This set of Numerical Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Gauss Jordan Method – 3”.
1. The modification of Gauss elimination method is called as ___________
a) Gauss Seidal
b) Gauss Jordan
c) Jacobi’s Method
d) Relaxation Method
View Answer
Explanation: The modified method of Gauss Elimination is called as Gauss Jordan method as it involves few changes in the procedure of Gauss Elimination.
2. Which of the following is not a step involved in Gauss Jordan Method?
a) Elimination of unknowns
b) Reduction of systems to diagonal matrix form
c) Performing elimination on equations above as well as below
d) Evaluation of cofactors
View Answer
Explanation: Elimination of unknowns, reduction of systems to diagonal matrix form and performing elimination on equations above as well as below are the steps involved in Gauss Jordan method.
3. The advantage of using Gauss Jordan method is which of the following?
a) More operations involved
b) Elimination is easier
c) Additional Calculations
d) No labour of back substitution
View Answer
Explanation: The advantage of using Gauss Jordan method is that it involves no labour of back substitution. Back substitution has to be done while solving linear equations formed during solving the problem.
4. Which is more efficient method?
a) Gauss Elimination
b) Gauss Jordan
c) Cramer’s Rule
d) Back substitution
View Answer
Explanation: The least number of operations to solve the simultaneous linear equations are done in Gauss Elimination. That’s why it is better.
5. Why Gauss Elimination is preferred over other methods?
a) Less number of operations are involved
b) Back substitution needed
c) Elimination of unknowns
d) Forms diagonal matrix form
View Answer
Explanation: Gauss Elimination is preferred over other methods because it involves less number of operations. There is no back substitution in Gauss Elimination.
6. In solving simultaneous equations by Gauss Jordan method, the coefficient matrix is reduced to ______ matrix.
a) Identity
b) Diagonal
c) Upper triangular
d) Lower triangular
View Answer
Explanation: In solving simultaneous equations by Gauss Jordan method, the coefficient matrix is reduced to diagonal matrix. After that, we are able to get to the solution of the unknowns using matrix multiplication.
7. Apply Gauss Jordan method to solve the following equations.
x + y + z = 9 2x – 3y + 4z = 13 3x + 4y + 5z = 40
a) x = 7, y = 2, z = 5
b) x = 7, y = 2, z = 8
c) x = 1, y = 3, z = 5
d) x = 5, y = 3, z = 6
View Answer
Explanation:
x + y + z = 9 …………….(i)
2x – 3y + 4z = 13 ………..(ii)
3x + 4y + 5z = 40 ………..(iii)
To eliminate x, operate (ii) – 2(i) and (iii) – 3(i),
x + y + z = 9 …………….(iv)
– 5y + 2z = -5 …………….(v)
y + 2z = 13 ………………(vi)
To eliminate y, operate (iv) + (1/5)(v) and (vi) – (1/5)(v),
x + (7/5) z = 8 …………..(vii)
-5y + 2z = -5 …………….(viii)
(12/5) z = 12……………..(ix)
To eliminate z, operate (vii) – (7/12) (ix) and (viii) – (5/6)(ix)
x = 1
-5y = -15
(12/5) z = 12
Hence, the solution is x = 1, y = 3, z = 5.
8. Apply Gauss Jordan method to solve the following equations.
10x – 7y + 3z + 5u = 6 -6x + 8y – z – 4u = 5 3x + y +4z + 11u = 2 5x – 9y – 2z + 4u = 7
a) u = 2, x = 5, y = 9, z = -7
b) u = 2, x = 7, y = 4, z = 9
c) u = 1, x = 5, y = 3, z = -6
d) u = 1, x = 5, y = 4, z = -7
View Answer
Explanation:
10x – 7y + 3z + 5u = 6 ……………..(i)
– 6x + 8y – z – 4u = 5 ……………..(ii)
3x + y +4z + 11u = 2 ……………….(iii)
5x – 9y – 2z + 4u = 7 ………………(iv)
To eliminate x, operate [(ii) – \((\frac{-6}{10})\) (i)], [(iii) – \((\frac{3}{10})\) (i)], [(iv) – \((\frac{5}{10})\)(i)]
10x – 7y + 3z + 5u = 6 ………………(v)
3.8y + 0.8z – u = 8.6 ……………….(vi)
3.1y + 3.1z + 9.5u = 0.2 …………….(vii)
-5.5y – 3.5z + 1.5u = 4 ……………..(viii)
To eliminate y, operate [(v) – \((\frac{-7}{3.8})\) (vi)], [(vii) – \((\frac{3.1}{3.8})\) (vi)], [(viii) – \((\frac{5.5}{3.8})\) (vi)],
10x + 4.4736z + 3.1578u = 21.8421 ………..(ix)
3.8y + 0.8z – u = 8.6 …………………..(x)
2.4473z + 10.3157u = -6.8157 …………….(xi)
-2.3421z + 0.0526u = 16.4473 …………….(xii)
To eliminate z, operate [(ix) – \((\frac{4.4736}{2.4473})\) (xi)], [(x) – \((\frac{0.8}{2.4473})\) (xi)], [(xii) – \((\frac{-2.3421}{2.4473})\)(xi)],
10x – 15.699u = 34.3010
3.8y – 4.3720u = 10.8279
2.4473z + 10.3157u = -6.8157
9.9247u = 9.9247
By back substitution,
u = 1
x = 5, y = 4, z = -7.
9. Apply Gauss Jordan method to solve the following equations.
x + y + z = 9 2x + y – z = 0 2x + 5y + 7z = 52
a) x = -3.5, y = 40.2, z = 7.625
b) x = -3.5, y = 52.5, z = 4.524
c) x = 2.3, y = 74.4, z = 8.524
d) x = 2.3, y = 85.7, z = 7.625
View Answer
Explanation: Here,
x + y + z = 9 ………………..(i)
2x + y – z = 0 ………………..(ii)
2x + 5y + 7z = 52 ……………..(iii)
To eliminate x, (iii) – 2(i), (ii) – 2(i)
x + y + z = 9 …………………(iv)
y + 5z = 43 ………………….(v)
y – 3z = -18 …………………(vi)
To eliminate y, (iv) – (vi) and (v) – (vi),
x + 4z = 27 …………………..(vii)
y + 5z = 43 …………………..(viii)
8z = 61 ………………………(ix)
To eliminate z, (ix) – 2(vii) and (ix) – (8/5) (viii),
-2x = 7
(-8/5) y = 61 – (104/5)
z = 7.625
Hence, x = -3.5, y = 40.2, z = 7.625.
10. Apply Gauss Jordan method to solve the following equations.
2x + y + z = 10 3x + 2y + 3z = 18 x + 4y + 9z = 16
a) x = 1, y = -9, z =3
b) x =6, y = 6, z =8
c) x = 7, y = -9, z = 5
d) x =7, y = 4, z =1
View Answer
Explanation: Here,
2x + y + z = 10 ………………(i)
3x + 2y + 3z = 18 …………….(ii)
x + 4y + 9z = 16 ……………..(iii)
To eliminate x (i) – 2(iii) and (ii) – 3(iii)
x + 4y + 9z = 16 ……………….(iv)
7y + 17z = 22 ………………….(v)
5y + 12z = 15 ………………….(vi)
To eliminate y, (v) – (7/5) (vi) and (vi) – (5/4) (iv),
3z – 5x = -20 …………………….(vii)
z = 5 ……………………………(viii)
5y + 12z = 15 …………………….(ix)
To eliminate z (ix) – 12(viii) and (vii) – 3(viii),
y = -9
x = 7
Hence, x = 7, y = -9, z = 5.
11. While using Gauss Jordan’s method, after all the elementary row operations if there are zeroes left on the main diagonal, then which of the following is correct?
a) System may have unique solution
b) System has no solution
c) System may have multiple numbers of finite solutions
d) System may have infinitely many solutions
View Answer
Explanation: After applying all the elementary row operations on the system, if the main diagonal still consists of zeroes that means that the system may have infinitely many solutions.
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