Numerical Analysis Interview Questions and Answers

This set of Numerical Analysis Interview Questions and Answers focuses on “Gauss Jordan Method – 2”.

1. Solve the given equations using Gauss Jordan method.

x + 2y + 6z = 66
3x + 4y + z = 78
6x - y - z = 57

a) x=4, y=3, z=2
b) x=13, y=6, z=9
c) x=12, y=9, z=6
d) x=4, y=2, z=3
View Answer

Answer: c
Explanation: By Gauss Jordan method we get
\(\begin{bmatrix}
1 & 2 & 6\\
3 & 4 & 1\\
6 & -1 & -1\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
66\\
78\\
57\\
\end{bmatrix} \)
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
\(\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
12\\
9\\
6\\
\end{bmatrix} \)
Hence x=12, y=9, z=6.

2. Solve the given equations using Gauss Jordan method.

3x+6y+18z=22
9x+12y+3z=26
18x-3y-3z=19

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a) x=1.33, y=1.33, z=0.67
b) x=1.33, y=0.67, z=0.67
c) x=1.33, y=1, z=0.67
d) x=1.33, y=1, z=1
View Answer

Answer: c
Explanation: By Gauss Jordan method we get
\(\begin{bmatrix}
3 & 6 & 18\\
9 & 12 & 3\\
18 & -3 & -3\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
22\\
26\\
19\\
\end{bmatrix} \)
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
\(\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
1.33\\
1\\
0.67\\
\end{bmatrix} \)
Hence x=1.33, y=1, z=0.67.

3. Solve the given equations using Gauss Jordan method.

3x+3y+3z=9
6x-9y+12z=13
9x+12y+15z=40

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a) x=0.33, y=1.67, z=1
b) x=1.67, y=1, z=0.33
c) x=1, y=0.33, z=1.67
d) x=0.33, y=1, z=1.67
View Answer

Answer: d
Explanation: By Gauss Jordan method we get
\(\begin{bmatrix}
3 & 3 & 3\\
6 & -9 & 12\\
9 & 12 & 15\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
9\\
13\\
40\\
\end{bmatrix} \)
By R₂-2R₁ and R₃-3R₁
5R₃ and -R₂
R₃-R₂ and R₂+(1/6) R₃, (1/12)R₃
(1/2)R₂
R₁-R₂-R₃
\(\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
0.33\\
1\\
1.67\\
\end{bmatrix} \)
Hence x=0.33, y=1, z=1.67.

4. Solve the given equations using Gauss Jordan method.

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

a) x=8.7, y=5.7, z=-1.3
b) x=8.7, y=-1.3, z=5.7
c) x=-1.3, y=5.7, z=8.7
d) x=5.7, y=8.7, z=-1.3
View Answer

Answer: a
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. Solve the given equations using Gauss Jordan method.

x+y+z=7.5
2x+3y+z=15
3x-2y+2z=4.5

a) x=3, y=1.5, z=1.5
b) x=3, y=1.5, z=1.5
c) x=1.5, y=3, z=1.5
d) x=1.5, y=3, z=3
View Answer

Answer: d
Explanation: By Gauss Jordan Elimination method we get,
x+y+z=7.5.…………….(i)
2x+3y+z=15….……….(ii)
3x-2y+2z=4.5…………(iii)
To eliminate x, subtract 2 times (i) from (ii) and 3 times (i) from (iii).
x+y+z=7.5…………….(iv)
y-z=0………………….(v)
-5y-z=-18……………(vi)
To eliminate y, subtract (v) from (iv) and add 5 times (v) to (vi).
x+2z=7.5…………….(vii)
y-z=0…………………(viii)
-6z=-18………………(ix)
To eliminate z add 1/3 times (ix) to (vii) and subtract 1/6 times (ix) from (viii).
Hence x=1.5, y=3, z=3.

6. Which of these methods is named after the mathematician Carl Friedrich Gauss?
a) Secant method
b) Newton Raphson method
c) Runge Kutta method
d) Gauss Jordan method
View Answer

Answer: d
Explanation: Gauss Jordan method is named after the mathematician Carl Friedrich Gauss. Its algorithm is used to solve linear equations.

7. Which of the following transformations are allowed in Gauss Jordan method?
a) Swapping a row
b) Swapping a column
c) Swapping two rows
d) Swapping two columns
View Answer

Answer: c
Explanation: Only row transformations are allowed in Gauss Jordan method. Swapping of two rows doesn’t affect the determinant value hence it is allowed here.

8. The Gauss Jordan method reduces a original matrix into a _____________
a) Identity matrix
b) Null matrix
c) Skew Hermitian matrix
d) Non-symmetric matrix
View Answer

Answer: a
Explanation: The Gauss Jordan method reduces a original matrix into a Row Echelon form. In Row Echelon form the matrix is identity.

9. Which of the following methods is used for obtaining the inverse of matrix?
a) Gauss Seidel method
b) Newton Raphson method
c) Gauss Jordan method
d) Secant Method
View Answer

Answer: c
Explanation: Gauss Jordan method deals with solving the matrix through row transformations. Hence it is most suitable to find the inverse of a matrix.

10. Solve the given equations using Gauss Jordan method.

2x+y-z=8
-3x-y+2z=-11
-2x+y+2z=-3

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

Answer: a
Explanation: By Gauss Jordan method we get,
\(\begin{bmatrix}
2 & 1 & -1\\
-3 & -1 & 2\\
-2 & 1 & 2\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
8\\
-11\\
-3\\
\end{bmatrix} \)
By,
R₁/2
3R₁+R₂ and 2R₁+R₃
2R₂
-(1/2)R₂+R₁ and -2R₂+R₃
R₃+R₂, -R₃+R₁ and –R₃
\(\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{bmatrix} \) \( \begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}\) = \( \begin{bmatrix}
2\\
3\\
-1\\
\end{bmatrix} \)
Hence x=2, y=3, z=-1.

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