Numerical Analysis Questions and Answers – Gauss Elimination Method – 1

This set of Numerical Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Gauss Elimination Method – 1”.

1. Solve the following equations by Gauss Elimination Method.

x+4y-z = -5
x+y-6z = -12
3x-y-z = 4

a) x = 1.64791, y = 1.14085, z = 2.08451
b) x = 1.65791, y = 1.14185, z = 2.08441
c) x = 1.64691, y = 1.14095, z = 2.08461
d) x = 1.64491, y = 1.15085, z = 2.09451
View Answer

Answer: a
Explanation: By Gauss Elimination method we get
\(\begin{bmatrix}
1 & 4 & -1\\
1 & 1 & -6\\
3 & -1 & -1\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
-5\\
-12\\
4\\
\end{bmatrix} \)

By R2-R1 and R3-3R1
\(\begin{bmatrix}
1 & 4 & -1\\
0 & -3 & -5\\
0 & -13 & 2\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
-5\\
-7\\
19\\
\end{bmatrix} \)

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R3-(-13/-3)*R2
\(\begin{bmatrix}
1.0000 & 4.0000 & -1.0000\\
0 & -3.0000 & -5.0000\\
0 & 0 & 23.6667\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
-5\\
-7\\
49.33333\\
\end{bmatrix} \)

x+4y-z = -5
-3y-5z = -7
23.6667z = 49.3333
Hence, z = 2.08451
-3y = -7+5z
Hence, y = -1.14085
x = -4y+z-5
Hence, x = 1.64791.

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2. Find the values of x, y, z in the following system of equations by gauss Elimination Method.

2x + y – 3z = -10
-2y + z = -2
z = 6

a) 2, 4, 6
b) 2, 7, 6
c) 3, 4, 6
d) 2, 4, 5
View Answer

Answer: a
Explanation: By Gauss Elimination method we get
\(\begin{bmatrix}
2 & 1 & -3\\
0 & -2 & 1\\
0 & 0 & 1\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
-10\\
-2\\
6\\
\end{bmatrix} \)
z = 6.
-2y+z = -2
Hence y = 4.
2x + y – 3z = -10
Hence z = 6.
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3. Solve the given system of equation by Gauss Elimination method.

3x + 4y – z = -6
-2y + 10z = -8
4y – 2z = -2

a) (-2, -1, -1)
b) (-1, -2, -1)
c) (-1, -1, -2)
d) (-1, -1, -1)
View Answer

Answer: d
Explanation: Here,
\(\begin{bmatrix}
3 & 4 & -1\\
0 & -2 & 10\\
0 & 4 & -2\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
-6\\
-8\\
-2\\
\end{bmatrix} \)
The Matrix almost represents a triangular Matrix. Multiplying Row2 by 2 and adding it to Row 3 we get a Upper Triangular Matrix with
x, y, z = (-1, -1, -1).
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4. The following system of equation has:

x – y – z = 4
2x – 2y – 2z = 8
5x – 5y – 5z = 20

a) Unique Solution
b) No solution
c) Infinitely many Solutions
d) Finite solutions
View Answer

Answer: c
Explanation: Multiplying Row 1 by -2, then adding Row 1 and Row 2 the 1st row of matrix gets reduced 0.
\(\begin{bmatrix}
0 & 0 & 0\\
2 & -2 & -2\\
5 & -5 & -5\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
0\\
8\\
20\\
\end{bmatrix} \)
Hence, there are infinitely many solutions.

5. Solve this system of equations and comment on the nature of the solution using Gauss Elimination method.

x + y + z = 0
-x – y + 3z = 3
-x – y – z = 2

a) Unique Solution
b) No solution
c) Infinitely many Solutions
d) Finite solutions
View Answer

Answer: b
Explanation: By Gauss Elimination method we add Row 1 and Row 3 to get the following matrix
\(\begin{bmatrix}
1 & 1 & 1\\
-1 & -1 & 3\\
0 & 0 & 0\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
0\\
3\\
2\\
\end{bmatrix} \)

Hence the matrix has no solution as 0 ≠ 2.

6. The aim of elimination steps in Gauss elimination method is to reduce the coefficient matrix to ____________
a) diagonal
b) identity
c) lower triangular
d) upper triangular
View Answer

Answer: d
Explanation: In Gauss elimination method we tend to reduce the given Matrix to an Upper Triangular matrix to solve for x, y, z.

7. Division by zero during in Gaussian elimination of the set of equations [A] * [X]=[C] signifies the coefficient matrix [A] is ____________
a) Invertible
b) Non Singular
c) Not determinable to be singular or non singular
d) Singular
View Answer

Answer: c
Explanation: Division by zero in Gauss Elimination method is not related to whether the matrix is singular or not.

8. In which of the following both sides of equation are multiplied by non-zero constant?
a) Gauss Elimination Method
b) Gaussian Inconsistent procedure
c) Gaussian consistent procedure
d) Gaussian substitute procedure
View Answer

Answer: a
Explanation: Gauss Elimination method employs both sides of equation to be multiplied by a non-zero constant. The matrix is then reduced to Upper Triangular Matrix to get values of the respective variables.

9. In Gaussian elimination method, original equations are transformed by using _____________
a) Column operations
b) Row Operations
c) Mathematical Operations
d) Subset Operation
View Answer

Answer: b
Explanation: Row Operations are used in Gauss Elimination method to reduce the Matrix to an Upper Triangular Matrix and thus solve for x, y, z.

10. The following information related the velocity and time of a vehicle. It governs a quadratic equation v(t)=at2+bt+c. Hence find the Matrix which represents the equation most accurately.

T s 0 14 15 20 30 35
V m/s 0 227.04 362.78 517.35 602.97 901.67

a) \(\begin{bmatrix}
176 & 14 & 1\\
225 & 15 & 1\\
400 & 20 & 1\\
\end{bmatrix} \) \( \begin{bmatrix}
a\\
b\\
c\\
\end{bmatrix}\) = \( \begin{bmatrix}
227.04\\
362.78\\
517.35\\
\end{bmatrix} \)
b) \(\begin{bmatrix}
225 & 15 & 1\\
400 & 20 & 1\\
900 & 30 & 1\\
\end{bmatrix} \) \( \begin{bmatrix}
a\\
b\\
c\\
\end{bmatrix}\) = \( \begin{bmatrix}
362.78\\
517.35\\
602.97\\
\end{bmatrix} \)
c) \(\begin{bmatrix}
0 & 0 & 1\\
225 & 15 & 1\\
400 & 20 & 1\\
\end{bmatrix} \) \( \begin{bmatrix}
a\\
b\\
c\\
\end{bmatrix}\) = \( \begin{bmatrix}
0\\
362.78\\
517.35\\
\end{bmatrix} \)
d) \(\begin{bmatrix}
400 & 20 & 1\\
900 & 30 & 1\\
1225 & 35 & 1\\
\end{bmatrix} \) \( \begin{bmatrix}
a\\
b\\
c\\
\end{bmatrix}\) = \( \begin{bmatrix}
517.35\\
602.97\\
901.67\\
\end{bmatrix} \)
View Answer

Answer: b
Explanation: The points closest to t=21 sec is 15, 20, 30
V(t0) = 362.78 m/s = a(15)2+b(15)+c
V(t1) = 517.35 m/s = a(20)2+b(20)+c
V(t2) = 602.97 m/s = a(30)2+b(30)+c
Hence,
225a+15b+c = 362.78
400a+20b+c = 517.35
900a+30b+c = 602.97
Which leads us to \(\begin{bmatrix}
225 & 15 & 1\\
400 & 20 & 1\\
900 & 30 & 1\\
\end{bmatrix} \) \( \begin{bmatrix}
a\\
b\\
c\\
\end{bmatrix}\) = \( \begin{bmatrix}
362.78\\
517.35\\
602.97\\
\end{bmatrix} \).

11. The Elimination process in Gauss Elimination method is also known as _____________
a) Forward Elimination
b) Backward Elimination
c) Sideways Elimination
d) Crossways Elimination
View Answer

Answer: a
Explanation: The Elimination process in Gauss Elimination Method is also known as Forward Elimination. N this method a Matrix is reduced to a Upper Triangular Matrix.

12. The reduced form of the Matrix in Gauss Elimination method is also called ____________
a) Column Echelon Form
b) Row-Column Echelon Form
c) Column-Row Echelon Form
d) Row Echelon Form
View Answer

Answer: d
Explanation: The reduced form of the Matrix in Gauss Elimination Method is called as Row Echelon Form. It is said so because only Row Operations are considered in Gauss Elimination method.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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