I) = 4, but det(A) + det(B) = 1 + 1 = 2.

The other three are standard properties. det(AB) = det(A)det(B) for square matrices. Since AA^-1 = I, its determinant is 1. And a matrix and its transpose have the same determinant.