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Linear Algebra and Matrices Previous Year Questions with Complete Solutions

Pravegaa PYQ Series · Volume 1: Mathematical Physics

MP02: Linear Algebra and Matrices

92 previous-year questions with complete solutions, from IIT JAM, GATE, CSIR NET-JRF, JEST and TIFR, arranged from IIT JAM and GATE through to TIFR and from the most recent paper backwards. Click Show solution under any question.

  • IIT JAM: 14
  • GATE: 14
  • CSIR NET: 25
  • JEST: 19
  • TIFR GS: 20
MP02.1IIT JAM 2026, Q10MCQ | 1 mark

Consider the matrix \[ A=\begin{pmatrix}0&1&1\\0&0&1\\1&0&0\end{pmatrix}. \] The value of \(\det(A^{-1})\) is:

  1. 4

  2. 1

  3. \(\dfrac14\)

  4. 0

Show solution

Expanding along the first column, only the \((3,1)\) entry contributes: \[ \det A=1\cdot(-1)^{3+1}\begin{vmatrix}1&1\\0&1\end{vmatrix}=1. \] Hence \[ \det(A^{-1})=\frac{1}{\det A}=1 \] \[ \boxed{\det(A^{-1})=1} \]

Answer: (B) \(1\)

MP02.2IIT JAM 2026, Q26MCQ | 2 marks

Consider a \(2\times2\) matrix \(A\). The determinant of \(A\) is \(-1\) and the \(\operatorname{trace}(A)=1\). Which of the following options gives the eigenvalues of \(A\)?

  1. \(1,0\)

  2. \(\dfrac{1+\sqrt5}{2},\dfrac{1-\sqrt5}{2}\)

  3. \(\dfrac{1+\sqrt5}{4},\dfrac{1-\sqrt5}{4}\)

  4. \(\dfrac{\sqrt5}{2},1-\dfrac{\sqrt5}{2}\)

Show solution

For a \(2\times2\) matrix the characteristic equation is \(\lambda^2-(\operatorname{tr}A)\lambda+\det A=0\): \[ \lambda^2-\lambda-1=0\quad\Longrightarrow\quad\lambda=\frac{1\pm\sqrt5}{2}. \] Check: sum \(=1\), product \(=(1-5)/4=-1\). \[ \boxed{\lambda=\frac{1\pm\sqrt5}{2}} \]

Answer: (B) \(\dfrac{1+\sqrt5}{2},\ \dfrac{1-\sqrt5}{2}\)

MP02.3IIT JAM 2025, Q13MCQ | 2 marks

If the system of linear equations \begin{align*} x+my+az&=0\\ 2x+ay+mz&=0\\ ax+2y-z&=0 \end{align*} with \(m\) and \(a\) as non-zero constants, admits a non-trivial solution, then which one of the following conditions is correct?

  1. \(m^2-a^2=3\)

  2. \(m^2-a^2=-3\)

  3. \(a^2-2m^2=-3\)

  4. \(m^2-2a^2=3\)

Show solution

A homogeneous system has a non-trivial solution only if the coefficient determinant vanishes: \[\begin{vmatrix}1&m&a\\2&a&m\\a&2&-1\end{vmatrix}=1(-a-2m)-m(-2-ma)+a(4-a^2)=0 .\] Expanding, \[-a-2m+2m+m^2a+4a-a^3=a\left(m^2+3-a^2\right)=0 .\] Since \(a\neq0\), \[\boxed{m^2-a^2=-3}\]

Answer: (B) \(m^2-a^2=-3\)

MP02.4IIT JAM 2024, Q3MCQ | 1 mark

Which of the following matrices is Hermitian as well as unitary?

  1. \(\begin{pmatrix}0&-i\\ i&0\end{pmatrix}\)

  2. \(\begin{pmatrix}0&i\\ i&0\end{pmatrix}\)

  3. \(\begin{pmatrix}1&-i\\ i&1\end{pmatrix}\)

  4. \(\begin{pmatrix}0&1+i\\ 1-i&0\end{pmatrix}\)

Show solution

Hermitian: \(A^\dagger=A\); unitary: \(A^\dagger A=I\).

For option (A), \(A=\begin{pmatrix}0&-i\\ i&0\end{pmatrix}\), \(A^\dagger=\begin{pmatrix}0&-i\\ i&0\end{pmatrix}=A\) (Hermitian), and \[ A^\dagger A=\begin{pmatrix}0&-i\\ i&0\end{pmatrix}\begin{pmatrix}0&-i\\ i&0\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}=I . \] (B) is not Hermitian (\(A^\dagger=-A\)); (C) gives \(A^\dagger A=2A\neq I\); (D) gives \(A^\dagger A=2I\). Hence \(\boxed{\text{(A)}}\).

Answer: (A) \(\begin{pmatrix}0&-i\\ i&0\end{pmatrix}\)

MP02.5IIT JAM 2023, Q15MCQ | 2 marks

Inverse of the matrix \(\begin{bmatrix}1&1&0\\2&3&0\\1&0&1\end{bmatrix}\) is

  1. \(\begin{bmatrix}1&-2&1\\-1&3&0\\0&0&1\end{bmatrix}\)

  2. \(\begin{bmatrix}3&-1&0\\-2&1&0\\-3&1&1\end{bmatrix}\)

  3. \(\begin{bmatrix}-1&-1&0\\2&3&0\\1&0&1\end{bmatrix}\)

  4. \(\begin{bmatrix}3&-2&-3\\-2&1&1\\0&0&1\end{bmatrix}\)

Show solution

\(A^{-1}=\dfrac{\operatorname{adj}A}{\det A}\). Expanding along the third column, \(\det A=1\cdot(1\cdot3-1\cdot2)=1\).

The cofactors are \(C_{11}=3,\ C_{12}=-2,\ C_{13}=-3,\ C_{21}=-1,\ C_{22}=1,\ C_{23}=1,\ C_{31}=0,\ C_{32}=0,\ C_{33}=1\), and the adjugate is the transpose of the cofactor matrix: \[ A^{-1}=\operatorname{adj}A=\begin{bmatrix}3&-1&0\\-2&1&0\\-3&1&1\end{bmatrix}. \] Check: \(AA^{-1}\) has rows \((3-2,\,-1+1,\,0)=(1,0,0)\), \((6-6,\,-2+3,\,0)=(0,1,0)\), \((3-3,\,-1+1,\,1)=(0,0,1)\). \[\boxed{\text{(B)}}\]

Answer: (B)

MP02.6IIT JAM 2023, Q55NAT | 2 marks

The sum of the eigenvalues \(\lambda_1\) and \(\lambda_2\) of matrix \(B=I+A+A^2\), where \(A=\begin{bmatrix}2&1\\-0.5&0.5\end{bmatrix}\) is (rounded off to two decimal places).

Show solution

The characteristic equation of \(A\) is \[ \begin{vmatrix}2-\lambda&1\\-0.5&0.5-\lambda\end{vmatrix}=(2-\lambda)(0.5-\lambda)+0.5=\lambda^2-2.5\lambda+1.5=0 \quad\Rightarrow\quad(\lambda-1)(\lambda-1.5)=0 , \] so \(A\) has eigenvalues \(1\) and \(1.5\). If \(A\vec x=a\vec x\) then \(B\vec x=(1+a+a^2)\vec x\), so the eigenvalues of \(B\) are \[ \lambda_1=1+1+1=3,\qquad \lambda_2=1+1.5+2.25=4.75 . \] \[ \lambda_1+\lambda_2=7.75 . \] (Check: \(A^2=\begin{bmatrix}3.5&2.5\\-1.25&-0.25\end{bmatrix}\), so \(\operatorname{Tr}B=2+2.5+3.25=7.75\).) \[\boxed{7.75}\]

Answer: 7.75 (range 7.70 to 7.80)

MP02.7IIT JAM 2022, Q31MSQ | 2 marks

Consider the \(2\times2\) matrix \(M=\begin{pmatrix}0&a\\a&b\end{pmatrix}\), where \(a,b>0\). Then,

  1. \(M\) is a real symmetric matrix

  2. One of the eigenvalues of \(M\) is greater than \(b\)

  3. One of the eigenvalues of \(M\) is negative

  4. Product of eigenvalues of \(M\) is \(b\)

Show solution

\(M^T=\begin{pmatrix}0&a\\a&b\end{pmatrix}=M\) with real entries, so \(M\) is real symmetric: (A) is true.

The characteristic equation is \[ \begin{vmatrix}-\lambda&a\\a&b-\lambda\end{vmatrix}=0\;\Rightarrow\;\lambda^2-b\lambda-a^2=0\;\Rightarrow\; \lambda_{1,2}=\frac{b\pm\sqrt{b^2+4a^2}}{2}. \] Since \(\sqrt{b^2+4a^2}>b\), \[ \lambda_1=\frac{b+\sqrt{b^2+4a^2}}{2}>\frac{b+b}{2}=b,\qquad \lambda_2=\frac{b-\sqrt{b^2+4a^2}}{2}<0 . \] So (B) and (C) are true. The product of the eigenvalues is \(\det M=-a^2\neq b\), so (D) is false. \[\boxed{\text{(A), (B), (C)}}\]

Answer: (A), (B), (C)

MP02.8IIT JAM 2021, Q2MCQ | 1 mark

Let \(M\) be a \(2\times2\) matrix. Its trace is \(6\) and its determinant has value \(8\). Its eigenvalues are

  1. \(2\) and \(4\)

  2. \(3\) and \(3\)

  3. \(2\) and \(6\)

  4. \(-2\) and \(-3\)

Show solution

The trace is the sum and the determinant is the product of the eigenvalues: \[ \lambda_1+\lambda_2=6,\qquad \lambda_1\lambda_2=8 \;\Rightarrow\;\lambda^2-6\lambda+8=0\;\Rightarrow\;(\lambda-2)(\lambda-4)=0 . \] \[ \boxed{\lambda_1=2,\;\lambda_2=4} \]

Answer: (A) 2 and 4

MP02.9IIT JAM 2020, Q31MSQ | 2 marks

If \(P\) and \(Q\) are Hermitian matrices which of the following is/are true?

(A matrix \(P\) is Hermitian if \(P=P^\dagger\), where the elements \(P^\dagger_{ij}=P^*_{ji}\))

  1. \(PQ+QP\) is always Hermitian

  2. \(i(PQ-QP)\) is always Hermitian

  3. \(PQ\) is always Hermitian

  4. \(PQ-QP\) is always Hermitian

Show solution

Using \((AB)^\dagger=B^\dagger A^\dagger\) with \(P^\dagger=P\), \(Q^\dagger=Q\): \begin{align*} (PQ+QP)^\dagger&=Q^\dagger P^\dagger+P^\dagger Q^\dagger=QP+PQ &&\text{(Hermitian)}\\ (PQ)^\dagger&=Q^\dagger P^\dagger=QP\neq PQ \text{ in general} &&\text{(not Hermitian)}\\ (PQ-QP)^\dagger&=QP-PQ=-(PQ-QP) &&\text{(anti-Hermitian)}\\ \left[i(PQ-QP)\right]^\dagger&=-i\left(QP-PQ\right)=i(PQ-QP) &&\text{(Hermitian)} \end{align*} \[ \boxed{\text{(A) and (B) are true}} \]

Answer: (A), (B)

MP02.10IIT JAM 2019, Q12MCQ | 2 marks

The eigenvalues of \(\begin{pmatrix}3&i&0\\-i&3&0\\0&0&6\end{pmatrix}\) are

  1. \(2,4\) and \(6\)

  2. \(2i,4i\) and \(6\)

  3. \(2i,4\) and \(8\)

  4. \(0,4\) and \(8\)

Show solution

The characteristic equation is \[ \begin{vmatrix}3-\lambda&i&0\\-i&3-\lambda&0\\0&0&6-\lambda\end{vmatrix}=0 \;\Rightarrow\;(6-\lambda)\left[(3-\lambda)^2-(i)(-i)\right]=0 \] \[ \Rightarrow\;(6-\lambda)\left[(\lambda-3)^2-1\right]=0 \;\Rightarrow\;(6-\lambda)(\lambda^2-6\lambda+8)=0 \;\Rightarrow\;(6-\lambda)(\lambda-2)(\lambda-4)=0. \] \[ \boxed{\lambda=2,\;4,\;6} \] (As expected for a Hermitian matrix, the eigenvalues are real.)

Answer: (A) \(2,4\) and \(6\)

MP02.11IIT JAM 2018, Q31MSQ | 2 marks

Let matrix \(M=\begin{pmatrix}4 & x\\ 6 & 9\end{pmatrix}\). If \(\det(M)=0\), then

  1. \(M\) is symmetric

  2. \(M\) is invertible

  3. one eigenvalue is \(13\)

  4. Its eigenvectors are orthogonal

Show solution

\[ \det M=36-6x=0\;\Rightarrow\;x=6,\qquad M=\begin{pmatrix}4 & 6\\ 6 & 9\end{pmatrix}. \] (A) \(M^{T}=M\), so \(M\) is symmetric. Correct.

(B) \(\det M=0\), so \(M\) is not invertible. Incorrect.

(C) The eigenvalues follow from \[ \begin{vmatrix}4-\lambda & 6\\ 6 & 9-\lambda\end{vmatrix}=0\;\Rightarrow\;(4-\lambda)(9-\lambda)-36=\lambda^2-13\lambda=0\;\Rightarrow\;\lambda=0,\ 13 . \] Correct.

(D) Eigenvectors of a real symmetric matrix belonging to distinct eigenvalues are orthogonal. Correct. \[ \boxed{(A),\ (C),\ (D)} \]

Answer: (A), (C), (D)

MP02.12IIT JAM 2017, Q8MCQ | 1 mark

For the three matrices given below, which one of the choices is correct? \[ \sigma_1=\begin{pmatrix}0&1\\1&0\end{pmatrix}\qquad \sigma_2=\begin{pmatrix}0&i\\-i&0\end{pmatrix}\qquad \sigma_3=\begin{pmatrix}1&0\\0&-1\end{pmatrix} \]

  1. \(\sigma_1\sigma_2=-i\sigma_3\)

  2. \(\sigma_1\sigma_2=i\sigma_3\)

  3. \(\sigma_1\sigma_2+\sigma_2\sigma_1=I\)

  4. \(\sigma_3\sigma_2=-i\sigma_1\)

Show solution

\[ \sigma_1\sigma_2=\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}0&i\\-i&0\end{pmatrix}=\begin{pmatrix}-i&0\\0&i\end{pmatrix}=-i\begin{pmatrix}1&0\\0&-1\end{pmatrix}=\boxed{-i\sigma_3} \] Checking the others: \(\sigma_2\sigma_1=\begin{pmatrix}i&0\\0&-i\end{pmatrix}\), so \(\sigma_1\sigma_2+\sigma_2\sigma_1=0\neq I\); and \(\sigma_3\sigma_2=\begin{pmatrix}0&i\\i&0\end{pmatrix}=i\sigma_1\neq-i\sigma_1\).

Answer: (A) \(\sigma_1\sigma_2=-i\sigma_3\)

MP02.13IIT JAM 2016, Q6MCQ | 1 mark

The eigenvalues of the matrix representing the following pair of linear equations \[ x+iy=0,\qquad ix+y=0 \] are

  1. \(1+i,\ 1+i\)

  2. \(1-i,\ 1-i\)

  3. \(1,\ i\)

  4. \(1+i,\ 1-i\)

Show solution

The coefficient matrix is \[ M=\begin{pmatrix}1&i\\ i&1\end{pmatrix},\qquad \det(M-\lambda I)=(1-\lambda)^2-i^2=(1-\lambda)^2+1=0 . \] Hence \(1-\lambda=\pm i\), i.e. \[ \boxed{\lambda=1+i,\ 1-i} \]

Answer: (D) \(1+i,\ 1-i\)

MP02.14IIT JAM 2016, Q11MCQ | 2 marks

For the given set of equations \[ x+y=1,\qquad y+z=1,\qquad x+z=1, \] which one of the following statements is correct?

  1. Equations are inconsistent

  2. Equations are consistent and a single non-trivial solution exists

  3. Equations are consistent and many solutions exist

  4. Equations are consistent and only a trivial solution exists.

Show solution

The coefficient matrix has determinant \[ \begin{vmatrix}1&1&0\\0&1&1\\1&0&1\end{vmatrix}=1(1-0)-1(0-1)+0=2\neq0 , \] so the non-homogeneous system has a unique solution. Adding the three equations, \(2(x+y+z)=3\), i.e. \(x+y+z=\tfrac32\), so \[ x=y=z=\tfrac12 . \] \[ \boxed{\text{consistent, single non-trivial solution}} \]

Answer: (B) Equations are consistent and a single non-trivial solution exists

MP02.15GATE 2026, Q37MCQ | 2 marks

Consider the Pauli matrices \(\sigma_x=\begin{pmatrix}0&1\\1&0\end{pmatrix}\), \(\sigma_y=\begin{pmatrix}0&-i\\ i&0\end{pmatrix}\), \(\sigma_z=\begin{pmatrix}1&0\\0&-1\end{pmatrix}\).

The value of \(\mathrm{Tr}\left(\sigma_z[\sigma_x,\sigma_y]\right)\) is

  1. \(2i\)

  2. \(i\)

  3. \(4i\)

  4. \(\dfrac i2\)

Show solution

\([\sigma_x,\sigma_y]=2i\sigma_z\), so \[\mathrm{Tr}(\sigma_z\cdot2i\sigma_z)=2i\,\mathrm{Tr}(\mathbb 1_2)=\boxed{4i}\]

Answer: (C) \(4i\)

MP02.16GATE 2025, Q61NAT | 2 marks

Consider the set \(\{1,x,x^2\}\). An orthonormal basis in \(x\in[-1,1]\) is formed from these three terms, where the normalization of a function \(f(x)\) is defined via \(\int_{-1}^{1}x^2[f(x)]^2dx=1\). If the orthonormal basis set is \(\left(\sqrt{\dfrac32},\;\sqrt{\dfrac52}\,x,\;\dfrac12\sqrt{\dfrac{21}{N}}\,(5x^2-3)\right)\), then the value of \(N\) (in integer) is

Show solution

The weight is \(w(x)=x^2\) and the third basis function is \(f(x)=\frac12\sqrt{\frac{21}{N}}(5x^2-3)\). (Check of the others: \(\int_{-1}^1x^2\cdot\frac32\,dx=1\), \(\int_{-1}^1x^2\cdot\frac52x^2\,dx=1\), and \(\int_{-1}^1x^2(5x^2-3)\,dx=2-2=0\).)

Normalization: \[1=\int_{-1}^{1}x^2f^2\,dx=\frac14\cdot\frac{21}{N}\int_{-1}^{1}x^2(5x^2-3)^2dx .\] \[\int_{-1}^{1}x^2(5x^2-3)^2dx=\int_{-1}^{1}\left(25x^6-30x^4+9x^2\right)dx=2\left(\frac{25}{7}-6+3\right)=\frac87 .\] \[1=\frac{21}{4N}\cdot\frac87=\frac6N\;\Rightarrow\;N=6 .\] \[\boxed{N=6}\]

Answer: 6 (range 6 to 6)

MP02.17GATE 2024, Q43MSQ | 2 marks

Consider two matrices: \(P=\begin{bmatrix}1&2\\0&1\end{bmatrix}\) and \(Q=\begin{bmatrix}1&0\\0&1\end{bmatrix}\). Which of the following statement is/are true?

  1. \(P\) and \(Q\) have same set of eigenvalues

  2. \(P\) and \(Q\) commute with each other

  3. \(P\) and \(Q\) have different sets of linearly independent eigenvectors

  4. \(P\) is diagonalizable

Show solution

Both matrices are upper triangular with diagonal entries \(1,1\), so both have eigenvalues \(\{1,1\}\): (A) is true. \(Q\) is the identity, so \(PQ=QP=P\): (B) is true.

For \(P\): \((P-I)\begin{pmatrix}a\\b\end{pmatrix}=\begin{pmatrix}2b\\0\end{pmatrix}=0\Rightarrow b=0\), so \(P\) has only one independent eigenvector \(\begin{pmatrix}1\\0\end{pmatrix}\). Every vector is an eigenvector of \(Q\), which has two independent eigenvectors, e.g. \(\begin{pmatrix}1\\0\end{pmatrix}\), \(\begin{pmatrix}0\\1\end{pmatrix}\). So (C) is true, and since \(P\) lacks a second independent eigenvector it is not diagonalizable: (D) is false. \[\boxed{\text{(A), (B), (C)}}\]

Answer: (A), (B), (C)

MP02.18GATE 2023, Q26MCQ | 1 mark

A \(4\times4\) matrix \(M\) has the property \(M^\dagger=-M\) and \(M^4=\mathbf 1\), where \(\mathbf 1\) is the \(4\times4\) identity matrix. Which one of the following is the CORRECT set of eigenvalues of the matrix \(M\)?

  1. \((1,1,-1,-1)\)

  2. \((i,i,-i,-i)\)

  3. \((i,i,i,-i)\)

  4. \((1,1,-i,-i)\)

Show solution

\(M^\dagger=-M\) means \(M\) is skew-Hermitian, so its eigenvalues are purely imaginary (or zero). \(M^4=\mathbf 1\) gives \(\lambda^4=1\), so \(\lambda\in\{\pm1,\pm i\}\). Together these give \(\lambda=\pm i\), which rules out (A) and (D). For a (real) skew-symmetric matrix the eigenvalues occur in complex-conjugate pairs \(\pm i\) and the trace vanishes, \(\operatorname{Tr}M=\sum\lambda=0\). Of (B) and (C), only \((i,i,-i,-i)\) satisfies this.

\(\boxed{(i,i,-i,-i)}\)

Answer: (B) \((i,i,-i,-i)\)

MP02.19GATE 2023, Q32MSQ | 1 mark

Which of the following options represent(s) linearly independent pair(s) of functions of a real variable \(x\)?

  1. \(e^{ix}\) and \(e^{-ix}\)

  2. \(x\) and \(e^x\)

  3. \(2^x\) and \(2^{-3+x}\)

  4. \(e^{ix}\) and \(\sin x\)

Show solution

Use the Wronskian \(W(f,g)=fg'-gf'\); the pair is linearly independent if \(W\not\equiv0\).

  • (A) \(W=e^{ix}(-ie^{-ix})-e^{-ix}(ie^{ix})=-2i\neq0\): independent.

  • (B) \(W=x\,e^x-e^x\cdot1=(x-1)e^x\not\equiv0\): independent.

  • (C) \(2^{-3+x}=\tfrac18\,2^x\) is a constant multiple of \(2^x\). Indeed \(W=2^x\cdot\tfrac18 2^x\ln2-\tfrac182^x\cdot2^x\ln2=0\): dependent.

  • (D) \(W=e^{ix}\cos x-\sin x\,(ie^{ix})=e^{ix}(\cos x-i\sin x)=1\neq0\): independent.

\(\boxed{\text{(A), (B), (D)}}\)

Answer: (A), (B), (D)

MP02.20GATE 2022, Q13MCQ | 1 mark

What is the maximum number of free independent real parameters specifying an \(n\) dimensional orthogonal matrix?

  1. \(n(n-2)\)

  2. \((n-1)^2\)

  3. \(\dfrac{n(n-1)}{2}\)

  4. \(\dfrac{n(n+1)}{2}\)

Show solution

For the \(2\times2\) case, take \(A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\). The condition \(A^TA=I\) gives \[ \begin{pmatrix}a^2+c^2 & ab+cd\\ ab+cd & b^2+d^2\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}, \] i.e. \(a^2+c^2=1\), \(b^2+d^2=1\) (two diagonal constraints) and \(ab+cd=0\) (one off-diagonal constraint). That leaves \(4-3=1\) free parameter (the rotation angle).

For an \(n\times n\) matrix there are \(n^2\) real entries. \(A^TA=I\) is a symmetric matrix equation: it imposes \(n\) diagonal (normalisation) constraints and \(^nC_2=\tfrac{n(n-1)}{2}\) off-diagonal (orthogonality) constraints. Hence the number of independent parameters is \[ n^2-n-\frac{n(n-1)}{2}=\frac{n(n-1)}{2}. \] \[\boxed{\dfrac{n(n-1)}{2}}\]

Answer: (C) \(\dfrac{n(n-1)}{2}\)

MP02.21GATE 2021, Q53MSQ | 2 marks

\(P\) and \(Q\) are two Hermitian matrices and there exists a matrix \(R\), which diagonalizes both of them, such that \(RPR^{-1}=S_1\) and \(RQR^{-1}=S_2\), where \(S_1\) and \(S_2\) are diagonal matrices. The correct statement(s) is (are)

(A) All the elements of both matrices \(S_1\) and \(S_2\) are real

(B) The matrix \(PQ\) can have complex eigenvalues.

(C) The matrix \(QP\) can have complex eigenvalues.

(D) The matrices \(P\) and \(Q\) commute

Show solution

(A) The diagonal entries of \(S_1\) and \(S_2\) are the eigenvalues of \(P\) and \(Q\). Hermitian matrices have real eigenvalues, and the off-diagonal elements are zero. True.

(D) \(P=R^{-1}S_1R\) and \(Q=R^{-1}S_2R\), so \[PQ=R^{-1}S_1S_2R,\qquad QP=R^{-1}S_2S_1R.\] Diagonal matrices commute (\(S_1S_2=S_2S_1\)), so \(PQ=QP\). True.

(B), (C) \(PQ\) and \(QP\) are similar to the diagonal matrix \(S_1S_2\), whose entries are products of real numbers. Their eigenvalues are therefore real. Both false. \[\boxed{\text{(A), (D)}}\]

Answer: (A), (D)

MP02.22GATE 2020, Q12MCQ | 1 mark

A real, invertible \(3\times3\) matrix \(M\) has eigenvalues \(\lambda_i,\ (i=1,2,3)\) and the corresponding eigenvectors are \(\ket{e_i},\ (i=1,2,3)\) respectively. Which one of the following is correct?

  1. \(M\ket{e_i}=\dfrac{1}{\lambda_i}\ket{e_i}\), for \(i=1,2,3\)

  2. \(M^{-1}\ket{e_i}=\dfrac{1}{\lambda_i}\ket{e_i}\), for \(i=1,2,3\)

  3. \(M^{-1}\ket{e_i}=\lambda_i\ket{e_i}\), for \(i=1,2,3\)

  4. The eigenvalues of \(M\) and \(M^{-1}\) are not related

Show solution

Since \(M\) is invertible, no \(\lambda_i\) is zero. Operating with \(M^{-1}\) on \(M\ket{e_i}=\lambda_i\ket{e_i}\): \[ \ket{e_i}=\lambda_i M^{-1}\ket{e_i}\quad\Rightarrow\quad \boxed{M^{-1}\ket{e_i}=\frac{1}{\lambda_i}\ket{e_i}} \] So \(M^{-1}\) has the same eigenvectors, with reciprocal eigenvalues.

Answer: (B)

MP02.23GATE 2020, Q36MCQ | 2 marks

The product of eigenvalues of \(\begin{pmatrix}0&0&1\\0&1&0\\1&0&0\end{pmatrix}\) is

  1. \(-1\)

  2. \(1\)

  3. \(0\)

  4. \(2\)

Show solution

The product of the eigenvalues equals the determinant. Expanding along the first row: \[ \lambda_1\lambda_2\lambda_3=\det A=1\cdot\begin{vmatrix}0&1\\1&0\end{vmatrix}=1\cdot(0-1)=\boxed{-1} \] (The eigenvalues are \(1,1,-1\).)

Answer: (A) \(-1\)

MP02.24GATE 2020, Q37MCQ | 2 marks

Let \(\ket{e_1}=\begin{pmatrix}1\\0\\0\end{pmatrix},\ \ket{e_2}=\begin{pmatrix}1\\1\\0\end{pmatrix},\ \ket{e_3}=\begin{pmatrix}1\\1\\1\end{pmatrix}\). Let \(S=\{\ket{e_1},\ket{e_2},\ket{e_3}\}\).

Let \(\mathbb{R}^3\) denote the three dimensional real vector space. Which one of the following is correct?

  1. \(S\) is an orthonormal set

  2. \(S\) is a linearly dependent set

  3. \(S\) is a basis for \(\mathbb{R}^3\)

  4. \(\displaystyle\sum_{i=1}^{3}\ket{e_i}\bra{e_i}=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}\)

Show solution

(A) \(\braket{e_1}{e_2}=1\neq0\), so \(S\) is not orthonormal.

(B) Suppose \(c_1\ket{e_1}+c_2\ket{e_2}+c_3\ket{e_3}=0\). Component-wise: \[ c_1+c_2+c_3=0,\qquad c_2+c_3=0,\qquad c_3=0\ \Rightarrow\ c_3=c_2=c_1=0 . \] So \(S\) is linearly independent, and (B) is false.

(C) Three linearly independent vectors in the three-dimensional space \(\mathbb{R}^3\) form a basis. True.

(D) Since the vectors are not orthonormal, the completeness relation fails: \[ \sum_i\ket{e_i}\bra{e_i}=\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix}+\begin{pmatrix}1&1&0\\1&1&0\\0&0&0\end{pmatrix}+\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}=\begin{pmatrix}3&2&1\\2&2&1\\1&1&1\end{pmatrix}\neq I . \] \[\boxed{S \text{ is a basis for } \mathbb{R}^3}\]

Answer: (C)

MP02.25GATE 2020, Q42MCQ | 2 marks

Which one of the following matrices does NOT represent a proper rotation in a plane?

  1. \(\begin{pmatrix}-\sin\theta&\cos\theta\\-\cos\theta&-\sin\theta\end{pmatrix}\)

  2. \(\begin{pmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{pmatrix}\)

  3. \(\begin{pmatrix}\sin\theta&\cos\theta\\-\cos\theta&\sin\theta\end{pmatrix}\)

  4. \(\begin{pmatrix}-\sin\theta&\cos\theta\\-\cos\theta&\sin\theta\end{pmatrix}\)

Show solution

A proper rotation is an orthogonal matrix (\(R^TR=I\)) with \(\det R=+1\). Each of (A), (B), (C) has orthonormal columns, and their determinants are \[ \text{(A)}\ \sin^2\theta+\cos^2\theta=1,\qquad \text{(B)}\ \cos^2\theta+\sin^2\theta=1,\qquad \text{(C)}\ \sin^2\theta+\cos^2\theta=1 . \] (Each is a rotation by a shifted angle, e.g. (C) is the rotation by \(\tfrac{\pi}{2}-\theta\).) For (D): \[ \det=(-\sin\theta)(\sin\theta)-(\cos\theta)(-\cos\theta)=\cos^2\theta-\sin^2\theta=\cos2\theta, \] which is not identically \(1\), and the columns \((-\sin\theta,-\cos\theta)^T\), \((\cos\theta,\sin\theta)^T\) are not orthogonal (their dot product is \(-2\sin\theta\cos\theta\)). So (D) is not a rotation. \[\boxed{\text{(D)}}\]

Answer: (D)

MP02.26GATE 2019, Q18MCQ | 1 mark

During a rotation, vectors along the axis of rotation remain unchanged. For the rotation matrix \(\begin{pmatrix}0&1&0\\0&0&-1\\-1&0&0\end{pmatrix}\), the unit vector along the axis of rotation is

  1. \(\dfrac13\left(2\hat i-\hat j+2\hat k\right)\)

  2. \(\dfrac{1}{\sqrt3}\left(\hat i+\hat j-\hat k\right)\)

  3. \(\dfrac{1}{\sqrt3}\left(\hat i-\hat j-\hat k\right)\)

  4. \(\dfrac13\left(2\hat i+2\hat j-\hat k\right)\)

Show solution

A vector along the axis is unchanged, \(A\vec x=\vec x\), i.e. it is the eigenvector with eigenvalue \(\lambda=1\): \[ \begin{pmatrix}0&1&0\\0&0&-1\\-1&0&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} \;\Rightarrow\; x_2=x_1,\quad -x_3=x_2,\quad -x_1=x_3 . \] The general eigenvector is \((k,k,-k)^T\), so the unit vector along the axis is \[ \hat n=\frac{1}{\sqrt3}\left(\hat i+\hat j-\hat k\right). \] \(\boxed{\tfrac{1}{\sqrt3}(\hat i+\hat j-\hat k)}\)

Answer: (B)

MP02.27GATE 2018, Q11MCQ | 1 mark

The eigenvalues of a Hermitian matrix are all

  1. real

  2. imaginary

  3. of modulus one

  4. real and positive

Show solution

If \(H=H^\dagger\) and \(H\ket{x}=\lambda\ket{x}\), then \(\lambda\braket{x}{x}=\bra{x}H\ket{x}=\bra{x}H^\dagger\ket{x}=\lambda^*\braket{x}{x}\), so \(\lambda=\lambda^*\). The eigenvalues are real but not necessarily positive (e.g. \(\sigma_z\) has eigenvalue \(-1\)). \[\boxed{\text{real}}\]

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