Matrix Inversion Questions and Answers – Cramer’s Rule

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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

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.

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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

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
View Answer

Answer: a
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

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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

Answer: b
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.

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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

Answer: c
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

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

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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

Answer: b
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

Answer: c
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

Answer: b
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

Answer: a
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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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn