Category: Engineering Mathematics Objective Questions

Linear Algebra Problems focuses on “Canonical Form or Sum of Squares Form”.

1. Which of the following is not a condition for a given real nonsingular quadratic form, Q = X^TAX, to be a negative definite quadratic form?
a) The number of positive square terms in the quadratic form is equal to zero
b) The rank of the matrix A is equal to the number of variables in the quadratic form (index)
c) All the eigen values of A are negative
d) The rank and index are equal
View Answer

Answer: d
Explanation: The quadratic form is said to be negative definitive if the rank is equal to index and the number of square terms is equal to zero or all the eigen values of the matrix are negative.

2. Signature of a quadratic form is the difference between the positive and negative terms in the canonical form.
a) True
b) False
View Answer

Answer: b
Explanation: Signature of a quadratic form is defined as ‘the difference between the number of positive and negative square terms in the canonical form.’

3. Determine the nature of the given matrix.
(begin{bmatrix}2 & 0 & 0 \ 1 & 2 & 1 \ 0 & 0 & 1end{bmatrix} )

a) Indefinite
b) Positive definite
c) Negative definite
d) Positive semi-definite
View Answer

4. What is the signature of the quadratic form, (Q = 7x_1^2+ 2x_2^2- 3x_3^2+x_{31}+ x_{23}=0)?
a) 2
b) -2
c) 1
d) -1
View Answer

Answer: c
Explanation: We know that signature is the difference between the number of positive and negative square terms of a quadratic form. Therefore, we have signature = 2 – 1 = 1.

5. What is the index of the quadratic form, (3x_1^2+ x_3^2+8x_{13}+ 9x_{21}+ 2x_{23}=0)?
a) 2
b) 4
c) 3
d) 1
View Answer

Answer: a
Explanation: The index of a quadratic form is the number of positive square terms.
Hence, from the given form, we have index = 2.

6. If A is a matrix, such that, A^k = 0, for positive integer k, then, A is known as Nilpotent matrix.
a) True
b) False
View Answer

Answer: a
Explanation: For a matrix A, A^k+1 = A, where k is a positive integer is known as periodic matrix.
Whereas, if A² = A, i.e. k=1, then it is known as idempotent matrix and if A^k = 0, then it is known as Nilpotent matrix.

7. Reduce the quadratic form to canonical form, (3x_1^2+ 2x_2^2+8x_{12}+8x_{23}+8x_{31}=0).
a) (begin{bmatrix}3 & 4 & 4 \4 & 0 & 4 \ 4 & 4 & 2end{bmatrix} )
b) (begin{bmatrix}3 & 4 & 4 \4 & 2 & 0 \ 4 & 4 & 0end{bmatrix} )
c) (begin{bmatrix}3 & 4 & 4 \4 & 2 & 4 \ 4 & 4 & 0end{bmatrix} )
d) (begin{bmatrix}3 & 4 & 0 \4 & 2 & 4 \ 4 & 4 & 0end{bmatrix} )
View Answer

Answer: c
Explanation: Given quadratic form is, (3x_1^2+ 2x_2^2+8x_{12}+8x_{23}+8x_{31}=0)
General form of the matrix can be written as, (begin{bmatrix}x_{11} & x_{12} & x_{13} \ x_{21} & x_{22} & x_{23} \ x_{31} & x_{32} & x_{33}end{bmatrix} )
Hence, the matrix form can be obtained by,
Placing the square term coefficients in the diagonal of the matrix such that, (x_{ii}= x_i^2,)
(begin{bmatrix}3 & x_{12} & x_{13} \ x_{21} & 2 & x_{23} \ x_{31} & x_{32} & 0end{bmatrix} )
Dividing the coefficients of terms x_ij between x_ij and x_ji positions, for example, the coefficient of x₁₂ is 8, hence the term x₁₂ =x₂₁= 8/2 = 4.
Therefore, the matrix form of the given quadratic equation is,
(begin{bmatrix}3 & 4 & 4 \ 4 & 2 & 4 \ 4 & 4 & 0end{bmatrix} )

8. Write the expression for spur of a matrix for a 3×3 matrix whose entries are in the form of a_ij.
a) a₁₁+ a₁₂+a₁₃
b) a₁₁+ a₂₁+a₃₁
c) a₁₂+ a₂₂+a₃₂
d) a₁₁+ a₂₂+a₃₃
View Answer

Answer: d
Explanation: The spur of a matrix is nothing but trace of the matrix. It is defined as, ‘the sum of the diagonal elements of the matrix’.

9. The Canonical form is also known as ‘sum of squares’ form.
a) False
b) True
View Answer

Answer: b
Explanation: The canonical form is also known as sum of squares form since after reducing to canonical form, we get the terms as sum of squares.

10. Which among the following is not a type of quadratic form?
a) Positive Semi-definite
b) Negative definite
c) Partial definite
d) Indefinite
View Answer

Answer: c
Explanation: Quadratic forms can be classified based on the nature of the eigen values of the matrix into 5 types:
i. Positive definite
ii. Negative definite
iii. Positive Semi-definite
iv. Negative Semi-definite
v. Indefinite

Global Education & Learning Series – Fourier Analysis and Partial Differential Equations.

To practice all areas of Linear Algebra Problems, here is complete set of 1000+ Multiple Choice Questions and Answers.

 

Fourier Analysis and Partial Differential Equations Multiple Choice Questions on “Solution of Second Order P.D.E.”.

1. What is the general form of second order non-linear partial differential equations (x and y being independent variables and z being a dependent variable)?
a) (F(x,y,z,frac{∂z}{∂x},frac{∂z}{∂y},frac{∂^2 z}{∂x^2},frac{∂^2 z}{∂y^2},frac{∂^2 z}{∂x∂y})=0)
b) (F(x,z,frac{∂z}{∂x},frac{∂z}{∂y},frac{∂^2 z}{∂x^2},frac{∂^2 z}{∂y^2})=0)
c) (F(y,z,frac{∂z}{∂x},frac{∂z}{∂y})=0)
d) F(x,y)=0
View Answer

Answer: a
Explanation: The most general second order partial differential equation in two independent variables x and y, and z as the dependent variable has the form,
(F(x,y,z,frac{∂z}{∂x},frac{∂z}{∂y},frac{∂^2 z}{∂x^2},frac{∂^2 z}{∂y^2},frac{∂^2 z}{∂x∂y})=0)

2. The solution of the general form of second order non-linear partial differential equation is obtained by Monge’s method.
a) False
b) True
View Answer

Answer: b
Explanation: Gaspard Monge (9 May 1746 – 28 July 1818) was a French mathematician, the inventor of descriptive geometry (the mathematical basis of technical drawing), and the father of differential geometry.

3. What is the reason behind the non-existence of any real function which satisfies the differential equation, (y’)² + 1 = 0?
a) Because for any real function, the left-hand side of the equation will be less than, or equal to one and thus cannot be zero
b) Because for any real function, the left-hand side of the equation becomes zero
c) Because for any real function, the left-hand side of the equation will be greater than, or equal to one and thus cannot be zero
d) Because for any real function, the left-hand side of the equation becomes infinity
View Answer

Answer: c
Explanation: Given: (y’)² + 1 = 0
Consider if y = 2x, then y’ = 2 and hence the left-hand side of the equation becomes 3 which is greater than 1. Therefore, the left-hand side of the equation will always be greater than, or equal to one and thus cannot be zero and hence the differential equation is not satisfied.

4. What is the order of the partial differential equation, (frac{∂^2 z}{∂x^2}-(frac{∂z}{∂y})^5+frac{∂^2 z}{∂x∂y}=0)?
a) Order-5
b) Order-1
c) Order-4
d) Order-2
View Answer

Answer: d
Explanation: The order of an equation is defined as the highest derivative present in the equation. Hence, in the given equation, (frac{∂^2 z}{∂x^2}-(frac{∂z}{∂y})^5+frac{∂^2 z}{∂x∂y}=0), the order is 2.

5. Which of the following is the condition for a second order partial differential equation to be hyperbolic?
a) b²-ac<0
b) b²-ac=0
c) b²-ac>0
d) b²-ac=<0
View Answer

Answer: c
Explanation: For a second order partial differential equation to be hyperbolic, the equation should satisfy the condition, b²-ac>0.

6. Which of the following represents the canonical form of a second order parabolic PDE?
a) (frac{∂^2 z}{∂η^2}+⋯=0 )
b) (frac{∂^2 z}{∂ζ∂η}+⋯=0)
c) (frac{∂^2 z}{∂α^2}+frac{∂^2 z}{∂β^2}…=0)
d) (frac{∂^2 z}{∂ζ^2}+⋯=0)
View Answer

Answer: a
Explanation: A second order linear partial differential equation can be reduced to so-called canonical form by an appropriate change of variables ξ = ξ(x, y), η = η(x, y).

7. The condition which a second order partial differential equation must satisfy to be elliptical is
b²-ac=0.
a) True
b) False
View Answer

Answer: b
Explanation: The condition for a second order partial differential equation to be elliptical is given by, b²-ac<0.

8. Which of the following statements is true?
a) Hyperbolic equations have three families of characteristic curves
b) Hyperbolic equations have one family of characteristic curves
c) Hyperbolic equations have no families of characteristic curves
d) Hyperbolic equations have two families of characteristic curves
View Answer

Answer: d
Explanation: The canonical variables ξ and η for a hyperbolic pde satisfy the equations,
(aζ_x+(b+sqrt{b^2-ac}) ζ_y=0 , and , aη_x+(b+sqrt{b^2-ac}) η_y=0)
The families of curves ξ = constant and η = constant are the characteristic curves. Hence, hyperbolic equations have two families of characteristic curves.

9. Which of the following represents the family of the characteristic curves for parabolic equations?
a) aζ_x+bζ_y=0
b) aζ_x+b=0
c) a+ζ_y=0
d) a(ζ_x+ζ_y)=0
View Answer

Answer: a
Explanation: Parabolic equations have only one family of characteristic curves. Instead of two equations like for hyperbolic equations, we have just the single equation, aζ_x+bζ_y=0 (or aη_x+bη_y=0).

10. The condition that a second order partial differential equation should satisfy to be parabolic is b²-ac=0.
a) True
b) False
View Answer

Answer: a
Explanation: If the second order partial differential equation satisfies the condition, b²-ac=0, then it is said to be parabolic in nature.

11. Elliptic equations have no characteristic curves.
a) True
b) False
View Answer

Answer: a
Explanation: For an elliptic equation, b² −ac < 0 so equations obtained contain complex coefficients and
have no real solutions. Hence, elliptic equations have no characteristic curves.

12. Singular solution of a differential equation is one that cannot be obtained from the general solution gotten by the usual method of solving the differential equation.
a) True
b) False
View Answer

Answer: a
Explanation: A differential equation is said to have a singular solution if in all points in the domain of the equation the uniqueness of the solution is violated. Hence, this solution cannot be obtained from the general solution.

13. In the formation of differential equation by elimination of arbitrary constants, after differentiating the equation with respect to independent variable, the arbitrary constant gets eliminated.
a) False
b) True
View Answer

Answer: a
Explanation: In the formation of differential equation by elimination of arbitrary constants, the first step is to differentiate the equation with respect to the dependent variable. Sometimes, the arbitrary constant gets eliminated after differentiation.

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here is complete set of 1000+ Multiple Choice Questions and Answers.

 

Engineering Mathematics Questions and Answers for Experienced people focuses on “Rolle’s Theorem – 2”.

1. Rolle’s Theorem tells about the
a) Existence of point c where derivative of a function becomes zero
b) Existence of point c where derivative of a function is positive
c) Existence of point c where derivative of a function is negative
d) Existence of point c where derivative of a function is either positive or negative
Answer: a
Explanation: Statement of Rolle’s Theorem is that, If function f(x) attains same value at point ‘a’ and ‘b’ [f(a) = f(b)], and continuous in closed interval [a, b] and differentiable in open interval (a, b), then there exists a point ‘c’ such that c∈(a,b) and f’(c) = 0.

2. Rolle’s Theorem is a special case of
a) Lebniz Theorem
b) Mean Value Theorem
c) Taylor Series of a function
d) Leibnit’x Theorem
Answer: b
Explanation: According to Mean Value Theorem, If a function exist at pt. ‘a’, ‘b’ and continuous in closed interval [a, b] and differentiable in open interval (a, b) then there exists a point ‘c’, such that c∈(a,b), Where,
f’(c)= [f(b)-f(a)]/(b-a).
Hence, By putting f(b) = f(a) in the statement of Mean Value Theorem, we get
f’(c) = [f(b)-f(a)]/(b-a) = 0. Which is a statement of Rolle’s Theorem.

3. Rolle’s theorem is applicable to the
a) Functions differentiable in closed interval [a, b] and continuous in open interval (a, b) only and having same value at point ‘a’ and ‘b’
b) Functions continuous in closed interval [a, b] only and having same value at point ‘a’ and ‘b’
c) Functions continuous in closed interval [a, b] and differentiable in open interval (a, b) only and having same value at point ‘a’ and ‘b’
d) Monotonically Increasing funtions
Answer: c
Explanation: Statement of Rolle’s Theorem is that, If function f(x) attains same value at point ‘a’ and ‘b’ [f(a) = f(b)], and continuous in closed interval [a, b] and differentiable in open interval (a, b), then there exists a point ‘c’ such that c∈(a,b) and f’(c) = 0.

4. Find the value of c(a point where slope of a atangent to curve is zero) if f(x) = Sin(x) is continuous over interval [0,π] and differentiable over interval (0, π) and c ∈(0,π)
a) π
b) ^π⁄₂
c) ^π⁄₆
d) ^π⁄₄
Answer: b
Explanation: Given, f(x)=Sin(x), x ∈ [0,π].
Now f(0) = f(π) = 0
f’(c) = Cos(c) = 0
c = ^π⁄₂.

5. Find the value of c if f(x) = x(x-3)e^3x, is continuous over interval [0,3] and differentiable over interval (0, 3) and c ∈(0,3)
a) 0.369
b) 2.703
c) 0
d) 3
Answer: b
Explanation: f(0) = 0
f(3) = 0
Hence, By rolle’s Theorem
f’(c) = (c-3) e^3c + c e^3c + 3c(c-3) e^3c = 0
Hence, c-3 + c + 3c² -9c = 0
3c² – 7c – 3 = 0
c = 2.703, -0.369
Now c ∈(0,3), hence, c = 2.703.

6. Find the value of c if f(x) = sin³(x)cos(x), is continuous over interval [0, ^π⁄₂] and differentiable over interval (0, ^π⁄₂) and c ∈(0, ^π⁄₂)
a) 0
b) ^π⁄₆
c) ^π⁄₃
d) ^π⁄₂
Answer: c
Explanation: f(x) = sin³(x)cos(x)
f(0) = 0
f(^π⁄₂) = 0
Hence,
f’(c) = 3sin²(c)cos(c)cos(c) – sin⁴(c) = 0
3sin²(c)cos²(c) – sin⁴(c) = 0
sin²(c)[3 cos²(c) – sin²(c)] = 0
either, sin²(c)=0 or 3 cos²(c) – sin²(c) = 0
Since sin²(c) cannot be zero because c cannot be 0
Hence, 3 cos²(c) – sin²(c)=0
tan²(c) = 3
tan(c) = √3
c = ^π⁄₃.

7. Find value of c where f(x) = sin(x) e^x tan(x), c ∈ (0,∞)
a) Tan^-1[-(2+c²)/(1+c²)
b) Tan^-1[-(2-c²)/(1+c²)]
c) Tan^-1[(2+c²)/(1+c²)]
d) Rolle’s Theorem is not applied, Cannot find the value of c
Answer: d
Explanation: Since, f(x) = e^xsin(x) tan(x) is not continuous over interval (0,∞), Hnece Rolle’s, theorem is not applied.

8. f(x) = 3Sin(2x), is continuous over interval [0,π] and differentiable over interval (0,π) and c ∈(0,π)
a) π
b) ^π⁄₂
c) ^π⁄₄
d) ^π⁄₈
Answer: b
Explanation: f(x) = 3Sin(2x)
f(0)=0
f(π)=0
Hence,
f’(c) = 6Cos(2c) = 0
c= ^π⁄₂.

9. Find the value of ‘a’ if f(x) = ax²+32x+4 is continuous over [-4, 0] and differentiable over (-4, 0) and satisfy the Rolle’s theorem. Hence find the point in interval (-2,0) at which its slope of a tangent is zero
a) 2, -2
b) 2, -1
c) 8, -1
d) 8, -2
Answer: d
Explanation: Since it satisfies Rolle’s Theorem,
f’(c) = 0 = 2ac+32 ………………(1)
and, f(0) = 4 hence by Rolle’s theorem
and, f(-4) = 4 = 16a-128+4 (because f(0)=f(-4) condition of rolle’s theorem)
⇒ a = 8
from, eq.(1)
⇒ c = -2.

10. Find the value of ‘a’ & ‘b’ if f(x) = ax² + bx + sin(x) is continuous over [0, π] and differentiable over (0, π) and satisfy the Rolle’s theorem at point c = ^π⁄₄.
a) 0.45,1.414
b) 0.45,-1.414
c) -0.45,1.414
d) -0.45,-1.414
Answer: b
Explanation: Since function f(x) is continuous over [0,π] and satisfy rolle’s theorem,
⇒ f(0) = f(π) = 0
⇒ f(π) = a π² + b π=0
⇒ a π+b=0 ………………….(1)
Since it satisfies rolle’s theorem at c = ^π⁄₄
f’(c) = 2ac + b + Cos(c) = 0
⇒ a(^π⁄₂) + b + ¹⁄_√2 = 0 ………………..(2)
From eq(1) and eq(2) we get,
⇒ a = 0.45
⇒ b = -1.414.

11. Find value of c(a point in f(x) where slope of tangent to curve is zero) where
f(x) = (begin{cases}Tan(x) & 0a) ^π⁄₄
b) Rolle’s Theorem is not applied, because function is not continuous in interval [0, ^π⁄₂]
c) Rolle’s Theorem is not applied, because function is not differential in interval (0, ^π⁄₂)
d) Function is both continuous and differentiable but Rolle’s theorem is not applicable as f(0) ≠ f(^π⁄₂)
Answer: b
Explanation: Continuity Check.
(lim_{xrightarrowπ/4-}f(x) = lim_{xrightarrowπ/4-}Tan(x) = 1)
(lim_{xrightarrowπ/4+}f(x) = lim_{xrightarrowπ/4+}Cos(x) = frac{1}{sqrt{2}})
(lim_{xrightarrowπ/4-}f(x) ≠ lim_{xrightarrowπ/4+}f(x))
Hence function is discontinuous in interval (0, ^π⁄₂).
Hence Rolle’s theorem cannot be applied.

12. Find value of c(a point in a curve where slope of tangent to curve is zero) where
f(x) = (begin{cases}x^2-x & 0a) 1.5
b) Rolle’s Theorem is not applied, because function is not continuous in interval [0,2]
c) Rolle’s Theorem is not applied, because function is not differential in interval (0,2)
d) Function is both continuous and differentiable but Rolle’s theorem is not applicable as f(0) ≠ f(2)
Answer: c
Explanation: Continuity Check
(lim_{xrightarrow1-}⁡f(x) = lim_{xrightarrow 1-}x^2-x = 0)
(lim_{xrightarrow1+}f(x) = lim_{xrightarrow 1+}3x^3-4x+1 = 0)
(lim_{xrightarrowπ/4-}f(x)= lim_{xrightarrow π/4+}f(x))
Hence function is Continuous.
Differentiability Check
(lim_{xrightarrow 1-}f'(x) = lim_{xrightarrow 1-}2x-1 = 1)
(lim_{xrightarrow 1+}f(x) = lim_{xrightarrow 1+}9x^2-4 = 5)
(lim_{xrightarrow π/4-}f(x) ≠ lim_{xrightarrow π/4+}f(x))
Hence function is not differentiable so Rolle’s Theorem cannot be applied.

13. f(x) = ln(10-x²), x=[-3,3], find the point in interval [-3,3] where slope of a tangent is zero,
a) 0
b) Rolle’s Theorem is not applied, because function is not continuous in interval [-3,3]
c) Rolle’s Theorem is not applied, because function is not differential in interval (-3,3)
d) 2
Answer: a
Explanation: Domain of f(x) = [-√10, +√10]
Hence given f(x) is continuous in interval [-3,3]
f’(x) = (frac{-2x}{10-x^2})
⇒ x ≠ ±√10
⇒ Domain of f’(x) = (-∞,∞)- ±√10
⇒ Hence f(x) is differential in interval (-3,3)
f’(c) = -2c/(10-c²) = 0
⇒ c=0.

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