Matrices and Determinants Questions and Answers

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.

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.

Answer: D

A matrix has rank zero if and only if all its elements are zero.

The rank is the largest number of linearly independent rows (or columns), equivalently the order of the largest non-vanishing minor. Any single non-zero element is itself a non-vanishing 1 by 1 minor, giving rank at least 1. So rank 0 means every element is zero: the null matrix.

A zero determinant with non-zero elements only says the rank is less than full. A matrix equal to its own inverse is invertible and has full rank, and a matrix with zero diagonal can still have full rank.