Matrices and Determinants Questions and Answers

Answer: A

The value is \(k = 8\).

A matrix is symmetric when it equals its transpose, so the off-diagonal entries must be equal:

\[ 2k + 5 = 3k - 3 \Rightarrow k = 8 \]

Then both off-diagonal entries are 21, and the diagonal entries (10 and 13) are unrestricted.

\(k = 5\) gives 15 and 12, \(k = -0.4\) gives 4.2 and -4.2 (equal and opposite, a skew-type condition on those entries), and the surd value does not make the two entries equal either.

Answer: B

A normalized eigenvector is \(\left(\frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}}\right)^{T}\).

The characteristic equation gives the eigenvalues:

\[ \begin{aligned} (5-\lambda)(3-\lambda)-3 &= 0 \\ \lambda^{2}-8\lambda+12 &= 0 \\ \lambda &= 2,\ 6 \end{aligned} \]

For \(\lambda = 2\): \(3x+3y=0\), so \(y=-x\) and the eigenvector is (1, -1), normalized to \((1/\sqrt{2}, -1/\sqrt{2})\). For \(\lambda = 6\): \(-x+3y=0\), giving (3, 1), normalized to \((3/\sqrt{10}, 1/\sqrt{10})\).

Option C has the right magnitudes for the second eigenvector but the wrong sign on its second component; check: \(A(3,-1)^{T} = (12, 0)^{T}\), not a multiple of (3, -1). Options A and D are unit vectors but not eigenvectors of A.

Answer: C

The system has an infinite number of solutions.

Subtract the third equation from the first:

\[ (x+2y+z)-(x-y+z) = 4-1 \Rightarrow 3y = 3,\ y = 1 \]

With y = 1, the first equation gives x + z = 2 and the second gives 2x + 2z = 4, which is the same condition. So y = 1 and x + z = 2, and every pair x, z summing to 2 works: the coefficient matrix has rank 2, equal to the rank of the augmented matrix, with three unknowns.

(1, 1, 1) and (2, 1, 0) are both solutions, but they are only two of infinitely many, so 'unique' and 'only the two solutions' are wrong. A linear system cannot have exactly two solutions, and the system is consistent, so 'no feasible solution' is wrong too.

Answer: D

The values are \(a = 4, b = 14\).

A non-trivial null space needs the coefficient matrix to be singular:

\[ \begin{aligned} \det &= 1(6a - 16) - 2(12 - 8) + 1(8 - 2a) \\ &= 4a - 16 = 0 \Rightarrow a = 4 \end{aligned} \]

With a = 4, subtract twice the first equation from the others: the second becomes \(2z = 2\), so z = 1, and the third becomes \(4z = b - 10\). For the system to be consistent, \(4 = b - 10\), so b = 14. Then x + 2y = 4 has infinitely many solutions.

With a = 8 the matrix is non-singular and the solution is unique. With a = 4 and b = 12 the two conditions on z clash (z = 1 and z = 1/2), so there is no solution at all.

Answer: A

The matrix has a negative eigenvalue when a is greater than \(\dfrac{3}{8}\).

For a 2 x 2 matrix, the eigenvalues multiply to the determinant and add to the trace:

\[ \begin{aligned} \lambda_{1} + \lambda_{2} &= 1 + 3 = 4 \\ \lambda_{1}\lambda_{2} &= 3 - 8a \end{aligned} \]

The eigenvalues are real here, since \((1 - 3)^{2} + 4 \times 8a = 4 + 32a \gt 0\). With a positive sum, one eigenvalue is negative exactly when the product is negative: \(3 - 8a \lt 0\), that is \(a \gt 3/8\).

For a = 1/4 or 1/5 the determinant is 1 or 1.4, still positive, so both eigenvalues are positive. 1/8 comes from setting the determinant to \(1 - 8a\), forgetting the factor 3.

Answer: C

The correct statement is \((AB)^{T} = B^{T}A^{T}\).

The transpose of a product reverses the order of the factors; this holds for any matrices whose product is defined, symmetric or not. For symmetric \(A\) and \(B\) it also equals \(BA\).

The others need extra conditions. \(A^{T} = A^{-1}\) defines an orthogonal matrix, \(A = A^{-1}\) an involution, and \(AB = BA\) holds only for commuting matrices; for example \(A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}\) and \(B = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\) are symmetric but \(AB \ne BA\).

Answer: A

P and Q are true and R is not, so the answer is P and Q, but not R.

M is upper triangular, so its eigenvalues are its diagonal entries 1, 2 and 3; P is true. Test the vectors:

\[ \begin{aligned} M\begin{bmatrix} 4 \\ 1 \\ 0 \end{bmatrix} &= \begin{bmatrix} 4 + 4 \\ 2 \\ 0 \end{bmatrix} = 2\begin{bmatrix} 4 \\ 1 \\ 0 \end{bmatrix} \\ M\begin{bmatrix} 4 \\ 2 \\ 0 \end{bmatrix} &= \begin{bmatrix} 12 \\ 4 \\ 0 \end{bmatrix} \end{aligned} \]

The first is twice the vector, so it is an eigenvector for eigenvalue 2 (Q true). (12, 4, 0) is not a multiple of (4, 2, 0), since 12/4 = 3 but 4/2 = 2, so R is false. Every option that includes R or excludes P is therefore wrong.

Answer: B

The trace of \(A^{1000}\) is \(2^{1000} + 1\).

The eigenvalues of A satisfy \(\lambda^{2} - (\text{tr}A)\lambda + \det A = 0\), with \(\text{tr}A = 5 - 4 = 1\) and \(\det A = -20 + 18 = -2\):

\[ \lambda^{2} - \lambda - 2 = 0 \Rightarrow \lambda = 2,\ -1 \]

The eigenvalues of \(A^{1000}\) are \(2^{1000}\) and \((-1)^{1000} = 1\), and the trace is their sum, \(2^{1000} + 1\).

\(2^{1000} - 1\) takes the second eigenvalue's power as -1 (an odd exponent would), and \(2^{1000}\) leaves it out; 1 counts only the second eigenvalue.

Answer: A

Both P and Q are TRUE.

P: a real symmetric matrix is orthogonally diagonalizable, \(A = QDQ^{T}\), so its rank equals the rank of the diagonal matrix D of eigenvalues, which is the number of nonzero eigenvalues counted with multiplicity. (For a non-symmetric matrix this can fail: a nonzero nilpotent matrix has rank at least 1 and only zero eigenvalues.)

Q: if \(AB = \mathbf{0}\), every column of B lies in the null space of A, so \(rank(B) \le n - rank(A)\) by the rank-nullity theorem, which gives \(rank(A) + rank(B) \le n\). This is also the case of Sylvester's inequality \(rank(AB) \ge rank(A) + rank(B) - n\).

Since both statements hold, the options calling either one false are wrong.

Answer: A

The eigenvalues of \(B\) are 4, 9, 16.

If \(Av = \lambda v\), then \(A^{2}v = \lambda^{2}v\), so any polynomial in \(A\) has the same eigenvector with the polynomial applied to the eigenvalue:

\[ Bv = (\lambda^{2} + 2\lambda + 1)v = (\lambda + 1)^{2} v \]

For \(\lambda = 1, 2, 3\) this gives \(2^{2}, 3^{2}, 4^{2} = 4, 9, 16\).

1, 2, 3 are the eigenvalues of \(A\) itself, 1, 4, 9 are those of \(A^{2}\) alone, and 4, 16, 25 squares \(\lambda + 1\) wrongly shifted for the last two.