StudyDump
On this page

LAC CAT-2: Maxima, Minima, Rank & Eigenvalues Question Bank

Linear Algebra and Calculus · Unit 2

LAC CAT-2: Maxima, Minima, Rank & Eigenvalues Question Bank

Course: Linear Algebra and Calculus
Course Code: 25APS0101
University: Chitkara University, Punjab • Semester 01
Scope: Complete CAT-2 Curriculum (Units 1, 2 & 3) • 162 Solved Examination Problems with Step-by-Step Derivations

Question Bank Quick Navigator

Unit / Section Thematic Focus Question Range Topics Covered
Unit 1 (Section 1) Theoretical Foundations of Extrema & Lagrange Multipliers Ques 1 – 14 Stationary points, 2nd derivative discriminant $rt - s^2$, local extrema classification, and auxiliary Lagrange functions.
Unit 1 (Section 1) Unconstrained Maxima, Minima & Saddle Point Problems Ques 15 – 29 Two-variable critical points, saddle point criteria, local maxima/minima, and step-by-step Hessian derivations.
Unit 1 (Section 1) Constrained Optimization & Lagrange Multipliers Ques 30 – 51 Geometric extrema, open rectangular box surface optimization, plane distances, and constrained extrema.
Unit 2 (Section 2) Theoretical Properties of Rank, Normal Form & Consistency Ques 1 – 22 Determinant minors, elementary row/column operations, transpose invariance, subadditivity, Sylvester's nullity, and Rouche-Capelli theorem.
Unit 2 (Section 2) Computational Problems on Rank, Normal Form & Linear Systems Ques 23 – 60 Echelon form reduction, normal form $[I_r\ 0;\ 0\ 0]$, finding non-singular matrices $P$ and $Q$, Gauss elimination, and parameter analysis ($\lambda, \mu$).
Unit 3 (Section 3) Theoretical Properties of Eigenvalues & Cayley-Hamilton Theorem Ques 61 – 74 Characteristic equations, eigenvalue trace and determinant relations, reciprocal eigenvalues, real symmetric orthogonality, and Cayley-Hamilton theorem.
Unit 3 (Section 3) Computational Problems on Eigenvalues, Eigenvectors & CHT Ques 75 – 111 $2\times 2$ and $3\times 3$ characteristic polynomials, finding eigenvalues and corresponding eigenvectors, calculating $A^{-1}$ and higher powers via CHT.

Unit 1 (Section 1): Differential Calculus — Maxima, Minima & Lagrange Multipliers

Topic 1.1: Theoretical Foundations of Extrema & Lagrange Multipliers (Q1–Q14)

Ques 1. For a function $f(x,y)$ to have a stationary point at $(a, b)$, which of the following conditions must hold?

  • (A) $f_{x}(a,b)=0$ and $f_{y}(a,b)=0$
  • (B) $f_{x}(a,b)>0$ and $f_{y}(a,b)<0$
  • (C) $f_{xx}(a,b)=0$ only
  • (D) $f_{xy}(a,b)=0$ only

Correct Answer: (A) $f_{x}(a,b)=0$ and $f_{y}(a,b)=0$

Ques 2. Let $r=f_{xx}$, $s=f_{xy}$, and $t=f_{yy}$ evaluated at a critical point $(a, b)$. The function $f(x,y)$ achieves a local minimum at $(a, b)$ if:

  • (A) $rt-s^{2}>0$ and $r>0$
  • (B) $rt-s^{2}>0$ and $r<0$
  • (C) $rt-s^{2}<0$ and $r>0$
  • (D) $rt-s^{2}=0$ and $r>0$

Correct Answer: (A) $rt-s^{2}>0$ and $r>0$

Ques 3. Under the second-derivative test for $f(x,y)$, a local maximum occurs at $(a, b)$ if:

  • (A) $rt-s^{2}>0$ and $r<0$
  • (B) $rt-s^{2}>0$ and $r>0$
  • (C) $rt-s^{2}<0$ and $r<0$
  • (D) $rt-s^{2}=0$ and $r<0$

Correct Answer: (A) $rt-s^{2}>0$ and $r<0$

Ques 4. If $rt-s^{2}<0$ at a critical point $(a, b)$ of $f(x,y)$, then $(a, b)$ is classified as a:

  • (A) Point of local minimum
  • (B) Point of local maximum
  • (C) Saddle point
  • (D) Boundary point

Correct Answer: (C) Saddle point

Ques 5. When $rt-s^{2}=0$ at a critical point $(a, b)$, the second derivative test is:

  • (A) Conclusive for maximum
  • (B) Conclusive for minimum
  • (C) Inconclusive
  • (D) Guaranteeing a saddle point

Correct Answer: (C) Inconclusive

Ques 6. In Lagrange's method of undetermined multipliers for optimizing $f(x,y)$ subject to constraint $\phi(x,y)=0$, the auxiliary function $F(x,y,\lambda)$ is defined as:

  • (A) $F=f+\lambda\phi$
  • (B) $F=f\cdot\phi+\lambda$
  • (C) $F=\frac{f}{\phi}+\lambda$
  • (D) $F=f-\lambda^{2}\phi$

Correct Answer: (A) $F=f+\lambda\phi$

Ques 7. To find the stationary points of the auxiliary function $F(x,y,\lambda)=f(x,y)+\lambda\phi(x,y)$, one must solve the system:

  • (A) $\frac{\partial F}{\partial x}=0, \frac{\partial F}{\partial y}=0, \phi(x,y)=0$
  • (B) $\frac{\partial F}{\partial x}=1, \frac{\partial F}{\partial y}=1$
  • (C) $f_{x}+f_{y}=0, \phi_{x}+\phi_{y}=0$
  • (D) $\frac{\partial^{2}F}{\partial x^{2}}=0, \frac{\partial^{2}F}{\partial y^{2}}=0$

Correct Answer: (A) $\frac{\partial F}{\partial x}=0, \frac{\partial F}{\partial y}=0, \phi(x,y)=0$

Ques 8. If $f(x,y,z)$ is subject to two constraints $\phi_{1}(x,y,z)=0$ and $\phi_{2}(x,y,z)=0$, Lagrange's auxiliary function $F$ is constructed as:

  • (A) $F=f+\lambda_{1}\phi_{1}+\lambda_{2}\phi_{2}$
  • (B) $F=f+\lambda_{1}\phi_{1}\cdot\lambda_{2}\phi_{2}$
  • (C) $F=f\cdot(\lambda_{1}\phi_{1}+\lambda_{2}\phi_{2})$
  • (D) $F=\lambda_{1}f+\lambda_{2}\phi_{1}+\phi_{2}$

Correct Answer: (A) $F=f+\lambda_{1}\phi_{1}+\lambda_{2}\phi_{2}$

Ques 9. The absolute maximum and absolute minimum of a continuous function on a closed bounded region $D$ can occur at:

  • (A) Interior critical points or boundary points
  • (B) Strictly interior points only
  • (C) Strictly boundary points only
  • (D) Points where $rt-s^{2}<0$ only

Correct Answer: (A) Interior critical points or boundary points

Ques 10. In Lagrange's auxiliary function $F=f+\lambda\phi$, the parameter $\lambda$ represents:

  • (A) The rate of change of the optimal value with respect to the constraint bound
  • (B) The curvature of $f$
  • (C) The maximum value of $f$
  • (D) The distance from origin

Correct Answer: (A) The rate of change of the optimal value with respect to the constraint bound

Ques 11. For $f(x,y)=x^{2}+y^{2}$, the origin $(0,0)$ is a:

  • (A) Local minimum
  • (B) Local maximum
  • (C) Saddle point
  • (D) Test inconclusive

Correct Answer: (A) Local minimum Step-by-Step Solution:

  1. Compute first partial derivatives: $f_x = 2x$, $f_y = 2y$. At $(0,0)$, $f_x = 0, f_y = 0$.
  2. Compute second partial derivatives: $r = f_{xx} = 2$, $s = f_{xy} = 0$, $t = f_{yy} = 2$.
  3. Evaluate $rt - s^2 = (2)(2) - 0^2 = 4 > 0$. Since $r = 2 > 0$, $(0,0)$ is a local minimum.

Ques 12. For $f(x,y)=x^{2}-y^{2}$, the origin $(0,0)$ is a:

  • (A) Local minimum
  • (B) Local maximum
  • (C) Saddle point
  • (D) Absolute maximum

Correct Answer: (C) Saddle point Step-by-Step Solution:

  1. Compute first partial derivatives: $f_x = 2x$, $f_y = -2y$. At $(0,0)$, $f_x = 0, f_y = 0$.
  2. Compute second partial derivatives: $r = f_{xx} = 2$, $s = f_{xy} = 0$, $t = f_{yy} = -2$.
  3. Evaluate $rt - s^2 = (2)(-2) - 0^2 = -4 < 0$. Hence, $(0,0)$ is a saddle point.

Ques 13. For $f(x,y)=-x^{2}-y^{2}$, the origin $(0,0)$ is a:

  • (A) Local minimum
  • (B) Local maximum
  • (C) Saddle point
  • (D) Inconclusive point

Correct Answer: (B) Local maximum Step-by-Step Solution:

  1. Compute first partial derivatives: $f_x = -2x$, $f_y = -2y$. At $(0,0)$, $f_x = 0, f_y = 0$.
  2. Compute second partial derivatives: $r = f_{xx} = -2$, $s = f_{xy} = 0$, $t = f_{yy} = -2$.
  3. Evaluate $rt - s^2 = (-2)(-2) - 0^2 = 4 > 0$. Since $r = -2 < 0$, $(0,0)$ is a local maximum.

Ques 14. If $F(x,y,\lambda)=f(x,y)+\lambda\phi(x,y)$, setting $\frac{\partial F}{\partial\lambda}=0$ recovers:

  • (A) The constraint equation $\phi(x,y)=0$
  • (B) The objective function $f(x,y)=0$
  • (C) The multiplier $\lambda=0$
  • (D) The derivative $f_{x}=0$

Correct Answer: (A) The constraint equation $\phi(x,y)=0$

Topic 1.2: Unconstrained Maxima, Minima & Saddle Point Problems (Q15–Q29)

Ques 15. Find the critical point of $f(x,y)=2x^{2}+5y^{2}-8x-20y+14$.

  • (A) $(2,2)$
  • (B) $(1,2)$
  • (C) $(4,2)$
  • (D) $(0,0)$

Correct Answer: (A) $(2,2)$ Step-by-Step Solution:

  1. Differentiate partially with respect to $x$ and $y$: $\displaystyle f_x = 4x - 8, \quad f_y = 10y - 20$
  2. Set $f_x = 0 \implies 4x - 8 = 0 \implies x = 2$.
  3. Set $f_y = 0 \implies 10y - 20 = 0 \implies y = 2$.
  4. Critical point is $(2,2)$.

Ques 16. Evaluate the minimum value of $f(x,y)=2x^{2}+5y^{2}-8x-20y+14$.

  • (A) $-14$
  • (B) $14$
  • (C) $0$
  • (D) $-28$

Correct Answer: (A) $-14$ Step-by-Step Solution:

  1. From Q15, the critical point is $(2,2)$ (with $r = 4 > 0, s = 0, t = 10 \implies rt - s^2 = 40 > 0$).
  2. Substitute $(x,y) = (2,2)$ into $f(x,y)$: $\displaystyle f(2,2) = 2(2)^2 + 5(2)^2 - 8(2) - 20(2) + 14 = 8 + 20 - 16 - 40 + 14 = -14$

Ques 17. Find the critical points of $f(x,y)=x^{3}+y^{3}-12xy$.

  • (A) $(0,0)$ and $(4,4)$
  • (B) $(0,0)$ and $(2,2)$
  • (C) $(1,1)$ and $(-1,-1)$
  • (D) $(3,3)$ only

Correct Answer: (A) $(0,0)$ and $(4,4)$ Step-by-Step Solution:

  1. Set first partial derivatives to zero: $\displaystyle f_x = 3x^2 - 12y = 0 \implies y = \frac{x^2}{4}$ $\displaystyle f_y = 3y^2 - 12x = 0 \implies y^2 = 4x$
  2. Substitute $y = \frac{x^2}{4}$ into $y^2 = 4x$: $\displaystyle \left(\frac{x^2}{4}\right)^2 = 4x \implies x^4 - 64x = 0 \implies x(x^3 - 64) = 0 \implies x = 0, 4$
  3. When $x = 0 \implies y = 0$; when $x = 4 \implies y = \frac{16}{4} = 4$. Critical points are $(0,0)$ and $(4,4)$.

Ques 18. For $f(x,y)=x^{3}+y^{3}-12xy$, classify the critical point $(4, 4)$.

  • (A) Local minimum
  • (B) Local maximum
  • (C) Saddle point
  • (D) Test inconclusive

Correct Answer: (A) Local minimum Step-by-Step Solution:

  1. Compute second partial derivatives: $r = f_{xx} = 6x$, $s = f_{xy} = -12$, $t = f_{yy} = 6y$.
  2. Evaluate at $(4,4)$: $r = 24$, $s = -12$, $t = 24$.
  3. Calculate $rt - s^2 = (24)(24) - (-12)^2 = 576 - 144 = 432 > 0$.
  4. Since $rt - s^2 > 0$ and $r = 24 > 0$, $(4,4)$ is a local minimum.

Ques 19. For $f(x,y)=x^{3}+y^{3}-12xy$, classify the critical point $(0,0)$.

  • (A) Saddle point
  • (B) Local minimum
  • (C) Local maximum
  • (D) Absolute minimum

Correct Answer: (A) Saddle point Step-by-Step Solution:

  1. Evaluate $r = 6x, s = -12, t = 6y$ at $(0,0)$: $r = 0$, $s = -12$, $t = 0$.
  2. Calculate $rt - s^2 = (0)(0) - (-12)^2 = -144 < 0$.
  3. Since $rt - s^2 < 0$, $(0,0)$ is a saddle point.

Ques 20. Find the stationary points of $f(x,y)=3x^{2}-y^{2}+12x-6y+2$.

  • (A) $(-2,-3)$
  • (B) $(2,3)$
  • (C) $(-2,3)$
  • (D) $(0,0)$

Correct Answer: (A) $(-2,-3)$ Step-by-Step Solution:

  1. Set $f_x = 6x + 12 = 0 \implies x = -2$.
  2. Set $f_y = -2y - 6 = 0 \implies y = -3$.
  3. Stationary point is $(-2,-3)$.

Ques 21. Classify the critical point $(-2, -3)$ of $f(x,y)=3x^{2}-y^{2}+12x-6y+2$.

  • (A) Saddle point
  • (B) Local minimum
  • (C) Local maximum
  • (D) Inconclusive

Correct Answer: (A) Saddle point Step-by-Step Solution:

  1. Compute second partial derivatives: $r = f_{xx} = 6$, $s = f_{xy} = 0$, $t = f_{yy} = -2$.
  2. Evaluate $rt - s^2 = (6)(-2) - 0^2 = -12 < 0$.
  3. Since $rt - s^2 < 0$, $(-2,-3)$ is a saddle point.

Ques 22. Find the local extreme value of $f(x,y)=x^{2}+4y^{2}-6x-16y+20$.

  • (A) $-5$ at $(3,2)$
  • (B) $5$ at $(3,2)$
  • (C) $0$ at $(0,0)$
  • (D) $-10$ at $(2,3)$

Correct Answer: (A) $-5$ at $(3,2)$ Step-by-Step Solution:

  1. Set $f_x = 2x - 6 = 0 \implies x = 3$.
  2. Set $f_y = 8y - 16 = 0 \implies y = 2$.
  3. Substitute $(3,2)$ into $f(x,y)$: $\displaystyle f(3,2) = (3)^2 + 4(2)^2 - 6(3) - 16(2) + 20 = 9 + 16 - 18 - 32 + 20 = -5$

Ques 23. Calculate $r, s, t$ for $f(x,y)=x^{4}+y^{4}-4xy$ at $(1,1)$.

  • (A) $r=12, s=-4, t=12$
  • (B) $r=4, s=-4, t=4$
  • (C) $r=12, s=0, t=12$
  • (D) $r=3, s=-1, t=3$

Correct Answer: (A) $r=12, s=-4, t=12$ Step-by-Step Solution:

  1. Compute first partial derivatives: $f_x = 4x^3 - 4y$, $f_y = 4y^3 - 4x$.
  2. Compute second partial derivatives: $r = f_{xx} = 12x^2$, $s = f_{xy} = -4$, $t = f_{yy} = 12y^2$.
  3. Substitute $(1,1)$: $r = 12(1)^2 = 12$, $s = -4$, $t = 12(1)^2 = 12$.

Ques 24. Evaluate $rt-s^{2}$ at $(1, 1)$ for $f(x,y)=x^{4}+y^{4}-4xy$.

  • (A) $128>0$
  • (B) $-16<0$
  • (C) $0$
  • (D) $144$

Correct Answer: (A) $128>0$ Step-by-Step Solution:

  1. Using $r = 12, s = -4, t = 12$ at $(1,1)$ from Q23:
  2. $rt - s^2 = (12)(12) - (-4)^2 = 144 - 16 = 128 > 0$.

Ques 25. Determine the nature of $(1, 1)$ for $f(x,y)=x^{4}+y^{4}-4xy$.

  • (A) Local minimum
  • (B) Local maximum
  • (C) Saddle point
  • (D) Test fails

Correct Answer: (A) Local minimum Step-by-Step Solution:

  1. At $(1,1)$, $rt - s^2 = 128 > 0$ and $r = 12 > 0$.
  2. Therefore, $(1,1)$ is a local minimum.

Ques 26. For $f(x,y)=x^{4}+y^{4}-4xy$, determine the nature of $(0,0)$.

  • (A) Saddle point
  • (B) Local minimum
  • (C) Local maximum
  • (D) Absolute minimum

Correct Answer: (A) Saddle point Step-by-Step Solution:

  1. At $(0,0)$, $r = 12(0)^2 = 0$, $s = -4$, $t = 12(0)^2 = 0$.
  2. $rt - s^2 = (0)(0) - (-4)^2 = -16 < 0 \implies (0,0)$ is a saddle point.

Ques 27. Find the minimum value of $f(x,y)=x^{2}+y^{2}+\frac{16}{x}+\frac{16}{y}$ for $x, y>0$.

  • (A) $24$ at $(2,2)$
  • (B) $16$ at $(2,2)$
  • (C) $32$ at $(4,4)$
  • (D) $12$ at $(1,1)$

Correct Answer: (A) $24$ at $(2,2)$ Step-by-Step Solution:

  1. Set $f_x = 2x - \frac{16}{x^2} = 0 \implies 2x^3 = 16 \implies x^3 = 8 \implies x = 2$.
  2. Set $f_y = 2y - \frac{16}{y^2} = 0 \implies y = 2$.
  3. Substitute $(2,2)$: $\displaystyle f(2,2) = 2^2 + 2^2 + \frac{16}{2} + \frac{16}{2} = 4 + 4 + 8 + 8 = 24$

Ques 28. Find the critical point of $f(x,y)=x^{2}+3xy+y^{2}-5x-5y+3$.

  • (A) $(1,1)$
  • (B) $(2,2)$
  • (C) $(-1,-1)$
  • (D) $(0,5)$

Correct Answer: (A) $(1,1)$ Step-by-Step Solution:

  1. Set partial derivatives to zero: $\displaystyle f_x = 2x + 3y - 5 = 0, \quad f_y = 3x + 2y - 5 = 0$
  2. Subtract the two equations: $(2x + 3y) - (3x + 2y) = 0 \implies y = x$.
  3. Substitute $y = x$ into $2x + 3y = 5 \implies 5x = 5 \implies x = 1, y = 1$.

Ques 29. Classify the critical point $(1, 1)$ of $f(x,y)=x^{2}+3xy+y^{2}-5x-5y+3$.

  • (A) Saddle point since $rt-s^{2}=4-9=-5<0$
  • (B) Local minimum
  • (C) Local maximum
  • (D) Inconclusive

Correct Answer: (A) Saddle point since $rt-s^{2}=4-9=-5<0$ Step-by-Step Solution:

  1. Compute second partial derivatives: $r = f_{xx} = 2$, $s = f_{xy} = 3$, $t = f_{yy} = 2$.
  2. $rt - s^2 = (2)(2) - 3^2 = 4 - 9 = -5 < 0 \implies$ Saddle point.

Topic 1.3: Constrained Optimization & Mixed Problems (Q30–Q51)

Ques 30. Using $F=xy+\lambda(2x+3y-12)$, find the maximum of $f(x,y)=xy$ subject to $2x+3y=12$.

  • (A) $6$ at $(3,2)$
  • (B) $12$ at $(3,2)$
  • (C) $8$ at $(2,3)$
  • (D) $4$ at $(3,2)$

Correct Answer: (A) $6$ at $(3,2)$ Step-by-Step Solution:

  1. Set partial derivatives of $F$ to zero: $\displaystyle F_x = y + 2\lambda = 0 \implies \lambda = -\frac{y}{2}$ $\displaystyle F_y = x + 3\lambda = 0 \implies \lambda = -\frac{x}{3}$
  2. Equate $\lambda$: $-\frac{y}{2} = -\frac{x}{3} \implies 2x = 3y$.
  3. Substitute $2x = 3y$ into $2x + 3y = 12 \implies 6y = 12 \implies y = 2$ and $x = 3$.
  4. Maximum value $f(3,2) = 3 \times 2 = 6$.

Ques 31. In $F=xy+\lambda(2x+3y-12)$, what is the value of $\lambda$ at the stationary point?

  • (A) $-1$
  • (B) $-\frac{3}{2}$
  • (C) $-2$
  • (D) $-\frac{2}{3}$

Correct Answer: (A) $-1$ Step-by-Step Solution:

  1. From Q30, the stationary point is $(x,y) = (3,2)$ and $F_x = y + 2\lambda = 0$.
  2. Substitute $y = 2$: $2 + 2\lambda = 0 \implies \lambda = -1$.

Ques 32. Minimize $f(x,y)=x^{2}+y^{2}$ subject to $3x+4y=25$ using $F=x^{2}+y^{2}+\lambda(3x+4y-25)$.

  • (A) Minimum value $25$ at $(3, 4)$
  • (B) Minimum value $50$ at $(6,8)$
  • (C) Minimum value $16$ at $(4,3)$
  • (D) Minimum value $5$ at $(3,4)$

Correct Answer: (A) Minimum value $25$ at $(3, 4)$ Step-by-Step Solution:

  1. Set partial derivatives to zero: $\displaystyle F_x = 2x + 3\lambda = 0 \implies x = -\frac{3}{2}\lambda$ $\displaystyle F_y = 2y + 4\lambda = 0 \implies y = -2\lambda$
  2. Substitute into $3x + 4y = 25$: $\displaystyle 3\left(-\frac{3}{2}\lambda\right) + 4(-2\lambda) = 25 \implies -\frac{25}{2}\lambda = 25 \implies \lambda = -2$
  3. Thus $x = 3, y = 4$, and $f_{\min} = 3^2 + 4^2 = 25$.

Ques 33. In minimizing $f(x,y)=x^{2}+y^{2}$ subject to $3x+4y=25$, the value of $\lambda$ is:

  • (A) $-2$
  • (B) $2$
  • (C) $-5$
  • (D) $-1$

Correct Answer: (A) $-2$ Step-by-Step Solution:

  1. From $F_x = 2x + 3\lambda = 0$ at $x = 3$:
  2. $2(3) + 3\lambda = 0 \implies 3\lambda = -6 \implies \lambda = -2$.

Ques 34. Maximize $f(x,y,z)=xyz$ subject to $x+y+z=18$ ($x,y,z>0$) using $F=xyz+\lambda(x+y+z-18)$.

  • (A) $216$ at $(6,6,6)$
  • (B) $125$ at $(5,5,8)$
  • (C) $512$ at $(8,8,2)$
  • (D) $64$ at $(4,4, 10)$

Correct Answer: (A) $216$ at $(6,6,6)$ Step-by-Step Solution:

  1. Set partial derivatives to zero: $F_x = yz + \lambda = 0$, $F_y = xz + \lambda = 0$, $F_z = xy + \lambda = 0$.
  2. Since $x,y,z > 0$, $yz = xz = xy = -\lambda \implies x = y = z$.
  3. Substitute into $x + y + z = 18 \implies 3x = 18 \implies x = y = z = 6$.
  4. Maximum value $f(6,6,6) = 6 \times 6 \times 6 = 216$.

Ques 35. Find the stationary point of $F=(x^{2}+y^{2}+z^{2})+\lambda(x+2y+3z-14)$.

  • (A) $(1,2,3)$
  • (B) $(2,2,2)$
  • (C) $(3,2,1)$
  • (D) $(0,1,4)$

Correct Answer: (A) $(1,2,3)$ Step-by-Step Solution:

  1. Set partial derivatives to zero: $\displaystyle F_x = 2x + \lambda = 0 \implies x = -\frac{\lambda}{2}$ $\displaystyle F_y = 2y + 2\lambda = 0 \implies y = -\lambda = 2x$ $\displaystyle F_z = 2z + 3\lambda = 0 \implies z = -\frac{3\lambda}{2} = 3x$
  2. Substitute into $x + 2y + 3z = 14$: $\displaystyle x + 2(2x) + 3(3x) = 14 \implies 14x = 14 \implies x = 1, y = 2, z = 3$

Ques 36. Maximize $f(x,y)=x^{3}y^{2}$ subject to $x+y=10$ ($x,y>0$) using $F=x^{3}y^{2}+\lambda(x+y-10)$.

  • (A) $3456$ at $(6,4)$
  • (B) $2000$ at $(5,5)$
  • (C) $1728$ at $(4,6)$
  • (D) $512$ at $(8,2)$

Correct Answer: (A) $3456$ at $(6,4)$ Step-by-Step Solution:

  1. Set partial derivatives to zero: $\displaystyle F_x = 3x^2y^2 + \lambda = 0, \quad F_y = 2x^3y + \lambda = 0$
  2. Equate $-\lambda$: $3x^2y^2 = 2x^3y \implies 3y = 2x \implies y = \frac{2}{3}x$.
  3. Substitute into $x + y = 10 \implies x + \frac{2}{3}x = 10 \implies \frac{5}{3}x = 10 \implies x = 6, y = 4$.
  4. Evaluate $f(6,4) = 6^3 \cdot 4^2 = 216 \times 16 = 3456$.

Ques 37. Optimize $f(x,y)=5x+12y$ subject to $x^{2}+y^{2}=169$ using $F=5x+12y+\lambda(x^{2}+y^{2}-169)$. The maximum value is:

  • (A) $169$ at $(5, 12)$
  • (B) $130$ at $(5,12)$
  • (C) $225$ at $(9,12)$
  • (D) $65$ at $(5,0)$

Correct Answer: (A) $169$ at $(5, 12)$ Step-by-Step Solution:

  1. Set partial derivatives to zero: $\displaystyle F_x = 5 + 2\lambda x = 0 \implies x = -\frac{5}{2\lambda}$ $\displaystyle F_y = 12 + 2\lambda y = 0 \implies y = -\frac{12}{2\lambda}$
  2. Substitute into $x^2 + y^2 = 169$: $\displaystyle \frac{25 + 144}{4\lambda^2} = 169 \implies 4\lambda^2 = 1 \implies \lambda = \pm\frac{1}{2}$
  3. For $\lambda = -\frac{1}{2}$, $(x,y) = (5,12)$ and $f_{\max} = 5(5) + 12(12) = 25 + 144 = 169$.

Ques 38. In optimizing $f(x,y)=5x+12y$ subject to $x^{2}+y^{2}=169$, the multiplier $\lambda$ for maximum is:

  • (A) $-\frac{1}{2}$
  • (B) $\frac{1}{2}$
  • (C) $-1$
  • (D) $-\frac{1}{13}$

Correct Answer: (A) $-\frac{1}{2}$ Step-by-Step Solution:

  1. From Q37, the maximum occurs at $(x,y) = (5,12)$.
  2. Substitute $x = 5$ into $F_x = 5 + 2\lambda x = 0 \implies 5 + 10\lambda = 0 \implies \lambda = -\frac{1}{2}$.

Ques 39. Dimensions of an open rectangular box of volume $V=108$ having minimum surface area $S=xy+2xz+2yz$ using $F=xy+2xz+2yz+\lambda(xyz-108)$ are:

  • (A) $x=6, y=6, z=3$
  • (B) $x=3, y=3, z=12$
  • (C) $x=4, y=4, z=6.75$
  • (D) $x=9, y=4, z=3$

Correct Answer: (A) $x=6, y=6, z=3$ Step-by-Step Solution:

  1. Set partial derivatives to zero: $\displaystyle F_x = y + 2z + \lambda yz = 0 \implies -\lambda = \frac{1}{z} + \frac{2}{y}$ $\displaystyle F_y = x + 2z + \lambda xz = 0 \implies -\lambda = \frac{1}{z} + \frac{2}{x}$ $\displaystyle F_z = 2x + 2y + \lambda xy = 0 \implies -\lambda = \frac{2}{y} + \frac{2}{x}$
  2. Equating these yields $x = y = 2z$.
  3. Substitute into $xyz = 108 \implies (2z)(2z)z = 108 \implies 4z^3 = 108 \implies z^3 = 27 \implies z = 3$, and $x = y = 6$.

Ques 40. Minimum surface area of the open box with $V=108$ is:

  • (A) $108$
  • (B) $72$
  • (C) $144$
  • (D) $216$

Correct Answer: (A) $108$ Step-by-Step Solution:

  1. Substitute the optimal dimensions $x = 6, y = 6, z = 3$ from Q39 into $S = xy + 2xz + 2yz$: $\displaystyle S_{\min} = (6)(6) + 2(6)(3) + 2(6)(3) = 36 + 36 + 36 = 108$

Ques 41. Maximum value of $f(x,y,z)=x+2y+3z$ subject to $x^{2}+y^{2}+z^{2}=14$ using $F=(x+2y+3z)+\lambda(x^{2}+y^{2}+z^{2}-14)$ is:

  • (A) $14$
  • (B) $28$
  • (C) $\sqrt{14}$
  • (D) $42$

Correct Answer: (A) $14$ Step-by-Step Solution:

  1. Set partial derivatives to zero: $F_x = 1 + 2\lambda x = 0$, $F_y = 2 + 2\lambda y = 0$, $F_z = 3 + 2\lambda z = 0$.
  2. Thus $\frac{x}{1} = \frac{y}{2} = \frac{z}{3} = k \implies x = k, y = 2k, z = 3k$.
  3. Substitute into $x^2 + y^2 + z^2 = 14 \implies k^2(1 + 4 + 9) = 14 \implies k = \pm 1$.
  4. For $k = 1$, $(x,y,z) = (1,2,3)$ and $f_{\max} = 1(1) + 2(2) + 3(3) = 14$.

Ques 42. Find the point on $x^{2}+y^{2}+z^{2}=14$ furthest from origin along vector $(1,2,3)$.

  • (A) $(1,2,3)$
  • (B) $(-1,-2,-3)$
  • (C) $(2,1,3)$
  • (D) $(3,2,1)$

Correct Answer: (A) $(1,2,3)$ Step-by-Step Solution:

  1. From Q41, maximizing the projection along $(1,2,3)$ on $x^2 + y^2 + z^2 = 14$ gives $(x,y,z) = (k, 2k, 3k)$ with $k = 1$.
  2. Hence the point is $(1,2,3)$.

Ques 43. Determine the critical points of $f(x,y)=x^{3}-3x+y^{2}-4y+5$.

  • (A) $(1,2)$ and $(-1,2)$
  • (B) $(0,0)$ and $(1,1)$
  • (C) $(2,1)$ and $(-2,1)$
  • (D) $(3,4)$

Correct Answer: (A) $(1,2)$ and $(-1,2)$ Step-by-Step Solution:

  1. Set $f_x = 3x^2 - 3 = 0 \implies x^2 = 1 \implies x = 1, -1$.
  2. Set $f_y = 2y - 4 = 0 \implies y = 2$.
  3. Critical points are $(1,2)$ and $(-1,2)$.

Ques 44. Classify $(1,2)$ for $f(x,y)=x^{3}-3x+y^{2}-4y+5$.

  • (A) Local minimum
  • (B) Local maximum
  • (C) Saddle point
  • (D) Inconclusive

Correct Answer: (A) Local minimum Step-by-Step Solution:

  1. Compute second partial derivatives: $r = f_{xx} = 6x$, $s = f_{xy} = 0$, $t = f_{yy} = 2$.
  2. At $(1,2)$: $r = 6(1) = 6$, $s = 0$, $t = 2$.
  3. $rt - s^2 = (6)(2) - 0 = 12 > 0$ and $r = 6 > 0 \implies$ Local minimum.

Ques 45. Classify $(-1,2)$ for $f(x,y)=x^{3}-3x+y^{2}-4y+5$.

  • (A) Saddle point
  • (B) Local minimum
  • (C) Local maximum
  • (D) Test fails

Correct Answer: (A) Saddle point Step-by-Step Solution:

  1. Evaluate $r = 6x, s = 0, t = 2$ at $(-1,2)$: $r = -6$, $s = 0$, $t = 2$.
  2. $rt - s^2 = (-6)(2) - 0 = -12 < 0 \implies$ Saddle point.

Ques 46. For $F=(x^{2}+3y^{2}+2z^{2})+\lambda(x+y+z-11)$, the ratios $x:y:z$ are:

  • (A) $1:\frac{1}{3}:\frac{1}{2}$
  • (B) $1:3:2$
  • (C) $3:1:2$
  • (D) $1:1:1$

Correct Answer: (A) $1:\frac{1}{3}:\frac{1}{2}$ Step-by-Step Solution:

  1. Set partial derivatives to zero: $\displaystyle F_x = 2x + \lambda = 0 \implies x = -\frac{\lambda}{2}$ $\displaystyle F_y = 6y + \lambda = 0 \implies y = -\frac{\lambda}{6}$ $\displaystyle F_z = 4z + \lambda = 0 \implies z = -\frac{\lambda}{4}$
  2. Ratio $x : y : z = \frac{1}{2} : \frac{1}{6} : \frac{1}{4} = 1 : \frac{1}{3} : \frac{1}{2}$.

Ques 47. Value of $x$ for stationary point of $F=(x^{2}+3y^{2}+2z^{2})+\lambda(x+y+z-11)$ is:

  • (A) $6$
  • (B) $3$
  • (C) $2$
  • (D) $1$

Correct Answer: (A) $6$ Step-by-Step Solution:

  1. From Q46, $y = \frac{x}{3}$ and $z = \frac{x}{2}$.
  2. Substitute into the constraint $x + y + z = 11$: $\displaystyle x + \frac{x}{3} + \frac{x}{2} = 11 \implies \frac{11x}{6} = 11 \implies x = 6$

Ques 48. For $f(x,y)=e^{-(x^{2}+y^{2})}$, the origin $(0,0)$ is a:

  • (A) Absolute maximum with value $1$
  • (B) Absolute minimum with value $0$
  • (C) Saddle point
  • (D) Inconclusive

Correct Answer: (A) Absolute maximum with value $1$ Step-by-Step Solution:

  1. Since $x^2 + y^2 \ge 0$ for all $(x,y) \in \mathbb{R}^2$, we have $-(x^2 + y^2) \le 0$.
  2. Because $e^u$ is strictly increasing, $f(x,y) = e^{-(x^2+y^2)} \le e^0 = 1$, achieving its absolute maximum of $1$ at $(0,0)$.

Ques 49. Function $f(x,y)=2x^{2}-4xy+y^{2}$ at origin $(0,0)$ has:

  • (A) Saddle point $(rt-s^{2}=4-16=-12<0)$
  • (B) Local minimum
  • (C) Local maximum
  • (D) Inconclusive

Correct Answer: (A) Saddle point $(rt-s^{2}=4-16=-12<0)$ Step-by-Step Solution:

  1. Compute second partial derivatives: $r = f_{xx} = 4$, $s = f_{xy} = -4$, $t = f_{yy} = 2$.
  2. Since $rt - s^2 = (4)(2) - (-4)^2 = 8 - 16 = -8 < 0$, the Hessian determinant is negative, confirming a saddle point at $(0,0)$.

Ques 50. Maximum of $f(x,y)=6x+8y$ subject to $x^{2}+y^{2}=100$ is:

  • (A) $100$
  • (B) $140$
  • (C) $50$
  • (D) $200$

Correct Answer: (A) $100$ Step-by-Step Solution:

  1. Using $F = 6x + 8y + \lambda(x^2 + y^2 - 100)$: $\displaystyle F_x = 6 + 2\lambda x = 0, \quad F_y = 8 + 2\lambda y = 0 \implies \frac{x}{6} = \frac{y}{8} = k$
  2. Substitute $x = 6k, y = 8k$ into $x^2 + y^2 = 100$: $\displaystyle (36 + 64)k^2 = 100 \implies 100k^2 = 100 \implies k = \pm 1$
  3. At $(6,8)$, $f_{\max} = 6(6) + 8(8) = 36 + 64 = 100$.

Ques 51. Minimum distance from origin to $2x+y+2z=18$ using $F=(x^{2}+y^{2}+z^{2})+\lambda(2x+y+2z-18)$ is:

  • (A) $6$
  • (B) $18$
  • (C) $36$
  • (D) $3$

Correct Answer: (A) $6$ Step-by-Step Solution:

  1. Set partial derivatives of $F$ to zero: $\displaystyle F_x = 2x + 2\lambda = 0 \implies x = -\lambda$ $\displaystyle F_y = 2y + \lambda = 0 \implies y = -\frac{\lambda}{2}$ $\displaystyle F_z = 2z + 2\lambda = 0 \implies z = -\lambda$
  2. Substitute into $2x + y + 2z = 18$: $\displaystyle -2\lambda - \frac{\lambda}{2} - 2\lambda = 18 \implies -\frac{9\lambda}{2} = 18 \implies \lambda = -4$
  3. Point is $(4, 2, 4)$, so minimum distance $d = \sqrt{4^2 + 2^2 + 4^2} = \sqrt{36} = 6$.

Unit 2 (Section 2): Linear Algebra — Matrix Rank, Inverses, Normal Form & Linear Systems

Topic 2.1: Theoretical Properties of Rank, Normal Form & Consistency (Q1–Q22)

Ques 1. The rank of a matrix $A$ is $r$ if:

  • (A) At least one minor of order $r$ is non-zero, and all minors of order $r+1$ are zero
  • (B) All minors of order $r$ are non-zero
  • (C) The matrix is square of order $r$
  • (D) The determinant of order $r$ is zero

Correct Answer: (A) At least one minor of order $r$ is non-zero, and all minors of order $r+1$ are zero

Ques 2. If $A$ is an $m\times n$ matrix, then $\text{rank}(A)$ satisfies:

  • (A) $\text{rank}(A)\le \min(m,n)$
  • (B) $\text{rank}(A)\ge \max(m,n)$
  • (C) $\text{rank}(A)=m+n$
  • (D) $\text{rank}(A)=m\cdot n$

Correct Answer: (A) $\text{rank}(A)\le \min(m,n)$

Ques 3. Which elementary transformation alters the rank of a matrix?

  • (A) Interchanging two rows
  • (B) Multiplying a row by a non-zero scalar
  • (C) Adding a non-zero scalar multiple of one row to another
  • (D) None of these

Correct Answer: (D) None of these

Ques 4. If $A^{T}$ is the transpose of matrix $A$, then:

  • (A) $\text{rank}(A^{T})=\text{rank}(A)$
  • (B) $\text{rank}(A^{T})\ne \text{rank}(A)$
  • (C) $\text{rank}(A^{T})=\frac{1}{\text{rank}(A)}$
  • (D) $\text{rank}(A^{T})=0$

Correct Answer: (A) $\text{rank}(A^{T})=\text{rank}(A)$

Ques 5. For matrices $A$ and $B$ compatible for addition, subadditivity of rank states:

  • (A) $\text{rank}(A+B)\le \text{rank}(A)+\text{rank}(B)$
  • (B) $\text{rank}(A+B)=\text{rank}(A)+\text{rank}(B)$
  • (C) $\text{rank}(A+B)\ge \text{rank}(A)+\text{rank}(B)$
  • (D) $\text{rank}(A+B)=\text{rank}(A)\cdot \text{rank}(B)$

Correct Answer: (A) $\text{rank}(A+B)\le \text{rank}(A)+\text{rank}(B)$

Ques 6. Sylvester's Law of Nullity for matrices $A_{m\times n}$ and $B_{n\times p}$ states:

  • (A) $\text{rank}(AB)\ge \text{rank}(A)+\text{rank}(B)-n$
  • (B) $\text{rank}(AB)\le \text{rank}(A)+\text{rank}(B)-n$
  • (C) $\text{rank}(AB)=\text{rank}(A)+\text{rank}(B)$
  • (D) $\text{rank}(AB)=\text{rank}(A)-\text{rank}(B)$

Correct Answer: (A) $\text{rank}(AB)\ge \text{rank}(A)+\text{rank}(B)-n$

Ques 7. An $n\times n$ square matrix $A$ is non-singular if and only if:

  • (A) $\text{rank}(A)=n$
  • (B) $\text{rank}(A)<n$
  • (C) $\text{rank}(A)=0$
  • (D) $\det(A)=0$

Correct Answer: (A) $\text{rank}(A)=n$

Ques 8. Every non-singular matrix $A$ of order $n$ can be reduced by elementary operations to:

  • (A) $I_{n}$
  • (B) $0_{n}$
  • (C) Diagonal matrix with zeros on main diagonal
  • (D) Upper triangular with zero diagonal

Correct Answer: (A) $I_{n}$

Ques 9. The normal form of an $m\times n$ matrix of rank $r$ is:

  • (A) $\begin{bmatrix}I_{r}&0\\ 0&0\end{bmatrix}$
  • (B) $\begin{bmatrix}0&I_{r}\\ I_{r}&0\end{bmatrix}$
  • (C) $\begin{bmatrix}I_{r}&I_{r}\\ 0&0\end{bmatrix}$
  • (D) $\begin{bmatrix}0&0\\ 0&0\end{bmatrix}$

Correct Answer: (A) $\begin{bmatrix}I_{r}&0\\ 0&0\end{bmatrix}$

Ques 10. The homogeneous system $AX=0$ with $n$ unknowns always possesses:

  • (A) At least the trivial solution $X=0$
  • (B) No solution
  • (C) Strictly non-trivial solutions only
  • (D) An inconsistent state

Correct Answer: (A) At least the trivial solution $X=0$

Ques 11. The system $AX=0$ has a non-trivial solution if and only if:

  • (A) $\text{rank}(A)<n$
  • (B) $\text{rank}(A)=n$
  • (C) $\det(A)\ne0$
  • (D) $\text{rank}(A)>n$

Correct Answer: (A) $\text{rank}(A)<n$

Ques 12. The system $AX=B$ is consistent if and only if:

  • (A) $\text{rank}(A)=\text{rank}([A|B])$
  • (B) $\text{rank}(A)\ne \text{rank}([A|B])$
  • (C) $\text{rank}(A)>\text{rank}([A|B])$
  • (D) $\text{rank}([A|B])=n+1$

Correct Answer: (A) $\text{rank}(A)=\text{rank}([A|B])$

Ques 13. The non-homogeneous system $AX=B$ with $n$ unknowns has a unique solution if and only if:

  • (A) $\text{rank}(A)=\text{rank}([A|B])=n$
  • (B) $\text{rank}(A)=\text{rank}([A|B])<n$
  • (C) $\text{rank}(A)<\text{rank}([A|B])$
  • (D) $\text{rank}(A)=0$

Correct Answer: (A) $\text{rank}(A)=\text{rank}([A|B])=n$

Ques 14. The system $AX=B$ with $n$ unknowns has infinitely many solutions if:

  • (A) $\text{rank}(A)=\text{rank}([A|B])=r<n$
  • (B) $\text{rank}(A)=\text{rank}([A|B])=n$
  • (C) $\text{rank}(A)\ne \text{rank}([A|B])$
  • (D) $\text{rank}(A)=n$ and $\text{rank}([A|B])=n+1$

Correct Answer: (A) $\text{rank}(A)=\text{rank}([A|B])=r<n$

Ques 15. If $\text{rank}(A)\ne \text{rank}([A|B])$, then $AX=B$ is:

  • (A) Inconsistent (no solution)
  • (B) Unique solution
  • (C) Infinitely many solutions
  • (D) Trivial solution

Correct Answer: (A) Inconsistent (no solution)

Ques 16. If $A$ is a square matrix of order $n$ and $\text{rank}(A)=r<n$, then $\det(A)$ is:

  • (A) $0$
  • (B) $1$
  • (C) Non-zero
  • (D) $r$

Correct Answer: (A) $0$

Ques 17. Performing row operations on $[A|I_{n}]$ until $A$ becomes $I_{n}$ transforms the right block into:

  • (A) $A^{-1}$
  • (B) $A^{T}$
  • (C) $\text{adj}(A)$
  • (D) $I_{n}$

Correct Answer: (A) $A^{-1}$

Ques 18. If $P, Q$ are non-singular such that $PAQ=\begin{bmatrix}I_{r}&0\\ 0&0\end{bmatrix}$, then $\text{rank}(A)$ is:

  • (A) $r$
  • (B) Order of $P$
  • (C) Order of $Q$
  • (D) $0$

Correct Answer: (A) $r$

Ques 19. Multiplying a matrix by a non-singular matrix:

  • (A) Preserves matrix rank
  • (B) Reduces rank
  • (C) Increases rank
  • (D) Makes rank zero

Correct Answer: (A) Preserves matrix rank

Ques 20. The rank of a non-zero column vector of size $n\times1$ is:

  • (A) $1$
  • (B) $n$
  • (C) $0$
  • (D) $n-1$

Correct Answer: (A) $1$

Ques 21. The rank of a zero matrix $0_{m\times n}$ is:

  • (A) $0$
  • (B) $1$
  • (C) $\min(m,n)$
  • (D) Undefined

Correct Answer: (A) $0$

Ques 22. If $A_{3\times4}$ and $B_{4\times3}$ have $\text{rank}(A)=3$ and $\text{rank}(B)=3$, maximum rank of $AB$ is:

  • (A) $3$
  • (B) $4$
  • (C) $12$
  • (D) $1$

Correct Answer: (A) $3$ Step-by-Step Solution:

  1. Product $AB$ has size $3 \times 3$.
  2. By rank inequality, $\text{rank}(AB) \le \min(\text{rank}(A), \text{rank}(B)) = \min(3, 3) = 3$.

Topic 2.2: Computational Problems on Rank, Normal Form & Linear Systems (Q23–Q60)

Ques 23. Find the rank of $A=\begin{bmatrix}2&4&6\\ 3&6&9\\ 5&10&15\end{bmatrix}$.

  • (A) $1$
  • (B) $2$
  • (C) $3$
  • (D) $0$

Correct Answer: (A) $1$ Step-by-Step Solution:

  1. Factor out common multiples from each row: $R_1 = 2[1 \;\; 2 \;\; 3]$, $R_2 = 3[1 \;\; 2 \;\; 3]$, $R_3 = 5[1 \;\; 2 \;\; 3]$.
  2. Apply $R_2 \to 2R_2 - 3R_1$ and $R_3 \to 2R_3 - 5R_1$: $\displaystyle A \sim \begin{bmatrix}2 & 4 & 6 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$
  3. Number of non-zero rows $= 1 \implies \text{rank}(A) = 1$.

Ques 24. Find the rank of $A=\begin{bmatrix}3&0&0\\ 1&4&0\\ 2&5&6\end{bmatrix}$.

  • (A) $3$
  • (B) $2$
  • (C) $1$
  • (D) $0$

Correct Answer: (A) $3$ Step-by-Step Solution:

  1. $A$ is a lower-triangular matrix.
  2. Compute its determinant by multiplying the main diagonal entries: $\det(A) = 3 \times 4 \times 6 = 72 \ne 0$.
  3. Since the $3 \times 3$ minor is non-zero, $\text{rank}(A) = 3$.

Ques 25. Find the rank of $A=\begin{bmatrix}1&2&4\\ 2&4&8\\ 3&6&12\end{bmatrix}$.

  • (A) $1$
  • (B) $2$
  • (C) $3$
  • (D) $0$

Correct Answer: (A) $1$ Step-by-Step Solution:

  1. Apply row operations $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 3R_1$: $\displaystyle A \sim \begin{bmatrix}1 & 2 & 4 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$
  2. Only $1$ non-zero row remains $\implies \text{rank}(A) = 1$.

Ques 26. Find the rank of $A=\begin{bmatrix}1&2&3\\ 2&-1&4\\ 3&1&7\end{bmatrix}$.

  • (A) $2$
  • (B) $3$
  • (C) $1$
  • (D) $0$

Correct Answer: (A) $2$ Step-by-Step Solution:

  1. Apply $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 3R_1$: $\displaystyle A \sim \begin{bmatrix}1 & 2 & 3 \\ 0 & -5 & -2 \\ 0 & -5 & -2\end{bmatrix}$
  2. Apply $R_3 \to R_3 - R_2$: $\displaystyle A \sim \begin{bmatrix}1 & 2 & 3 \\ 0 & -5 & -2 \\ 0 & 0 & 0\end{bmatrix}$
  3. There are $2$ non-zero rows in echelon form $\implies \text{rank}(A) = 2$.

Ques 27. For what value of $k$ does $A=\begin{bmatrix}2&4&6\\ 1&k&3\\ 3&6&9\end{bmatrix}$ have rank $1$?

  • (A) $k=2$
  • (B) $k=4$
  • (C) $k=0$
  • (D) $k=1$

Correct Answer: (A) $k=2$ Step-by-Step Solution:

  1. For $\text{rank}(A) = 1$, all rows must be proportional to $R_1 = [2 \;\; 4 \;\; 6]$.
  2. Since $R_2 = [1 \;\; k \;\; 3] = \frac{1}{2}R_1$, we must have $k = \frac{4}{2} = 2$.

Ques 28. Determine the rank of $A=\begin{bmatrix}1&3&5&7\\ 2&6&10&14\\ 3&9&15&21\end{bmatrix}$.

  • (A) $1$
  • (B) $2$
  • (C) $3$
  • (D) $4$

Correct Answer: (A) $1$ Step-by-Step Solution:

  1. Apply $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 3R_1$: $\displaystyle A \sim \begin{bmatrix}1 & 3 & 5 & 7 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{bmatrix}$
  2. Only $1$ non-zero row remains $\implies \text{rank}(A) = 1$.

Ques 29. Find the rank of $A=\begin{bmatrix}0&1&2\\ 3&0&0\\ 0&0&4\end{bmatrix}$.

  • (A) $3$
  • (B) $2$
  • (C) $1$
  • (D) $0$

Correct Answer: (A) $3$ Step-by-Step Solution:

  1. Interchange $R_1 \leftrightarrow R_2$: $\displaystyle A \sim \begin{bmatrix}3 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{bmatrix}$
  2. This is in upper-triangular echelon form with $3$ non-zero pivots ($3, 1, 4$), so $\text{rank}(A) = 3$.

Ques 30. Find the rank of $A=\begin{bmatrix}1&1&2&3\\ 0&2&4&6\\ 0&0&0&0\end{bmatrix}$.

  • (A) $2$
  • (B) $3$
  • (C) $1$
  • (D) $4$

Correct Answer: (A) $2$ Step-by-Step Solution:

  1. The matrix is already in row echelon form.
  2. Counting the non-zero rows gives $\text{rank}(A) = 2$.

Ques 31. Reduce $A=\begin{bmatrix}2&0\\ 0&3\end{bmatrix}$ to normal form.

  • (A) $I_{2}$
  • (B) $\begin{bmatrix}1&0\\ 0&0\end{bmatrix}$
  • (C) $0_{2}$
  • (D) $I_{1}$

Correct Answer: (A) $I_{2}$ Step-by-Step Solution:

  1. Apply $R_1 \to \frac{1}{2}R_1$ and $R_2 \to \frac{1}{3}R_2$: $\displaystyle A \sim \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} = I_2$

Ques 32. Find non-singular $P, Q$ such that $PAQ=I_{2}$ for $A=\begin{bmatrix}2&1\\ 5&3\end{bmatrix}$. Rank of $A$ is:

  • (A) $2$
  • (B) $1$
  • (C) $0$
  • (D) $3$

Correct Answer: (A) $2$ Step-by-Step Solution:

  1. Compute $\det(A) = (2)(3) - (1)(5) = 6 - 5 = 1 \ne 0$.
  2. Since $A$ is a $2 \times 2$ non-singular matrix, $\text{rank}(A) = 2$.

Ques 33. Determine consistency: $x+y+z=4$, $2x+3y+4z=11$, $x+2y+4z=11$.

  • (A) Consistent with unique solution
  • (B) Consistent with infinite solutions
  • (C) Inconsistent
  • (D) Trivial solution only

Correct Answer: (A) Consistent with unique solution Step-by-Step Solution:

  1. Form the augmented matrix and apply $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - R_1$: $\displaystyle \begin{bmatrix}1 & 1 & 1 & \mid & 4 \\ 2 & 3 & 4 & \mid & 11 \\ 1 & 2 & 4 & \mid & 11\end{bmatrix} \sim \begin{bmatrix}1 & 1 & 1 & \mid & 4 \\ 0 & 1 & 2 & \mid & 3 \\ 0 & 1 & 3 & \mid & 7\end{bmatrix}$
  2. Apply $R_3 \to R_3 - R_2$: $\displaystyle \sim \begin{bmatrix}1 & 1 & 1 & \mid & 4 \\ 0 & 1 & 2 & \mid & 3 \\ 0 & 0 & 1 & \mid & 4\end{bmatrix}$
  3. Since $\text{rank}(A) = \text{rank}([A|B]) = 3 = n$, the system is consistent with a unique solution.

Ques 34. Solve for $z$ in: $x+y+z=4$, $2x+3y+4z=11$, $x+2y+4z=11$.

  • (A) $z=2$
  • (B) $z=4$
  • (C) $z=3$
  • (D) $z=0$

Correct Answer: (B) $z=4$ (Note: (A) $z=2$ if using the bank's uniform Option-A answer key) Step-by-Step Solution:

  1. From the row reduction in Q33:
    • Row 2 gives $y + 2z = 3$
    • Row 3 gives $y + 3z = 7$
  2. Subtracting Row 2 from Row 3 yields $z = 7 - 3 = 4$.

Ques 35. Find $y$ in: $x+y+z=4$, $2x+3y+4z=11$, $x+2y+4z=11$.

  • (A) $y=3$
  • (B) $y=1$
  • (C) $y=-5$
  • (D) $y=0$

Correct Answer: (C) $y=-5$ (Note: (A) $y=3$ if using the bank's uniform Option-A answer key) Step-by-Step Solution:

  1. Substitute $z = 4$ into Row 2 ($y + 2z = 3$): $\displaystyle y + 2(4) = 3 \implies y = 3 - 8 = -5$

Ques 36. Find $x$ in: $x+y+z=4$, $2x+3y+4z=11$, $x+2y+4z=11$.

  • (A) $x=-1$
  • (B) $x=1$
  • (C) $x=0$
  • (D) $x=5$

Correct Answer: (D) $x=5$ (Note: (A) $x=-1$ if using the bank's uniform Option-A answer key) Step-by-Step Solution:

  1. Substitute $y = -5$ and $z = 4$ into Equation 1 ($x + y + z = 4$): $\displaystyle x + (-5) + 4 = 4 \implies x - 1 = 4 \implies x = 5$

Ques 37. For what value of $a$ is $x+y+z=2$, $x+2y+3z=5$, $x+3y+az=9$ INCONSISTENT?

  • (A) $a=5$
  • (B) $a\ne5$
  • (C) $a=0$
  • (D) $a=3$

Correct Answer: (A) $a=5$ Step-by-Step Solution:

  1. Form $[A|B]$ and apply $R_2 \to R_2 - R_1$, $R_3 \to R_3 - R_1$: $\displaystyle \begin{bmatrix}1 & 1 & 1 & \mid & 2 \\ 0 & 1 & 2 & \mid & 3 \\ 0 & 2 & a-1 & \mid & 7\end{bmatrix}$
  2. Apply $R_3 \to R_3 - 2R_2$: $\displaystyle \begin{bmatrix}1 & 1 & 1 & \mid & 2 \\ 0 & 1 & 2 & \mid & 3 \\ 0 & 0 & a-5 & \mid & 1\end{bmatrix}$
  3. The system is inconsistent when $a - 5 = 0 \implies a = 5$ (which makes $\text{rank}(A) = 2 \ne \text{rank}([A|B]) = 3$).

Ques 38. System $x+y+z=6$, $x+2y+3z=10$, $2x+3y+4z=16$: find $\text{rank}(A)$ and $\text{rank}([A|B])$.

  • (A) $\text{rank}(A)=2, \text{rank}([A|B])=2$
  • (B) $\text{rank}(A)=3, \text{rank}([A|B])=3$
  • (C) $\text{rank}(A)=2, \text{rank}([A|B])=3$
  • (D) $\text{rank}(A) = 1, \text{rank}([A|B]) = 1$

Correct Answer: (A) $\text{rank}(A)=2, \text{rank}([A|B])=2$ Step-by-Step Solution:

  1. Notice that Equation 3 is the exact sum of Equation 1 and Equation 2: $(x+y+z) + (x+2y+3z) = 2x+3y+4z = 6+10 = 16$.
  2. Applying $R_3 \to R_3 - R_1 - R_2$ reduces the third row completely to zeros in both $A$ and $[A|B]$.
  3. Since the first two rows are non-proportional, $\text{rank}(A) = 2$ and $\text{rank}([A|B]) = 2$.

Ques 39. Determine $c$ such that $x+y+z=5$, $x+2y+3z=9$, $x+2y+cz=12$ is INCONSISTENT.

  • (A) $c=3$
  • (B) $c\ne3$
  • (C) $c=0$
  • (D) $c=1$

Correct Answer: (A) $c=3$ Step-by-Step Solution:

  1. Subtract Equation 2 from Equation 3 ($R_3 \to R_3 - R_2$): $\displaystyle (x + 2y + cz) - (x + 2y + 3z) = 12 - 9 \implies (c - 3)z = 3$
  2. If $c = 3$, this becomes $0z = 3$ (impossible), making the system inconsistent.

Ques 40. Determine $c$ such that system in above question has UNIQUE solution.

  • (A) $c\ne3$
  • (B) $c=3$
  • (C) $c=2$
  • (D) $c=0$

Correct Answer: (A) $c\ne3$ Step-by-Step Solution:

  1. From Q39, row reduction gives pivots $1, 1,$ and $(c - 3)$.
  2. For a unique solution, $\text{rank}(A) = 3$, which requires the third pivot to be non-zero: $c - 3 \ne 0 \implies c \ne 3$.

Ques 41. Find rank of homogeneous matrix $A=\begin{bmatrix}1&1&1\\ 2&2&2\\ 3&3&3\end{bmatrix}$.

  • (A) $1$
  • (B) $3$
  • (C) $2$
  • (D) $0$

Correct Answer: (A) $1$ Step-by-Step Solution:

  1. Apply $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 3R_1$: $\displaystyle A \sim \begin{bmatrix}1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$
  2. Number of non-zero rows is $1 \implies \text{rank}(A) = 1$.

Ques 42. Number of independent non-trivial solutions for $AX=0$ with $A=\begin{bmatrix}1&1&1\\ 2&2&2\\ 3&3&3\end{bmatrix}$ is:

  • (A) $3-1=2$
  • (B) $1$
  • (C) $0$
  • (D) $3$

Correct Answer: (A) $3-1=2$ Step-by-Step Solution:

  1. Number of unknowns $n = 3$, and from Q41, $r = \text{rank}(A) = 1$.
  2. Number of linearly independent solutions (nullity) $= n - r = 3 - 1 = 2$.

Ques 43. Find rank of $A=\begin{bmatrix}1&2&3\\ 2&4&7\\ 3&6&10\end{bmatrix}$.

  • (A) $2$
  • (B) $3$
  • (C) $1$
  • (D) $0$

Correct Answer: (A) $2$ Step-by-Step Solution:

  1. Apply $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 3R_1$: $\displaystyle A \sim \begin{bmatrix}1 & 2 & 3 \\ 0 & 0 & 1 \\ 0 & 0 & 1\end{bmatrix}$
  2. Apply $R_3 \to R_3 - R_2$: $\displaystyle A \sim \begin{bmatrix}1 & 2 & 3 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{bmatrix}$
  3. Number of non-zero rows $= 2 \implies \text{rank}(A) = 2$.

Ques 44. Find rank of $A=\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&0\end{bmatrix}$.

  • (A) $2$
  • (B) $3$
  • (C) $1$
  • (D) $0$

Correct Answer: (A) $2$ Step-by-Step Solution:

  1. The matrix is already in canonical normal form $\begin{bmatrix}I_2 & 0 \\ 0 & 0\end{bmatrix}$ with $2$ non-zero rows.
  2. Therefore, $\text{rank}(A) = 2$.

Ques 45. Find $k$ if $A=\begin{bmatrix}1&2\\ 3&k\end{bmatrix}$ has rank $1$.

  • (A) $k=6$
  • (B) $k=5$
  • (C) $k=0$
  • (D) $k=1$

Correct Answer: (A) $k=6$ Step-by-Step Solution:

  1. For a non-zero $2 \times 2$ matrix to have rank $1$, its determinant must be zero: $\displaystyle \det(A) = (1)(k) - (2)(3) = 0 \implies k - 6 = 0 \implies k = 6$

Ques 46. Find rank of $A=\begin{bmatrix}1&2&3\\ 0&0&0\\ 0&0&0\end{bmatrix}$.

  • (A) $1$
  • (B) $3$
  • (C) $2$
  • (D) $0$

Correct Answer: (A) $1$ Step-by-Step Solution:

  1. $A$ is in row echelon form with exactly $1$ non-zero row $\implies \text{rank}(A) = 1$.

Ques 47. Find rank of $A=\begin{bmatrix}1&0\\ 0&1\\ 1&1\end{bmatrix}$.

  • (A) $2$
  • (B) $3$
  • (C) $1$
  • (D) $0$

Correct Answer: (A) $2$ Step-by-Step Solution:

  1. Apply $R_3 \to R_3 - R_1 - R_2$: $\displaystyle A \sim \begin{bmatrix}1 & 0 \\ 0 & 1 \\ 0 & 0\end{bmatrix}$
  2. There are $2$ non-zero rows $\implies \text{rank}(A) = 2$.

Ques 48. System $2x+3y=5$, $4x+6y=10$ has:

  • (A) Infinitely many solutions
  • (B) Unique solution
  • (C) No solution
  • (D) Trivial solution only

Correct Answer: (A) Infinitely many solutions Step-by-Step Solution:

  1. Check the ratio of coefficients: $\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}$, $\frac{c_1}{c_2} = \frac{5}{10} = \frac{1}{2}$.
  2. Since $\text{rank}(A) = \text{rank}([A|B]) = 1 < 2$, the system has infinitely many solutions.

Ques 49. System $2x+3y=5$, $4x+6y=12$ has:

  • (A) No solution
  • (B) Unique solution
  • (C) Infinitely many solutions
  • (D) $x=0, y=0$

Correct Answer: (A) No solution Step-by-Step Solution:

  1. Check the ratio of coefficients: $\frac{2}{4} = \frac{3}{6} \ne \frac{5}{12}$.
  2. Since $\text{rank}(A) = 1 \ne \text{rank}([A|B]) = 2$, the system is inconsistent (no solution).

Ques 50. System $x+2y=3$, $3x+y=4$ has:

  • (A) Unique solution
  • (B) No solution
  • (C) Infinite solutions
  • (D) Inconsistent

Correct Answer: (A) Unique solution Step-by-Step Solution:

  1. Compute determinant of coefficient matrix: $\det\begin{bmatrix}1 & 2 \\ 3 & 1\end{bmatrix} = 1 - 6 = -5 \ne 0$.
  2. Since $\text{rank}(A) = 2 = n$, the system has a unique solution.

Ques 51. Rank of identity matrix $I_{4}$ is:

  • (A) $4$
  • (B) $1$
  • (C) $0$
  • (D) $2$

Correct Answer: (A) $4$

Ques 52. Rank of diagonal matrix $\text{diag}(5,0,3,0)$ is:

  • (A) $2$
  • (B) $4$
  • (C) $1$
  • (D) $3$

Correct Answer: (A) $2$ Step-by-Step Solution:

  1. Count the non-zero diagonal elements in $\text{diag}(5, 0, 3, 0)$: there are two ($5$ and $3$).
  2. Hence $\text{rank} = 2$.

Ques 53. If $\text{rank}(A_{3\times3})=3$, then system $AX=0$ has:

  • (A) Trivial solution $X=0$ only
  • (B) Infinitely many solutions
  • (C) No solution
  • (D) $2$ independent parameters

Correct Answer: (A) Trivial solution $X=0$ only

Ques 54. Normal form of $A=\begin{bmatrix}0&2\\ 3&0\end{bmatrix}$ is:

  • (A) $I_{2}$
  • (B) $\begin{bmatrix}1&0\\ 0&0\end{bmatrix}$
  • (C) $O_{2}$
  • (D) $I_{1}$

Correct Answer: (A) $I_{2}$ Step-by-Step Solution:

  1. Swap $R_1 \leftrightarrow R_2$ to get $\begin{bmatrix}3 & 0 \\ 0 & 2\end{bmatrix}$, then divide $R_1$ by $3$ and $R_2$ by $2$ to get $\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} = I_2$.

Ques 55. For $A=\begin{bmatrix}1&2&3\\ 2&4&6\end{bmatrix}$, rank is:

  • (A) $1$
  • (B) $2$
  • (C) $3$
  • (D) $0$

Correct Answer: (A) $1$ Step-by-Step Solution:

  1. Apply $R_2 \to R_2 - 2R_1$: $\begin{bmatrix}1 & 2 & 3 \\ 0 & 0 & 0\end{bmatrix} \implies \text{rank}(A) = 1$.

Ques 56. If $A$ is $3\times3$ with $\det(A)=5$, rank of $A$ is:

  • (A) $3$
  • (B) $2$
  • (C) $1$
  • (D) $0$

Correct Answer: (A) $3$

Ques 57. If $A$ is $3\times3$ with $\det(A)=0$ and at least one $2\times2$ minor non-zero, rank is:

  • (A) $2$
  • (B) $3$
  • (C) $1$
  • (D) $0$

Correct Answer: (A) $2$

Ques 58. System $x+y=2$, $x-y=0$ has solution:

  • (A) $x=1, y=1$
  • (B) $x=2, y=0$
  • (C) $x=0, y=0$
  • (D) No solution

Correct Answer: (A) $x=1, y=1$ Step-by-Step Solution:

  1. Add the two equations: $(x + y) + (x - y) = 2 + 0 \implies 2x = 2 \implies x = 1$.
  2. From $x - y = 0 \implies y = x = 1$.

Ques 59. System $x+y=2$, $2x+2y=5$ is:

  • (A) Inconsistent
  • (B) Consistent with unique solution
  • (C) Consistent with infinite solutions
  • (D) Homogeneous

Correct Answer: (A) Inconsistent Step-by-Step Solution:

  1. Apply $R_2 \to R_2 - 2R_1$: $(2x + 2y) - 2(x + y) = 5 - 2(2) \implies 0 = 1$, which is inconsistent.

Ques 60. Rank of $A=\begin{bmatrix}1&0&0\\ 0&0&0\\ 0&0&0\end{bmatrix}$ is:

  • (A) $1$
  • (B) $0$
  • (C) $3$
  • (D) $2$

Correct Answer: (A) $1$

Unit 3 (Section 3): Linear Algebra — Eigenvalues, Eigenvectors & Cayley-Hamilton Theorem

Topic 3.1: Theoretical Properties of Eigenvalues & Cayley-Hamilton Theorem (Q61–Q74)

Ques 61. The characteristic equation of a square matrix $A$ is defined by:

  • (A) $\det(A-\lambda I)=0$
  • (B) $\det(A+\lambda I)=1$
  • (C) $(A-\lambda I)X=1$
  • (D) $\text{trace}(A-\lambda I)=0$

Correct Answer: (A) $\det(A-\lambda I)=0$

Ques 62. A scalar $\lambda$ is an eigenvalue of $A$ if and only if there exists a non-zero vector $X$ such that:

  • (A) $AX=\lambda X$
  • (B) $AX=\lambda I$
  • (C) $A+X=\lambda X$
  • (D) $AX=X$

Correct Answer: (A) $AX=\lambda X$

Ques 63. The sum of the eigenvalues of a square matrix $A$ equals:

  • (A) The trace of $A$ (sum of main diagonal elements)
  • (B) The determinant of $A$
  • (C) The rank of $A$
  • (D) The nullity of $A$

Correct Answer: (A) The trace of $A$ (sum of main diagonal elements)

Ques 64. The product of the eigenvalues of a square matrix $A$ equals:

  • (A) $\det(A)$
  • (B) $\text{trace}(A)$
  • (C) $\text{rank}(A)$
  • (D) $0$

Correct Answer: (A) $\det(A)$

Ques 65. If $\lambda$ is an eigenvalue of an invertible matrix $A$, then an eigenvalue of $A^{-1}$ is:

  • (A) $\frac{1}{\lambda}$
  • (B) $-\lambda$
  • (C) $\lambda^{2}$
  • (D) $\lambda$

Correct Answer: (A) $\frac{1}{\lambda}$

Ques 66. If $\lambda$ is an eigenvalue of $A$, then an eigenvalue of $A^{k}$ $(k\in\mathbb{N})$ is:

  • (A) $\lambda^{k}$
  • (B) $k\lambda$
  • (C) $\frac{\lambda}{k}$
  • (D) $\lambda+k$

Correct Answer: (A) $\lambda^{k}$

Ques 67. The matrices $A$ and $A^{T}$ have:

  • (A) Exactly the same eigenvalues
  • (B) Reciprocal eigenvalues
  • (C) Negative eigenvalues
  • (D) Completely different eigenvalues

Correct Answer: (A) Exactly the same eigenvalues

Ques 68. Eigenvectors corresponding to distinct eigenvalues of a real symmetric matrix are:

  • (A) Orthogonal
  • (B) Linearly dependent
  • (C) Equal
  • (D) Parallel

Correct Answer: (A) Orthogonal

Ques 69. A matrix $A$ has at least one eigenvalue equal to zero if and only if:

  • (A) $A$ is singular $(\det(A)=0)$
  • (B) $A$ is non-singular
  • (C) $A$ is identity matrix
  • (D) $\text{trace}(A)=0$

Correct Answer: (A) $A$ is singular $(\det(A)=0)$

Ques 70. The Cayley-Hamilton Theorem states that every square matrix:

  • (A) Satisfies its own characteristic equation
  • (B) Is diagonalizable
  • (C) Has distinct eigenvalues
  • (D) Is symmetric

Correct Answer: (A) Satisfies its own characteristic equation

Ques 71. If the characteristic polynomial of $A_{3\times3}$ is $\lambda^{3}-c_{1}\lambda^{2}+c_{2}\lambda-c_{3}=0$, by CHT:

  • (A) $A^{3}-c_{1}A^{2}+c_{2}A-c_{3}I=0$
  • (B) $A^{3}+c_{1}A^{2}+c_{2}A+c_{3}I=0$
  • (C) $A^{3}-c_{1}A^{2}+c_{2}A-c_{3}=0$
  • (D) $A^{2}-c_{1}A+c_{2}I=0$

Correct Answer: (A) $A^{3}-c_{1}A^{2}+c_{2}A-c_{3}I=0$

Ques 72. A matrix $A_{n\times n}$ is diagonalizable if and only if:

  • (A) It has $n$ linearly independent eigenvectors
  • (B) $\det(A)=0$
  • (C) All eigenvalues are zero
  • (D) It is upper triangular

Correct Answer: (A) It has $n$ linearly independent eigenvectors

Ques 73. Eigenvalues of a triangular (upper or lower) or diagonal matrix are simply:

  • (A) Its main diagonal elements
  • (B) All zeros
  • (C) All ones
  • (D) Sum of row elements

Correct Answer: (A) Its main diagonal elements

Ques 74. If $\lambda$ is an eigenvalue of $A$ with eigenvector $X$, then an eigenvalue of $A^{2}+3A-2I$ is:

  • (A) $\lambda^{2}+3\lambda-2$
  • (B) $2\lambda+3$
  • (C) $\lambda^{2}+3$
  • (D) $\frac{\lambda^{2}+3\lambda}{2}$

Correct Answer: (A) $\lambda^{2}+3\lambda-2$

Topic 3.2: Computational Problems on Eigenvalues, Eigenvectors & CHT (Q75–Q111)

Ques 75. Find the eigenvalues of $A=\begin{bmatrix}3&0\\ 0&7\end{bmatrix}$.

  • (A) $3,7$
  • (B) $0,10$
  • (C) $21,1$
  • (D) $3,3$

Correct Answer: (A) $3,7$ Step-by-Step Solution:

  1. Since $A$ is a diagonal matrix, its eigenvalues are its main diagonal elements: $\lambda_1 = 3, \lambda_2 = 7$.

Ques 76. Find the eigenvalues of $A=\begin{bmatrix}4&1\\ 0&5\end{bmatrix}$.

  • (A) $4,5$
  • (B) $1,9$
  • (C) $20,0$
  • (D) $4,4$

Correct Answer: (A) $4,5$ Step-by-Step Solution:

  1. Since $A$ is an upper-triangular matrix, its eigenvalues are the main diagonal entries: $\lambda_1 = 4, \lambda_2 = 5$.

Ques 77. Find the trace of $A=\begin{bmatrix}2&3&5\\ 1&4&2\\ 3&1&6\end{bmatrix}$.

  • (A) $12$
  • (B) $10$
  • (C) $8$
  • (D) $15$

Correct Answer: (A) $12$ Step-by-Step Solution:

  1. Sum the main diagonal entries: $\text{trace}(A) = a_{11} + a_{22} + a_{33} = 2 + 4 + 6 = 12$.

Ques 78. The sum of eigenvalues of $A=\begin{bmatrix}2&3&5\\ 1&4&2\\ 3&1&6\end{bmatrix}$ is:

  • (A) $12$
  • (B) $24$
  • (C) $6$
  • (D) $0$

Correct Answer: (A) $12$ Step-by-Step Solution:

  1. Sum of eigenvalues $= \text{trace}(A) = 2 + 4 + 6 = 12$.

Ques 79. Find the product of eigenvalues of $A=\begin{bmatrix}2&0\\ 1&3\end{bmatrix}$.

  • (A) $6$
  • (B) $5$
  • (C) $2$
  • (D) $3$

Correct Answer: (A) $6$ Step-by-Step Solution:

  1. Product of eigenvalues $= \det(A) = (2)(3) - (0)(1) = 6$.

Ques 80. Find the characteristic equation of $A=\begin{bmatrix}1&2\\ 3&4\end{bmatrix}$.

  • (A) $\lambda^{2}-5\lambda-2=0$
  • (B) $\lambda^{2}-5\lambda+2=0$
  • (C) $\lambda^{2}+5\lambda-2=0$
  • (D) $\lambda^{2}-4\lambda-1=0$

Correct Answer: (A) $\lambda^{2}-5\lambda-2=0$ Step-by-Step Solution:

  1. Compute $\text{trace}(A) = 1 + 4 = 5$ and $\det(A) = (1)(4) - (2)(3) = 4 - 6 = -2$.
  2. Substitute into $\lambda^2 - \text{trace}(A)\lambda + \det(A) = 0$: $\displaystyle \lambda^2 - 5\lambda - 2 = 0$

Ques 81. Find the eigenvalues of $A=\begin{bmatrix}0&1\\ -2&-3\end{bmatrix}$.

  • (A) $-1,-2$
  • (B) $1,2$
  • (C) $0,-3$
  • (D) $3,-1$

Correct Answer: (A) $-1,-2$ Step-by-Step Solution:

  1. $\text{trace}(A) = 0 + (-3) = -3$ and $\det(A) = (0)(-3) - (1)(-2) = 2$.
  2. Characteristic equation: $\lambda^2 + 3\lambda + 2 = 0 \implies (\lambda + 1)(\lambda + 2) = 0 \implies \lambda = -1, -2$.

Ques 82. If eigenvalues of $A$ are $2, 3$, find eigenvalues of $A^{2}$.

  • (A) $4,9$
  • (B) $2,3$
  • (C) $5,6$
  • (D) $8,27$

Correct Answer: (A) $4,9$ Step-by-Step Solution:

  1. Square each eigenvalue of $A$: $\lambda_1^2 = 2^2 = 4$ and $\lambda_2^2 = 3^2 = 9$.

Ques 83. If eigenvalues of $A$ are $1, 4$, find eigenvalues of $A^{-1}$.

  • (A) $1, \frac{1}{4}$
  • (B) $-1,-4$
  • (C) $1,16$
  • (D) $4,1$

Correct Answer: (A) $1, \frac{1}{4}$ Step-by-Step Solution:

  1. Take the reciprocal of each eigenvalue of $A$: $\frac{1}{1} = 1$ and $\frac{1}{4}$.

Ques 84. If eigenvalues of $A$ are $2,5$, find eigenvalues of $3A$.

  • (A) $6,15$
  • (B) $2,5$
  • (C) $5,8$
  • (D) $6,5$

Correct Answer: (A) $6,15$ Step-by-Step Solution:

  1. Multiply each eigenvalue of $A$ by $3$: $3 \times 2 = 6$ and $3 \times 5 = 15$.

Ques 85. If eigenvalues of $A$ are $1, 2, 3$, find $\det(A)$.

  • (A) $6$
  • (B) $1$
  • (C) $5$
  • (D) $0$

Correct Answer: (A) $6$ Step-by-Step Solution:

  1. $\det(A) = \lambda_1 \lambda_2 \lambda_3 = 1 \times 2 \times 3 = 6$.

Ques 86. If eigenvalues of $A$ are $1, 2, 3$, find $\text{trace}(A)$.

  • (A) $6$
  • (B) $5$
  • (C) $11$
  • (D) $1$

Correct Answer: (A) $6$ Step-by-Step Solution:

  1. $\text{trace}(A) = \lambda_1 + \lambda_2 + \lambda_3 = 1 + 2 + 3 = 6$.

Ques 87. If $A=\begin{bmatrix}1&4\\ 2&3\end{bmatrix}$, find its eigenvalues.

  • (A) $5,-1$
  • (B) $1,3$
  • (C) $2,4$
  • (D) $4,-2$

Correct Answer: (A) $5,-1$ Step-by-Step Solution:

  1. $\text{trace}(A) = 1 + 3 = 4$ and $\det(A) = (1)(3) - (4)(2) = -5$.
  2. Characteristic equation: $\lambda^2 - 4\lambda - 5 = 0 \implies (\lambda - 5)(\lambda + 1) = 0 \implies \lambda = 5, -1$.

Ques 88. An eigenvector of $A=\begin{bmatrix}1&4\\ 2&3\end{bmatrix}$ corresponding to $\lambda=5$ is:

  • (A) $\begin{bmatrix}1\\ 1\end{bmatrix}$
  • (B) $\begin{bmatrix}1\\ -1\end{bmatrix}$
  • (C) $\begin{bmatrix}2\\ 1\end{bmatrix}$
  • (D) $\begin{bmatrix}0\\ 1\end{bmatrix}$

Correct Answer: (A) $\begin{bmatrix}1\\ 1\end{bmatrix}$ Step-by-Step Solution:

  1. Set up $(A - 5I)X = 0$: $\displaystyle \begin{bmatrix}1-5 & 4 \\ 2 & 3-5\end{bmatrix}\begin{bmatrix}x_1 \\ x_2\end{bmatrix} = \begin{bmatrix}-4 & 4 \\ 2 & -2\end{bmatrix}\begin{bmatrix}x_1 \\ x_2\end{bmatrix} = \begin{bmatrix}0 \\ 0\end{bmatrix}$
  2. Both rows give $-x_1 + x_2 = 0 \implies x_1 = x_2$. Choosing $x_1 = 1$ gives $X = \begin{bmatrix}1 \\ 1\end{bmatrix}$.

Ques 89. An eigenvector of $A=\begin{bmatrix}1&4\\ 2&3\end{bmatrix}$ corresponding to $\lambda=-1$ is:

  • (A) $\begin{bmatrix}2\\ -1\end{bmatrix}$
  • (B) $\begin{bmatrix}1\\ 1\end{bmatrix}$
  • (C) $\begin{bmatrix}1\\ 2\end{bmatrix}$
  • (D) $\begin{bmatrix}-1\\ 2\end{bmatrix}$

Correct Answer: (A) $\begin{bmatrix}2\\ -1\end{bmatrix}$ Step-by-Step Solution:

  1. Set up $(A - (-1)I)X = (A + I)X = 0$: $\displaystyle \begin{bmatrix}1+1 & 4 \\ 2 & 3+1\end{bmatrix}\begin{bmatrix}x_1 \\ x_2\end{bmatrix} = \begin{bmatrix}2 & 4 \\ 2 & 4\end{bmatrix}\begin{bmatrix}x_1 \\ x_2\end{bmatrix} = \begin{bmatrix}0 \\ 0\end{bmatrix}$
  2. Thus $2x_1 + 4x_2 = 0 \implies x_1 = -2x_2$. Choosing $x_2 = -1$ gives $x_1 = 2$, so $X = \begin{bmatrix}2 \\ -1\end{bmatrix}$.

Ques 90. If $A^{2}-4A+3I=0$, express $A^{-1}$ using CHT.

  • (A) $A^{-1}=\frac{1}{3}(4I-A)$
  • (B) $A^{-1}=4I-A$
  • (C) $A^{-1}=3I-4A$
  • (D) $A^{-1}=\frac{1}{4}(A-3I)$

Correct Answer: (A) $A^{-1}=\frac{1}{3}(4I-A)$ Step-by-Step Solution:

  1. Isolate the identity term: $3I = 4A - A^2$.
  2. Multiply both sides by $A^{-1}$: $3A^{-1} = 4I - A \implies A^{-1} = \frac{1}{3}(4I - A)$.

Ques 91. If $A^{2}-5A+6I=0$, evaluate $A^{3}$.

  • (A) $19A-30I$
  • (B) $5A-6I$
  • (C) $25A-36I$
  • (D) $30A-19I$

Correct Answer: (A) $19A-30I$ Step-by-Step Solution:

  1. Rewrite the given equation as $A^2 = 5A - 6I$.
  2. Multiply both sides by $A$: $A^3 = 5A^2 - 6A$.
  3. Substitute $A^2 = 5A - 6I$: $\displaystyle A^3 = 5(5A - 6I) - 6A = 25A - 30I - 6A = 19A - 30I$

Ques 92. Characteristic equation of $A=\begin{bmatrix}2&1&1\\ 0&1&0\\ 1&1&2\end{bmatrix}$ is:

  • (A) $(\lambda-1)^{2}(\lambda-3)=0$
  • (B) $(\lambda-2)^{3}=0$
  • (C) $\lambda^{3}-5\lambda^{2}+7\lambda-3=0$
  • (D) Both A and C

Correct Answer: (D) Both A and C Step-by-Step Solution:

  1. Expand $\det(A - \lambda I) = 0$ along the second row: $\displaystyle \begin{vmatrix}2-\lambda & 1 & 1 \\ 0 & 1-\lambda & 0 \\ 1 & 1 & 2-\lambda\end{vmatrix} = (1-\lambda)[(2-\lambda)^2 - 1] = 0$
  2. Simplify: $(1-\lambda)(\lambda^2 - 4\lambda + 3) = (1-\lambda)(\lambda-1)(\lambda-3) = 0 \implies (\lambda-1)^2(\lambda-3) = 0$.
  3. Expanding $(\lambda^2 - 2\lambda + 1)(\lambda - 3) = 0$ gives $\lambda^3 - 5\lambda^2 + 7\lambda - 3 = 0$.

Ques 93. Eigenvalues of matrix in Question mentioned above are:

  • (A) $1,1,3$
  • (B) $2,2,2$
  • (C) $1,2,3$
  • (D) $0,1,4$

Correct Answer: (A) $1,1,3$ Step-by-Step Solution:

  1. From Q92, the factored characteristic equation is $(\lambda - 1)^2(\lambda - 3) = 0$.
  2. Roots are $\lambda = 1, 1, 3$.

Ques 94. If eigenvalues of $A_{3\times3}$ are $1, 2, 4$, characteristic equation is:

  • (A) $\lambda^{3}-7\lambda^{2}+14\lambda-8=0$
  • (B) $\lambda^{3}-6\lambda^{2}+11\lambda-6=0$
  • (C) $\lambda^{3}-8\lambda^{2}+14\lambda-7=0$
  • (D) $\lambda^{3}-7\lambda+8=0$

Correct Answer: (A) $\lambda^{3}-7\lambda^{2}+14\lambda-8=0$ Step-by-Step Solution:

  1. Form $(\lambda - 1)(\lambda - 2)(\lambda - 4) = 0$.
  2. Compute symmetric sums of roots:
    • $S_1 = 1 + 2 + 4 = 7$
    • $S_2 = (1)(2) + (2)(4) + (4)(1) = 2 + 8 + 4 = 14$
    • $S_3 = (1)(2)(4) = 8$
  3. Characteristic equation: $\lambda^3 - 7\lambda^2 + 14\lambda - 8 = 0$.

Ques 95. By CHT, for matrix in Question mentioned above, $A^{3}=$

  • (A) $7A^{2}-14A+8I$
  • (B) $6A^{2}-11A+6I$
  • (C) $8A^{2}-14A+7I$
  • (D) $7A-8I$

Correct Answer: (A) $7A^{2}-14A+8I$ Step-by-Step Solution:

  1. Substitute $A$ into $\lambda^3 - 7\lambda^2 + 14\lambda - 8 = 0$ by CHT: $\displaystyle A^3 - 7A^2 + 14A - 8I = 0 \implies A^3 = 7A^2 - 14A + 8I$

Ques 96. Inverse of matrix with characteristic equation $\lambda^{3}-7\lambda^{2}+14\lambda-8=0$ is:

  • (A) $\frac{1}{8}(A^{2}-7A+14I)$
  • (B) $\frac{1}{7}(A^{2}-8A+14I)$
  • (C) $A^{2}-7A+14I$
  • (D) $\frac{1}{14}(A^{2}-7A+8I)$

Correct Answer: (A) $\frac{1}{8}(A^{2}-7A+14I)$ Step-by-Step Solution:

  1. By CHT, $A^3 - 7A^2 + 14A = 8I$.
  2. Multiply both sides by $A^{-1}$: $\displaystyle A^2 - 7A + 14I = 8A^{-1} \implies A^{-1} = \frac{1}{8}(A^2 - 7A + 14I)$

Ques 97. If $A=\begin{bmatrix}1&0&0\\ 0&2&0\\ 0&0&3\end{bmatrix}$, eigenvalues of $A^{3}-2A+I$ are:

  • (A) $0,5,22$
  • (B) $1,8,27$
  • (C) $1,4,9$
  • (D) $0,2,6$

Correct Answer: (A) $0,5,22$ Step-by-Step Solution:

  1. The eigenvalues of diagonal matrix $A$ are $\lambda = 1, 2, 3$.
  2. Evaluate $P(\lambda) = \lambda^3 - 2\lambda + 1$ for each $\lambda$:
    • For $\lambda = 1$: $1^3 - 2(1) + 1 = 0$
    • For $\lambda = 2$: $2^3 - 2(2) + 1 = 5$
    • For $\lambda = 3$: $3^3 - 2(3) + 1 = 22$

Ques 98. If $\det(A-\lambda I)=-\lambda^{3}+6\lambda^{2}-11\lambda+6$, product of eigenvalues is:

  • (A) $6$
  • (B) $-6$
  • (C) $11$
  • (D) $1$

Correct Answer: (A) $6$ Step-by-Step Solution:

  1. Product of eigenvalues $= \det(A)$.
  2. Set $\lambda = 0$ in $\det(A - \lambda I)$: $\det(A) = -0^3 + 6(0)^2 - 11(0) + 6 = 6$.

Ques 99. If $A=\begin{bmatrix}3&4\\ 4&-3\end{bmatrix}$, trace of $A$ is:

  • (A) $0$
  • (B) $6$
  • (C) $-6$
  • (D) $25$

Correct Answer: (A) $0$ Step-by-Step Solution:

  1. $\text{trace}(A) = 3 + (-3) = 0$.

Ques 100. Determinant of $A=\begin{bmatrix}3&4\\ 4&-3\end{bmatrix}$ is:

  • (A) $-25$
  • (B) $25$
  • (C) $0$
  • (D) $-7$

Correct Answer: (A) $-25$ Step-by-Step Solution:

  1. $\det(A) = (3)(-3) - (4)(4) = -9 - 16 = -25$.

Ques 101. Eigenvalues of $A=\begin{bmatrix}3&4\\ 4&-3\end{bmatrix}$ are:

  • (A) $5,-5$
  • (B) $3,-3$
  • (C) $25,-25$
  • (D) $0,5$

Correct Answer: (A) $5,-5$ Step-by-Step Solution:

  1. Using $\text{trace}(A) = 0$ and $\det(A) = -25$ from Q99–Q100:
  2. Characteristic equation: $\lambda^2 - 0\lambda - 25 = 0 \implies \lambda^2 = 25 \implies \lambda = 5, -5$.

Ques 102. Eigenvalues of $A=\begin{bmatrix}0&1\\ 1&0\end{bmatrix}$ are:

  • (A) $1,-1$
  • (B) $1,1$
  • (C) $0,0$
  • (D) $2,-2$

Correct Answer: (A) $1,-1$ Step-by-Step Solution:

  1. $\det(A - \lambda I) = \begin{vmatrix}-\lambda & 1 \\ 1 & -\lambda\end{vmatrix} = \lambda^2 - 1 = 0 \implies \lambda = 1, -1$.

Ques 103. Eigenvalues of $A=\begin{bmatrix}0&-1\\ 1&0\end{bmatrix}$ are:

  • (A) $i,-i$
  • (B) $1,-1$
  • (C) $0,0$
  • (D) $1,1$

Correct Answer: (A) $i,-i$ Step-by-Step Solution:

  1. $\det(A - \lambda I) = \begin{vmatrix}-\lambda & -1 \\ 1 & -\lambda\end{vmatrix} = \lambda^2 + 1 = 0 \implies \lambda = i, -i$.

Ques 104. If $A$ is $2\times2$ with $\text{trace}(A)=8$ and $\det(A)=12$, its eigenvalues are:

  • (A) $2,6$
  • (B) $3,5$
  • (C) $4,4$
  • (D) $1,12$

Correct Answer: (A) $2,6$ Step-by-Step Solution:

  1. Characteristic equation: $\lambda^2 - \text{trace}(A)\lambda + \det(A) = \lambda^2 - 8\lambda + 12 = 0$.
  2. Factoring gives $(\lambda - 2)(\lambda - 6) = 0 \implies \lambda = 2, 6$.

Ques 105. If $A$ is $2\times2$ with $\text{trace}(A)=5$ and $\det(A)=6$, its eigenvalues are:

  • (A) $2,3$
  • (B) $1,6$
  • (C) $-1,-6$
  • (D) $0,5$

Correct Answer: (A) $2,3$ Step-by-Step Solution:

  1. Characteristic equation: $\lambda^2 - 5\lambda + 6 = 0 \implies (\lambda - 2)(\lambda - 3) = 0 \implies \lambda = 2, 3$.

Ques 106. If eigenvalues of $A$ are $0, 2, 5$, then rank of $A$ is at most:

  • (A) $2$
  • (B) $3$
  • (C) $1$
  • (D) $0$

Correct Answer: (A) $2$ Step-by-Step Solution:

  1. Since $\lambda = 0$ is an eigenvalue of $A_{3 \times 3}$, $\det(A) = 0 \cdot 2 \cdot 5 = 0$.
  2. With $1$ zero eigenvalue and $2$ non-zero distinct eigenvalues, $\text{rank}(A) = 3 - 1 = 2$.

Ques 107. Eigenvalues of $I_{3}$ are:

  • (A) $1,1,1$
  • (B) $3,0,0$
  • (C) $0,0,0$
  • (D) $1,2,3$

Correct Answer: (A) $1,1,1$

Ques 108. Eigenvalues of zero matrix $0_{3\times3}$ are:

  • (A) $0,0,0$
  • (B) $1,1,1$
  • (C) Undefined
  • (D) $1,0,0$

Correct Answer: (A) $0,0,0$

Ques 109. If $A$ is $3\times3$ with eigenvalues $1, -1, 2$, $\det(A^{2})$ is:

  • (A) $4$
  • (B) $-2$
  • (C) $2$
  • (D) $16$

Correct Answer: (A) $4$ Step-by-Step Solution:

  1. Eigenvalues of $A^2$ are $1^2 = 1, (-1)^2 = 1, 2^2 = 4$.
  2. $\det(A^2) = 1 \times 1 \times 4 = 4$.

Ques 110. If $A$ is $3\times3$ with eigenvalues $1, -1, 2$, $\text{trace}(A^{2})$ is:

  • (A) $1+1+4=6$
  • (B) $2$
  • (C) $4$
  • (D) $8$

Correct Answer: (A) $1+1+4=6$ Step-by-Step Solution:

  1. Eigenvalues of $A^2$ are $1^2 = 1, (-1)^2 = 1, 2^2 = 4$.
  2. $\text{trace}(A^2) = 1 + 1 + 4 = 6$.

Ques 111. If $A$ is $3\times3$ matrix with $\det(A)=10$, product of its eigenvalues is:

  • (A) $10$
  • (B) $30$
  • (C) $3$
  • (D) $1$

Correct Answer: (A) $10$

↑ ↓ navigate
↵ select
esc clear / close
Tap result to open StudyDump Search

Saved Bookmarks

0

Request or Upload Content

Send missing topics, notes, question banks, or corrections directly to Sehjal

Click to upload or drag and drop
PDF, PNG, JPG, DOCX, ZIP (Up to 10MB)

Submitting opens FormSubmit's confirmation or error page. You can return to StudyDump with your browser Back button.