This set of Numerical Methods Multiple Choice Questions & Answers (MCQs) focuses on “Cramer’s Rule”.
1. Cramer’s Rule fails for ___________
a) Determinant > 0
b) Determinant < 0
c) Determinant = 0
d) Determinant = non-real
Answer: c
Explanation: This is because Cramer’s rule involves division by determinant which should never be equal to 0 leading to not defined numbers.
2. Cramer’s Rule is not suitable for which type of problems?
a) Small systems with 4 unknowns
b) Systems with 2 unknowns
c) Large systems
d) Systems with 3 unknowns
Answer: c
Explanation: Generally, in large systems, excessive multiplicative operations are required which becomes very cumbersome to solve.
3. Apply Cramer’s rule to solve the following equations.
3x + y + 2z = 3
2x – 3y –z = -3
X +2y +z = 4
a) X = 1, y = 2, z = -1
b) X = 2, y = 1, z = -1
c) X = 2, y = -1, z = 1
d) X = 1, y = -1, z = 2
Answer: a
Explanation:
∆ = (begin{pmatrix}3&1&2\2&-3&-1\1&2&1end{pmatrix}) = 8
X = (1/∆)(begin{pmatrix}3&1&2\-3&-3&-1\4&2&1end{pmatrix}) = (1/8)8 = 1
Y = (1/∆)(begin{pmatrix}3&3&2\2&-3&-1\1&4&1end{pmatrix}) = (1/8)16 = 2
Z = (1/∆)(begin{pmatrix}3&1&3\2&-3&-3\1&2&4end{pmatrix}) = (1/8)(-8) = -1
Hence, x = 1, y = 2, z = -1.
4. Apply Cramer’s rule to solve the following equations.
x + 3y + 6z = 2
3x – y + z = 9
X – 4y + 2z = 7
a) X = 1, y = 2, z = -1
b) X = 2, y = – 1, z = -0.5
c) X = 1, y = 2, z = -0.5
d) X = 2, y = 2, z = -1
Answer: b
Explanation:
∆ = (begin{pmatrix}1&3&6\3&-1&4\1&-4&2end{pmatrix}) = -58
X = (1/∆) = (begin{pmatrix}2&3&6\9&-1&4\7&-4&2end{pmatrix}) = -116/-58 = 2
Y = (1/∆) = (begin{pmatrix}1&2&6\3&9&4\1&7&2end{pmatrix}) = 58/-58 = -1
Z = (1/∆) = (begin{pmatrix}1&3&2\3&-1&9\1&-4&7end{pmatrix}) = -29/-58 = 0.5
Hence, x = 2, y = -1, z = -0.5.
5. Apply Cramer’s rule to solve the following equations.
x + y + z = 6.6
x – y + z = 2.2
x + 2y + 3z = 15.2
a) x = 1.5, y = 2.2, z = -0.5
b) x = 1.5, y = 2.2, z = -0.5
c) x = 1.2, y = 2, z = 3.2
d) x = 1.2, y = 2.2, z = -3.2
Answer: d
Explanation:
∆ = (begin{pmatrix}1&1&1\1&-1&1\1&2&3end{pmatrix}) = -4
X = (1/∆) = (begin{pmatrix}6.6&1&1\2.2&-1&1\15.2&2&3end{pmatrix}) = -4.8/-4 = 1.2
Y = (1/∆) = (begin{pmatrix}1&6.6&1\1&2.2&1\1&15.2&3end{pmatrix}) = -8.8/-4 = 2.2
Z = (1/∆) = (begin{pmatrix}1&1&6.6\1&-1&2.2\1&2&15.2end{pmatrix}) = -12.8/-4 = 3.2
Hence, x = 1.2, y = 2.2, z = 3.2.
6. Apply Cramer’s rule to solve the following equations.
x + y + z =3
x + 2y + 3z = 4
x + 4y + 9z = 1
a) x = -0.5, y = 6, z = -2.5
b) x = -0.5, y = 4, z = -2.5
c) x = 4.5, y = 6, z = 1
d) x = 4.5, y = 6, z = 2
Answer: a
Explanation:
∆ = (begin{pmatrix}1&1&1\1&2&3\1&4&9end{pmatrix}) = 2
X = (1/∆) = (begin{pmatrix}3&1&1\4&2&3\1&4&9end{pmatrix}) = -0.5
Y = (1/∆) = (begin{pmatrix}1&3&1\1&4&3\1&1&9end{pmatrix}) = 6
Z = (1/∆) = (begin{pmatrix}1&1&3\1&2&4\1&4&1end{pmatrix}) = -2.5
Hence, X = -0.5, y = 6, z = -2.5.
7. Apply Cramer’s rule to solve the following equations.
2x – y + z = 3
3x + 2y + 4z = 19
6x + 7y – z = 17
a) X = 0.456, y = 1.5442, z = 3.154
b) X = 0.437, y = 1.5312, z = 3.656
c) X = 0.356, y =2.547, z = 5.474
d) X = 0.356, y = 1.722, z = 9.424
Answer: b
Explanation:
∆ = (begin{pmatrix}2&-1&1\3&4&3\6&7&1end{pmatrix}) = -64
x = (1/∆) = (begin{pmatrix}3&-1&1\19&4&3\17&7&1end{pmatrix}) = -28/-64 = 0.437
y = (1/∆) = (begin{pmatrix}2&3&1\3&19&3\6&17&1end{pmatrix}) = -98/-64 = 1.5312
z = (1/∆) = (begin{pmatrix}2&-1&3\3&4&19\6&7&17end{pmatrix}) = -234/-64 = 3.656
Hence, X = 0.437, y = 1.5312, z = 3.656.
8. Apply Cramer’s rule to solve the following equations.
3x + y + z = 8
2x – 3y -2z = -5
7x + 2y – 5z = 0
a) X = 1, y =4, z = 2.5
b) X = 4.562, y =4, z = 3.1
c) X = 0.2179, y =1, z = 2.5
d) X = 4.2, y =4, z = 3.145
Answer: c
Explanation:
∆ = (begin{pmatrix}3&1&1\2&-3&-2\7&2&-5end{pmatrix}) = 78
x = (1/∆) = (begin{pmatrix}8&1&1\-5&-3&-2\0&2&-5end{pmatrix}) = 117/78 = 0.2179
y = (1/∆) = (begin{pmatrix}3&8&1\2&-5&-2\7&0&-5end{pmatrix}) = 78/78 = 1
z = (1/∆) = (begin{pmatrix}3&1&8\2&-3&-5\7&2&5end{pmatrix}) = 195/78 = 2.5
Hence, X = 0.2179, y =1, z = 2.5.
9. Apply Cramer’s rule to solve the following equations.
2x + y + z = 10
3x + 2y + 3z = 18
X + 4y +9z = 16
a) X = -9, y = 1, z = 5
b) X = 7, y = -9, z = 5
c) X = 7, y = 1, z = 5
d) X = 9, y = 1, z = 3
Answer: b
Explanation:
∆ = (begin{pmatrix}2&1&1\3&2&3\1&4&9end{pmatrix}) = -2
x = (1/∆) = (begin{pmatrix}10&1&1\18&2&3\16&4&9end{pmatrix}) = -14/-2 = 7
y = (1/∆) = (begin{pmatrix}2&10&1\3&18&3\1&16&9end{pmatrix}) = 18/-2 = -9
z = (1/∆) = (begin{pmatrix}2&1&10\3&2&18\1&4&16end{pmatrix}) = -10/-2 = 5
Hence, X = 7, y = -9, z = 5.
10. Apply Cramer’s rule to solve the following equations.
2x – y + 3z = 9
x + y + z = 6
x – y + z = 2
a) x = 1, y = 2, z = 3
b) x = 2, y = 2, z = 3
c) x = 2, y = 3, z = 7
d) x = 1, y = 3, z = 8
Answer: a
Explanation: ∆ = (begin{pmatrix}2&-1&3\1&1&1\1&-1&1end{pmatrix}) = -2
x = (1/∆) = (begin{pmatrix}9&-1&3\6&1&1\2&-1&1end{pmatrix}) = -2/-2 = 1
y = (1/∆) = (begin{pmatrix}2&9&3\1&6&1\1&2&1end{pmatrix}) = -4/-2 = 2
z = (1/∆) = (begin{pmatrix}2&-1&9\1&1&6\1&-1&2end{pmatrix}) = -6/-2 = 3
Hence, X = 1, y = 2, z = 3.