Numerical Analysis Questions and Answers – Gauss Jordan Method – 1

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

1. Solve the equations using Gauss Jordan method.

x + 2y + 6z = 22
3x + 4y + z = 26
6x - y - z = 19

a) x = 4, y = 3, z = 2
b) x = 4, y = 3, z = 2
c) x = 4, y = 3, z = 2
d) x = 4, y = 3, z = 2
View Answer

Answer: a
Explanation: By Gauss Jordan method we get
\(\begin{bmatrix}
1 & 2 & 6\\
3 & 4 & 1\\
6 & -1 & -1\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
22\\
26\\
19\\
\end{bmatrix} \)

By R2-3R1 and R3-6R1

\(\begin{bmatrix}
1 & 2 & 6\\
0 & -2 & -17\\
0 & -13 & 37\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
22\\
-40\\
-113\\
\end{bmatrix} \)

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-2R3 and -13R2

\(\begin{bmatrix}
1 & 2 & 6\\
0 & 26 & 221\\
0 & 26 & 74\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
22\\
520\\
226\\
\end{bmatrix} \)

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R3-R2 and R2/13, R3/(-147)

\(\begin{bmatrix}
1 & 2 & 6\\
0 & 2 & 17\\
0 & 0 & 1\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
22\\
40\\
2\\
\end{bmatrix} \)

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R1-R2 and R1+11R3, R2-117R3

\(\begin{bmatrix}
1 & 0 & 0\\
0 & 2 & 0\\
0 & 0 & 1\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
4\\
6\\
2\\
\end{bmatrix} \)

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(1/2)R2

\(\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
4\\
3\\
2\\
\end{bmatrix} \)

Hence x = 4, y = 3, z = 2.

2. Solve the equations using Gauss Jordan method.

x + 2y + 6z = 44
3x + 4y + z = 52
6x - y - z = 38

a) x = 8, y = 6, z = 4
b) x = 8, y = 4, z = 6
c) x = 4, y = 8, z = 6
d) x = 8, y = 6, z = 2
View Answer

Answer: a
Explanation: By Gauss Elimination Jordan method we get
\(\begin{bmatrix}
1 & 2 & 6\\
3 & 4 & 1\\
6 & -1 & -1\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
44\\
52\\
38\\
\end{bmatrix} \)

By R2-3R1 and R3-6R1
-2R3 and -13R2
R3-R2 and R2/13, R3/(-147)
R1-R2 and R1+11R3, R2-117R3
(1/2)R2
\(\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
8\\
6\\
4\\
\end{bmatrix} \)
Hence x = 8, y = 6, z = 4.

3. Solve the equations using Gauss Jordan method.

x+y+z=9
2x-3y+4z=13
3x+4y+5z=40

a) x=1, y=3, z=4
b) x=1, y=3, z=5
c) x=1, y=3, z=7
d) x=1, y=3, z=2
View Answer

Answer: b
Explanation: By Gauss Jordan method we get
\(\begin{bmatrix}
1 & 1 & 1\\
2 & -3 & 4\\
3 & 4 & 5\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
9\\
13\\
40\\
\end{bmatrix} \)

By R2-2R1 and R3-3R1

\(\begin{bmatrix}
1 & 2 & 6\\
0 & -5 & 2\\
0 & 1 & 2\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
9\\
-5\\
13\\
\end{bmatrix} \)

5R3 and -R2

\(\begin{bmatrix}
1 & 1 & 1\\
0 & 5 & -2\\
0 & 5 & 10\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
9\\
5\\
65\\
\end{bmatrix} \)

R3-R2 and R2+(1/6) R3, (1/12)R3

\(\begin{bmatrix}
1 & 1 & 1\\
0 & 5 & 0\\
0 & 0 & 1\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
9\\
15\\
5\\
\end{bmatrix} \)

(1/2)R2

\(\begin{bmatrix}
1 & 1 & 1\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
9\\
3\\
5\\
\end{bmatrix} \)

R1-R2-R3

\(\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
1\\
3\\
5\\
\end{bmatrix} \)

Hence x =1, y=3, z=5.

4. Solve the equations using Gauss Jordan method.

2x-3y+z=-1
x+4y+5z=25
3x-4y+z=2

a) x=1, y=3, z=4
b) x=1, y=3, z=5
c) x=1, y=3, z=7
d) x=1, y=3, z=2
View Answer

Answer: b
Explanation: By Gauss Jordan method we get
\(\begin{bmatrix}
1 & 4 & 5\\
2 & -3 & 1\\
3 & -4 & 1\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
25\\
-1\\
2\\
\end{bmatrix} \)

By R2-2R1 and R3-3R1

\(\begin{bmatrix}
1 & 4 & 5\\
0 & -11 & -9\\
0 & -16 & -14\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
25\\
-51\\
-73\\
\end{bmatrix} \)

-R3 and -R2, 16R2 and 11R1

\(\begin{bmatrix}
1 & 4 & 5\\
0 & 176 & 144\\
0 & 176 & 154\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
25\\
816\\
803\\
\end{bmatrix} \)

R3-R2 and (1/16)R2

\(\begin{bmatrix}
1 & 4 & 5\\
0 & 11 & 9\\
0 & 0 & 10\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
25\\
51\\
-13\\
\end{bmatrix} \)

R2-(9/10)R3 and R1-(1/2)R3

\(\begin{bmatrix}
1 & 4 & 0\\
0 & 11 & 0\\
0 & 0 & 10\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
63/2\\
627/10\\
-13\\
\end{bmatrix} \)

R1-(4/11)R2, (1/11)R1 and (1/10)R3

\(\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
8.7\\
5.7\\
-1.3\\
\end{bmatrix} \)

Hence x = 8.7, y = 5.7, z = -1.3.

5. In Gauss Jordan method which of the following transformations are allowed?
a) Diagonal transformation
b) Column transformation
c) Row transformation
d) Square transformation
View Answer

Answer: c
Explanation: In Gauss Jordan method the transformations carried out on the augmented matrix are row transformations. The matrix is reduced to Row Echelon form using Row transformations.

6. The augmented matrix in Gauss Jordan method is reduced to ______________
a) Row Echelon form
b) Column Echelon form
c) Matrix Echelon form
d) Augmented form
View Answer

Answer: a
Explanation: The matrix is reduced to Row Echelon form using Row transformations. In Gauss Jordan method the transformations carried out on the augmented matrix are row transformations.

7. Solve the equations using Gauss Jordan method.

x+2y+6z = 12
3x+4y+z = 24
6x-y-z = 36

a) x = 48/7, y = 8/7, z = 4/7
b) x = 4/7, y = 48/7, z = 4/7
c) x = 44/7, y = 8/7, z = 4/7
d) x = 4/7, y = 8/7, z = 44/7
View Answer

Answer: c
Explanation: By Gauss Elimination Jordan method we get
\(\begin{bmatrix}
1 & 2 & 6\\
3 & 4 & 1\\
6 & -1 & -1\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
12\\
24\\
36\\
\end{bmatrix} \)

By R2-3R1 and R3-6R1
-2R3 and -13R2
R3-R2 and R2/13, R3/(-147)
R1-R2 and R1+11R3, R2-117R3
(1/2)R2

\(\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
5\\
9\\
2\\
\end{bmatrix} \)
Hence x=5, y=9, z=2.

8. Solve the equations using Gauss Jordan method.

x + 2y + 6z = 15
3x + 4y + z = 16
6x - y - z = 20

a) x=3.735, y=0.795, z=1.612
b) x=3.735, y=3.735, z=1.612
c) x=3.735, y=1.612, z=3.735
d) x=1.612, y=0.795, z=3.735
View Answer

Answer: a
Explanation: By Gauss Jordan method we get
\(\begin{bmatrix}
1 & 2 & 6\\
3 & 4 & 1\\
6 & -1 & -1\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
15\\
16\\
20\\
\end{bmatrix} \)

By R2-3R1 and R3-6R1
-2R3 and -13R2
R3-R2 and R2/13, R3/(-147)
R1-R2 and R1+11R3, R2-117R3
(1/2)R2

\(\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
3.735\\
0.795\\
1.612\\
\end{bmatrix} \)

Hence x=3.735, y=0.795, z=1.612.

Sanfoundry Global Education & Learning Series – Numerical Methods.

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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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