Matrix Operations and Inverses Questions and Answers

Answer: C

\(A^{8} = \begin{pmatrix} 625 & 0 \\ 0 & 625 \end{pmatrix}\).

Square A first:

\[ \begin{aligned} A^{2} &= \begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix}\begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix} \\ &= \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} = 5I \end{aligned} \]

So \(A^{8} = (A^{2})^{4} = 5^{4}I = 625I\).

25I is A⁴, 125I is A⁶ and 3125I is A¹⁰: each comes from taking the wrong number of factors of 5I.

Answer: C

Both statements are correct.

For \(A\) of size \(m \times n\) and \(B\) of size \(n \times m\), \(AB\) is \(m \times m\) and \(BA\) is \(n \times n\), so both traces exist. Each is the same double sum:

\[ \begin{aligned} tr(AB) &= \sum_{i=1}^{m} \sum_{j=1}^{n} a_{ij} b_{ji} \\ tr(BA) &= \sum_{j=1}^{n} \sum_{i=1}^{m} b_{ji} a_{ij} \end{aligned} \]

The terms are identical, only summed in a different order, so \(tr(AB) = tr(BA)\) even though the two products have different sizes. Statement 2 is the special case \(m = n\), so it is correct too. The options calling either statement wrong confuse \(tr(AB) = tr(BA)\) with \(AB = BA\), which does fail in general.

Answer: C

\(A^{50}\) is the matrix with first column \((1, 25, 25)\) and the identity elsewhere.

Square \(A\) first:

\[ A^{2} = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} = I + N, \]

where \(N\) has 1 in positions (2,1) and (3,1) and zeros elsewhere. Then \(N^{2} = 0\), so by the binomial theorem

\[ A^{50} = (I + N)^{25} = I + 25N. \]

That gives entries 25 below the diagonal in the first column. The option with 50 is \(I + 50N\), which forgets that only \(A^{2}\), not \(A\), has the form \(I + N\); 48 and 24 come from miscounting the power in the same way.

Answer: B

The correct statements are Q and R only.

\(M\) equals its conjugate transpose: the diagonal entries 2, 5, 3 are real, and the off-diagonal pairs \(3 + 2i\) and \(3 - 2i\), \(-4\) and \(-4\), \(6i\) and \(-6i\) are conjugates. So \(M\) is Hermitian, and then \((iM)^{*} = -iM^{*} = -iM\), so \(iM\) is skew-Hermitian (Q).

A Hermitian matrix has real eigenvalues (R). The eigenvalues of \(iM\) are \(i\) times those of \(M\), so they are purely imaginary, and real only if they are 0. So S could hold only if every eigenvalue of \(M\) were 0, which for a Hermitian matrix means \(M = 0\); \(M\) is not zero, so S fails. P is false because \(M\) is not skew-Hermitian: its diagonal is not purely imaginary.

Answer: C

The correct option is \(Z = Z^{-1}\) for some \(Z\).

Property (iii) says \(Z^{2} = Z\): the matrix is idempotent, and (ii) says it is not the zero matrix. The identity matrix \(I\) satisfies all three properties, and \(I^{-1} = I\), so there is a \(Z\) equal to its own inverse.

That one example disposes of the first two options. \(I\) is invertible, so it is false that the inverse exists for no \(Z\). And \(I^{transpose} = I = I^{-1}\), so the transpose is not different from the inverse for every \(Z\).

The last option fails for \(Z = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\), which is non-zero and idempotent: with \(h = (0, 1)^{transpose}\), \(h^{transpose}Zh = 0\), which is not greater than 0.

Answer: B

\(B^{-1}\) is obtained by interchanging the second and third rows of \(A^{-1}\), and then adding \(-3\) times the fifth row to the third row.

Column operations are right multiplication by elementary matrices. Let \(P\) swap columns 2 and 3, and \(E = I + 3e_3e_5^{T}\) add 3 times column 3 to column 5. Then

\[ \begin{aligned} B &= APE \\ B^{-1} &= E^{-1}P^{-1}A^{-1} \\ &= (I - 3e_3e_5^{T})\,P\,A^{-1} \end{aligned} \]

Left multiplication is a row operation, applied from the right-hand factor outwards: first \(P\) swaps rows 2 and 3 of \(A^{-1}\); then \(I - 3e_3e_5^{T}\) replaces row 3 by row 3 minus 3 times row 5.

The first option gets the swap right but carries the column operation over unchanged. Inverting it flips the sign to \(-3\), and moving from columns to rows exchanges the roles of the third and fifth lines. The two options that add before swapping have the order reversed: inverting a product reverses the order of its factors.

Answer: C

Both P and Q are correct.

P: \(X^{T}AX\) is a \(1 \times 1\) matrix, so it equals its own transpose. Using \(A^{T} = -A\):

\[ X^{T}AX = (X^{T}AX)^{T} = X^{T}A^{T}X = -X^{T}AX \]

so \(X^{T}AX = 0\) for every real \(X\).

Q: take determinants of \(A^{T} = -A\): \(\det A = \det(-A) = (-1)^{n}\det A\). An orthogonal matrix has \(\det A = \pm 1 \neq 0\), so \((-1)^{n} = 1\) and \(n\) is even. The matrix \(\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}\) shows the case \(n = 2\) does occur. Every odd-order skew-symmetric matrix is singular, so it cannot be orthogonal.

Answer: C

\(A^{2}\) is the permutation matrix \(\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}\).

Index rows and columns by \(j, k = 0, 1, 2\). Then \(A_{jk} = \omega^{jk}/\sqrt{3}\), and since \(\omega^{3} = 1\):

\[ \begin{aligned} (A^{2})_{jk} &= \frac{1}{3}\sum_{m=0}^{2} \omega^{jm}\omega^{mk} \\ &= \frac{1}{3}\sum_{m=0}^{2} \omega^{m(j + k)} \end{aligned} \]

The sum \(1 + \omega^{r} + \omega^{2r}\) is 3 when \(r\) is a multiple of 3 and 0 otherwise (because \(1 + \omega + \omega^{2} = 0\)). So \((A^{2})_{jk} = 1\) exactly when \(j + k \equiv 0 \pmod 3\): the pairs (0, 0), (1, 2) and (2, 1). That gives a 1 in the top-left corner and the swap of the last two coordinates.

A is the discrete Fourier matrix, which is unitary but not an involution, so \(A^{2} \ne I\); \(A^{2}\) has real 0/1 entries, so it is not A; and the matrix swapping the first two coordinates puts its 1s in the wrong places.

Answer: D

The false statement is if both A and B have the same rank, then they are congruent.

For real symmetric matrices, Sylvester's law of inertia says congruence is determined by the rank and the signature (the numbers of positive and negative eigenvalues). Equal rank is not enough: \(I_{2}\) and \(-I_{2}\) both have rank 2, but \(P^{t}I_{2}P\) is positive semi-definite for every P, so it can never equal \(-I_{2}\).

The others are true. If \(B = P^{t}AP\), then transposing gives \(B^{t} = P^{t}A^{t}P\), and inverting gives \(B^{-1} = Q^{t}A^{-1}Q\) with \(Q = (P^{t})^{-1}\); and congruence is an equivalence relation, so two matrices congruent to I are congruent to each other.

Answer: B

\(A^{35} = \begin{pmatrix} 1 & 35 \\ 0 & 1 \end{pmatrix}\).

Write \(A = I + N\) with \(N = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\). Since \(N^{2} = 0\) and \(I\) commutes with \(N\), the binomial theorem gives

\[ A^{n} = I + nN = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix} \]

For \(n = 35\), the top-right entry is 35. The 70 in option A is the entry of \(A^{70}\), and options C and D change the diagonal, which stays 1 because \(A\) is upper triangular with 1s on the diagonal.