# Matrix Inversion Questions and Answers – Gauss Elimination Method – 2

This set of Numerical Methods Multiple Choice Questions & Answers focuses on “Gauss Elimination Method – 2”.

1. Which of the following step is not involved in Gauss Elimination Method?
a) Elimination of unknowns
b) Reduction to an upper triangular system
c) Finding unknowns by back substitution
d) Evaluation of cofactors

Explanation: Elimination of unknowns, reduction to an upper triangular system and finding unknowns by back substitution are the primary steps involved in Gauss Elimination.

2. Gauss Elimination Method is well adopted for which of the application?
a) Computer operations
b) Network circuit problems
c) MATLAB operations
d) Telecommunication operations

Explanation: Gauss Elimination Method is well adopted for computer operations. It is an effective and fast process.

3. What are the coefficients of the equation obtained during the elimination called?
a) Joints
b) Pivots
c) Calculated coefficients
d) Operative coefficients

Explanation: The coefficients of the equation obtained during the elimination are called as pivots.

4. How the transformation of coefficient matrix A to upper triangular matrix is done?
a) Elementary row transformations
b) Elementary column transformations
c) Successive multiplication
d) Successive division

Explanation: The transformation of coefficient matrix A to upper triangular matrix is done through elementary row transformations. We will not lead towards the correct answer if we’ll perform transformations in rows as well as columns.

5. How many types of pivoting are there?
a) 2
b) 3
c) 4
d) 5

Explanation: There are two types of pivoting, namely, partial and complete pivoting.
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6. The modified procedure of complete pivoting is called as ____________
a) Partial
c) Reduced
d) Modified

Explanation: The modified procedure of complete pivoting is called as Partial Pivoting.

7. Apply Gauss Elimination method to solve the following equations.

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

a) x = 1.6479, y = -1.1408, z = 2.0845
b) x = 4.0461, y = -1.1408, z = 3.254
c) x = 7.2478, y = -2.586, z = 8.265
d) x = 2.8471, y = 5.5123, z = 2.0845

Explanation: x + 4y – z = -5 ………(i)
x + y – 6z = -12 …………………..(ii)
3x – y – z = 4 …………………….(iii)

To eliminate x, operate (ii) – (i) and (iii) – 3(i),

-3y – 5z = -7 ………………………(iv)
-13 + 2z = 19 ………………………(v)
To eliminate y, (v) – (13/3)(iv),
(71/3) z = (148/3)
Now by back substitution,
Z = $$\frac{148}{71}$$ = 2.0845
Y = $$\frac{7}{3} – \frac{5}{3}(\frac{148}{71})$$
= $$\frac{-81}{71}$$ = -1.1408
X = -5 -4$$(\frac{-81}{71}) + (\frac{148}{71})$$
= $$\frac{117}{71}$$ = 1.6479
Hence x = 1.6479, y = -1.1408, z = 2.0845.

8. Apply Gauss Elimination 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 = 1, z = -7, y = 4, x = 5
b) u = 1, z = -7, y = 4, x = 5
c) u = 1, z = -7, y = 4, x = 5
d) u = 1, z = -7, y = 4, x = 5

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)] and [(iv) – $$\frac{5}{10}$$(i)]

3.8y + 0.8z – u = 8.6 ……………………(v)
3.1y + 3.1z + 9.5u = 0.2 …………………(vi)
-5.5y – 3.5z + 1.5u = 4 ………………….(vii)

To eliminate y, operate [(vi) – $$\frac{3.1}{3.8}$$(v)], [(vii) – $$\frac{5.5}{3.8}$$(v)],

2.447z + 10.31u = -6.815 …………………(viii)
-2.342z + 0.052u = 16.44 …………………(ix)

To eliminate z operate [(ix) – $$(\frac{-2.342}{2.447})$$ (viii)] ,
9.924u = 9.924
By back substitution,
u = 1, z = -7, y = 4, x = 5.

9. Apply Gauss Elimination method to solve the following equations.

2x + y + z = 10
3x + 2y + 3z = 18
X + 4y + 9z = 16

a) X = 7, y = -4, z = 5
b) X = 7, y = -9, z = 5
c) X = 5, y = 1, z = -8
d) X = 5, y = 1, z = -3

Explanation:
2x + y + z = 10 ………………….(i)
3x + 2y + 3z = 18 ………………..(ii)
x + 4y + 9z = 16 …………………(iii)

To eliminate x, operate (i) – 2(iii), (ii) – 3(iii)

7y + 17z = 22 ……………………(iv)
5y + 12z = 15 ……………………(v)

To eliminate y, operate [(iv) – $$\frac{7}{5}$$(v)]
0.2z = 1
By back substitution,
z = 5
7y = 22 – 85
y= -9
x = 16 + 36 – 45
x = 7
Hence, x = 7, y = -9, z = 5.

10. Apply Gauss Elimination method to solve the following equations.

2x – y + 3z = 9
x + y + z = 6
x – y + z = 2

a) X = 5, y = 14, z = 5
b) X = -13, y = 4, z = 15
c) X = -13, y = 1, z = -8
d) X = 13, y = 1, z = -8

Explanation: 2x – y + 3z = 9 ……….(i)
x + y + z = 6 ……………………(ii)
x – y + z = 2 ……………………(iii)

To eliminate x, operate (ii) – (iii)
y = 4
Now, operate (i) – 2(ii),
-3y + z = 3
Now by back substitution,
Z = 15
X + 4 +15 = 6
X = -13.

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