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

View Answer

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

View Answer

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

View Answer

Explanation:

∆ = \(\begin{pmatrix}3&1&2\\2&-3&-1\\1&2&1\end{pmatrix}\) = 8

X = (1/∆)\(\begin{pmatrix}3&1&2\\-3&-3&-1\\4&2&1\end{pmatrix}\) = (1/8)8 = 1

Y = (1/∆)\(\begin{pmatrix}3&3&2\\2&-3&-1\\1&4&1\end{pmatrix}\) = (1/8)16 = 2

Z = (1/∆)\(\begin{pmatrix}3&1&3\\2&-3&-3\\1&2&4\end{pmatrix}\) = (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

View Answer

Explanation:

∆ = \(\begin{pmatrix}1&3&6\\3&-1&4\\1&-4&2\end{pmatrix}\) = -58

X = (1/∆) = \(\begin{pmatrix}2&3&6\\9&-1&4\\7&-4&2\end{pmatrix}\) = -116/-58 = 2

Y = (1/∆) = \(\begin{pmatrix}1&2&6\\3&9&4\\1&7&2\end{pmatrix}\) = 58/-58 = -1

Z = (1/∆) = \(\begin{pmatrix}1&3&2\\3&-1&9\\1&-4&7\end{pmatrix}\) = -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

View Answer

Explanation:

∆ = \(\begin{pmatrix}1&1&1\\1&-1&1\\1&2&3\end{pmatrix}\) = -4

X = (1/∆) = \(\begin{pmatrix}6.6&1&1\\2.2&-1&1\\15.2&2&3\end{pmatrix}\) = -4.8/-4 = 1.2

Y = (1/∆) = \(\begin{pmatrix}1&6.6&1\\1&2.2&1\\1&15.2&3\end{pmatrix}\) = -8.8/-4 = 2.2

Z = (1/∆) = \(\begin{pmatrix}1&1&6.6\\1&-1&2.2\\1&2&15.2\end{pmatrix}\) = -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

View Answer

Explanation:

∆ = \(\begin{pmatrix}1&1&1\\1&2&3\\1&4&9\end{pmatrix}\) = 2

X = (1/∆) = \(\begin{pmatrix}3&1&1\\4&2&3\\1&4&9\end{pmatrix}\) = -0.5

Y = (1/∆) = \(\begin{pmatrix}1&3&1\\1&4&3\\1&1&9\end{pmatrix}\) = 6

Z = (1/∆) = \(\begin{pmatrix}1&1&3\\1&2&4\\1&4&1\end{pmatrix}\) = -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

View Answer

Explanation:

∆ = \(\begin{pmatrix}2&-1&1\\3&4&3\\6&7&1\end{pmatrix}\) = -64

x = (1/∆) = \(\begin{pmatrix}3&-1&1\\19&4&3\\17&7&1\end{pmatrix}\) = -28/-64 = 0.437

y = (1/∆) = \(\begin{pmatrix}2&3&1\\3&19&3\\6&17&1\end{pmatrix}\) = -98/-64 = 1.5312

z = (1/∆) = \(\begin{pmatrix}2&-1&3\\3&4&19\\6&7&17\end{pmatrix}\) = -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

View Answer

Explanation:

∆ = \(\begin{pmatrix}3&1&1\\2&-3&-2\\7&2&-5\end{pmatrix}\) = 78

x = (1/∆) = \(\begin{pmatrix}8&1&1\\-5&-3&-2\\0&2&-5\end{pmatrix}\) = 117/78 = 0.2179

y = (1/∆) = \(\begin{pmatrix}3&8&1\\2&-5&-2\\7&0&-5\end{pmatrix}\) = 78/78 = 1

z = (1/∆) = \(\begin{pmatrix}3&1&8\\2&-3&-5\\7&2&5\end{pmatrix}\) = 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

View Answer

Explanation:

∆ = \(\begin{pmatrix}2&1&1\\3&2&3\\1&4&9\end{pmatrix}\) = -2

x = (1/∆) = \(\begin{pmatrix}10&1&1\\18&2&3\\16&4&9\end{pmatrix}\) = -14/-2 = 7

y = (1/∆) = \(\begin{pmatrix}2&10&1\\3&18&3\\1&16&9\end{pmatrix}\) = 18/-2 = -9

z = (1/∆) = \(\begin{pmatrix}2&1&10\\3&2&18\\1&4&16\end{pmatrix}\) = -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

View Answer

Explanation: ∆ = \(\begin{pmatrix}2&-1&3\\1&1&1\\1&-1&1\end{pmatrix}\) = -2

x = (1/∆) = \(\begin{pmatrix}9&-1&3\\6&1&1\\2&-1&1\end{pmatrix}\) = -2/-2 = 1

y = (1/∆) = \(\begin{pmatrix}2&9&3\\1&6&1\\1&2&1\end{pmatrix}\) = -4/-2 = 2

z = (1/∆) = \(\begin{pmatrix}2&-1&9\\1&1&6\\1&-1&2\end{pmatrix}\) = -6/-2 = 3

Hence, X = 1, y = 2, z = 3.

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