This set of Numerical Analysis Multiple Choice Questions & Answers (MCQs) 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
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
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
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
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
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
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
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
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
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
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
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
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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