Numerical Methods - Free MCQ Practice Test with solutions, GATE MA Engineering

Option C is correct.

Performing Gaussian elimination on the given system leads (after the first step R₂ → R₂ - 2R₁, R₃ → R₃ - R₁) to the two-row equations -2y - 10z = -14 and 2y + z = 5.

Adding these two equations gives -9z = -9, so z = 1.

Substituting z = 1 into 2y + z = 5 gives 2y + 1 = 5, so y = 2.

Substituting y = 2 and z = 1 into the first equation yields x = -3.

Thus the sum is x + y + z = -3 + 2 + 1 = 0.

The upper triangular coefficient matrix obtained during elimination has diagonal entries 1, -2 and -9, so its trace is 1 + (-2) + (-9) = -10.

Therefore the required difference is (x + y + z) - trace = 0 - (-10) = 10.

Concept:
Euler's Method to generate a numerical solution to an initial value problem of the form:
y' = f (x, y), y(x₀) = y₀
y_{n + 1} = y_n + h f(x_n, y_n)
Calculation:
We have,
x₀ =  0, y₀ = 1, h = 0.1
x₁ = x₀ + h
For n = 0
⇒ y₁ = y₀ + hf(x_n, y_n)
⇒ y₁ = 1 - 0.1 ×  (0 - 1)²
⇒ y₁ = 1 - 0.1
⇒ y₁ = 0. 9
For n = 1
⇒ y₂ = y₁ + hf(x₁ - y₁²)
⇒ y₂ = 0.9 + 0.1[0.1 - (0.9)²]
⇒ y₂ = 0.9 + 0.1[0.1 - 0. 81]
⇒ y₂ = 0.9 - 0.1 ×  0.71
⇒ y₂ = 0.9 - 0.071
⇒ y₂ = 0. 829
⇒ y₂ = 0.83
∴ The approximately. two decimal place value is 0.83 

Concept:
The following table shows the different methods of numerical integration and degree of polynomials for which they will produce results of minimum error or zero error:

From the above table, it is clear that both Trapezoidal Rule polynomials of degree ≤ 1
Alternate Method
We know,
While deriving the formula for numerical integrations f(x) is assumed as -

• Quadratic polynomial → Simpson's 1/3 Rule
• Cubic polynomial → Simpson's 3/8 Rule
• Linear polynomial → Trapezoidal Rule

Concept:
The trapezoidal rule states that:

Calculation:

Total number of intervals = 4

Concept:
Bisection method:
Used to find the root for a function. Root of a function f(x) = a such that f(a)= 0
Property: if a function f(x) is continuous on the interval [a…b] and sign of f(a) ≠  sign of f(b). There is a value c belongs to [a…b] such that f(c) = 0, means c is a root in between [a….b]
Note:
Bisection method cut the interval into 2 halves and check which half contains a root of the equation.
1) Suppose interval [a…b] .
2) Cut interval in the middle to find m : m = (a + b)/2
3) sign of f(m) not matches with f(a) proceed the search in the new interval.
Calculation:
The bisection method is applied to a given problem with [1, 9]
After 1 iteration

Now since f(x₁) > 0, x₂ replaces x₁
Now, x₀ = 1 and x₁ = 5
And after 2^nd iteration 

Now since f(x₁) f(x₂) > 0, x₂ replaces x₁ and x₀ = 1 and x₁ = 3 and after 3^rd iteration

Now, f(x₂) = f(2) = 2⁴ – 2³ – 2² – 4 = 0
So the method converges exactly to the root in 3 iterations.

Concept:
Gauss Jordan method,
In this method, an augmented matrix is formed by the coefficient of x, y, and z. then by using row transformation, it is converted into a diagonal matrix i.e. all non-diagonal elements will be zero.
Calculation:
x + 2y + 3z = 11
2x - 2y - z = 2
3x - y + 2z = 12

AX = B

R₂ → R₂ - 2R₁
R₃ → R₃ - 3R₁

R₃ → 6R₃ + R₂
R₁ → 3R₁ + R₂

R₁ → R₁ + 2R₃
R₂ → R₂ - 7R₃

Now, again converting this matrix into equation,
3x = 9 ⇒ x = 3
-6y = 6 ⇒ y = 1
-z = -2 ⇒ z = 2
x + y + z = 3 + 2 + 1 = 6

Bracketing Methods:

• All bracketing methods always converge, whereas open methods (may sometimes diverge).
• We must start with an initial interval [a,b], where f(a) and f(b) have opposite signs.
• Since the graph y = f(x) of a continuous function is unbroken, it will cross the abscissa at a zero x = 'a' that lies somewhere within the interval [a,b].
• One of the ways to test a numerical method for solving the equation f(x) = 0 is to check its performance on a polynomial whose roots are known.

Bisection method:
Used to find the root for a function. Root of a function f(x) = a such that f(a)= 0
Property: if a function f(x) is continuous on the interval [a…b] and sign of f(a) ≠ sign of f(b). There is a value c belongs to [a…b] such that f(c) = 0, means c is a root in between [a….b]
Note:
Bisection method cut the interval into 2 halves and check which half contains a root of the equation.
1) Suppose interval [a, b] .
2) Cut interval in the middle to find m : m = (a + b)/2
3) sign of f(m) not matches with f(a), proceed the search in new interval.

Concept:
Trapezoidal rule states that for a function y = f(x)

x_n = x₀ + nh, where n = Number of sub-intervals
h = step-size

For a trapezoidal rule, a number of sub-intervals must be a multiple of 1.
Calculation:

Here: x₀ = 4, x₁ = 3, x₂ = 12, h = 1
From equation (1);

Key Points:
Apart from the trapezoidal rule, other numerical integration methods are:
Simpson’s one-third rule:
For applying this rule, the number of subintervals must be a multiple of 2.

Simpson’s three-eighths rule:
For applying this rule, the number of subintervals must be a multiple of 3.

Trapezoidal rule
It integrates a linear function exactly and produces errors for polynomial functions of degree 2 or higher.

Here, the interval is divided into 'n' number of intervals (even or odd) of equal width 'h' b is the upper limit, a is the lower limit, h is the step size
It fits for a 1-degree (linear) polynomial.
According to Simpson's 1/3 rule:

It fits for 2-degree (quadratic) polynomial.
According to Simpson's 3/8 rule

It fits for 3-degree (cubic) polynomial.

Concept:
If data points are given as a function of f, then the various order divided differences are as follows,
Zeroth-order divided difference:
f[x₀] = f(x₀);
First-order divided difference:

Second-order divided difference:

Calculation:
Given f(x) = x²;
Using the second-order divided difference formula, we get 

∴ the second-order divided difference of x² is 1.