Linear Algebra Questions and Answers – Eigenvalues and Vectors of a Matrix

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This set of Linear Algebra Multiple Choice Questions & Answers (MCQs) focuses on “Eigenvalues and Vectors of a Matrix”.

1. Find the Eigen values for the following 2×2 matrix.
A=\(\begin{bmatrix}1&8\\2&1\end{bmatrix}\).
a) -3
b) 2
c) 6
d) 4
View Answer

Answer: a
Explanation: We know that for any given matrix
[A-λI]X=0 and |A-λI|=0
[A-λI]=\(\begin{bmatrix}1&8\\2&1\end{bmatrix} -\lambda \begin{bmatrix}1&0\\0&1\end{bmatrix}\)
[A-λI]=\(\begin{bmatrix}1-\lambda &8\\2&1-\lambda \end{bmatrix}\)
|A-λI|=(1-λ)(1-λ)-16=0
(1-λ)2=16
(1-λ)=±4
λ=-3 or λ=5.
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2. Find the Eigenvalue for the given matrix.
A=\(\begin{bmatrix}4&1&3\\1&3&1\\2&0&5\end{bmatrix}\).
a) 13
b) -3
c) 7.1
d) 8.3
View Answer

Answer: c
Explanation: We know that for any given matrix
[A-λI]X=0 and |A-λI|=0
[A-λI]=\(\begin{bmatrix}4&1&3\\1&3&1\\2&0&5\end{bmatrix}-\lambda \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)
The characteristic polynomial is given by-
λ3-(Sum of diagonal elements) λ2+(Sum of minor of diagonal element)λ-|A|=0
λ3-12λ2+40λ-39=0
λ=7.1 or λ=3.

3. Find the Eigen vector for value of λ=-2 for the given matrix.
A=\(\begin{bmatrix}3&5\\3&1\end{bmatrix}\).
a) \(\begin{bmatrix}0\\-1\end{bmatrix}\)
b) \(\begin{bmatrix}1\\-1\end{bmatrix}\)
c) \(\begin{bmatrix}-1\\-1\end{bmatrix}\)
d) \(\begin{bmatrix}1\\0\end{bmatrix}\)
View Answer

Answer: b
Explanation: We know that for any given matrix
[A-λI]X=0 and |A-λI|=0
Given that, λ=-2
[A-λI]=\(\begin{bmatrix}3&5\\3&1\end{bmatrix}-(-2)\begin{bmatrix}1&0\\0&1\end{bmatrix}\)
[A-λI]=\(\begin{bmatrix}3+2&5\\3&1+2\end{bmatrix}\)
[A-λI]=\(\begin{bmatrix}5&5\\3&3\end{bmatrix}\)
Since, [A-λI]X=0
\(\begin{bmatrix}5&5\\3&3\end{bmatrix}
\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix}\)
Thus,
5x+5y=0 and 3x+3y=0
Let x=t,
Then, y=-t
X = \(\begin{bmatrix}t\\-t\end{bmatrix} = \begin{bmatrix}1\\-1\end{bmatrix}\)

4. Find the Eigen vector for value of λ=3 for the given matrix.
A=\(\begin{bmatrix}3&10&5\\-2&-3&-4\\3&5&7\end{bmatrix}\).
a) \(\begin{bmatrix}-1\\-1\\2\end{bmatrix}\)
b) \(\begin{bmatrix}-1\\1\\2\end{bmatrix}\)
c) \(\begin{bmatrix}-1\\-1\\-2\end{bmatrix}\)
d) \(\begin{bmatrix}-1\\-2\\2\end{bmatrix}\)
View Answer

Answer: a
Explanation: We know that for any given matrix
[A-λI]X=0 and |A-λI|=0
Given that, λ=3
[A-λI]=\(\begin{bmatrix}3&10&5\\-2&-3&-4\\3&5&7\end{bmatrix}-(3)\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)
[A-λI]=\(\begin{bmatrix}3-3&10&5\\-2&-3-3&-4\\3&5&7-3\end{bmatrix}\)
[A-λI]=\(\begin{bmatrix}0&10&5\\-2&-6&-4\\3&5&4\end{bmatrix}\)
Since, [A-λI]X=0
\(\begin{bmatrix}0&10&5\\-2&-6&-4\\3&5&4\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}\)
Using Row transformation

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(1) Interchanging R1 and R2/2
\(\begin{bmatrix}-1&-3&-2\\0&10&5\\3&5&4\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}\)
(2) R3=R3+3R1
\(\begin{bmatrix}0&10&5\\0&10&5\\0&-4&-2\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}\)
(3) R2=R2/5 and R3=R3+2R2
\(\begin{bmatrix}-1&-3&-2\\0&2&1\\0&0&0\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}\)
Thus,
-x-3y-2z=0 and 2y+z=0
Let z=t,then y=\(\frac{-t}{2}\) and x=\(\frac{-t}{2}\)

X = \(\begin{bmatrix}-1\\-1\\2\end{bmatrix}\)

5. Find the Eigen value and the Eigen Vector for the given matrix.
A=\(\begin{bmatrix}3&4&2\\1&6&2\\1&4&4\end{bmatrix}\).
a) 3, \(\begin{bmatrix}1\\1\\1\end{bmatrix}\)
b) 9, \(\begin{bmatrix}1\\1\\1\end{bmatrix}\)
c) 9, \(\begin{bmatrix}1\\0\\1\end{bmatrix}\)
d) 2, \(\begin{bmatrix}1\\0\\1\end{bmatrix}\)
View Answer

Answer: b
Explanation: We know that for any given matrix
[A-λI]X=0 and |A-λI|=0
[A-λI]=\(\begin{bmatrix}3&4&2\\1&6&2\\1&4&4\end{bmatrix}-\lambda \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)
The characteristic polynomial is given by-
λ3-(Sum of diagonal elements) λ2+(Sum of minor of diagonal element)λ-|A|=0

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λ3-13λ2+40λ-36=0
λ=9 or λ=2
For λ=9,
[A-λI]=\(\begin{bmatrix}3&4&2\\1&6&2\\1&4&4\end{bmatrix}-9\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)
[A-λI]=\(\begin{bmatrix}3-9&4&2\\1&6-9&2\\1&4&4-9\end{bmatrix}\)
[A-λI]=\(\begin{bmatrix}-6&4&2\\1&-3&2\\1&4&-5\end{bmatrix}\)
[A-λI]X=0
\(\begin{bmatrix}-6&4&2\\1&-3&2\\1&4&-5\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}\)

(1) Interchanging R1 and R2
\(\begin{bmatrix}1&-3&2\\-6&4&2\\1&4&-5\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}\)

(2) R2=R2+6R1 and R3=R3-R1
\(\begin{bmatrix}1&-3&2\\0&-14&14\\0&7&-7\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}\)

(3) R3=R3+R2/2
\(\begin{bmatrix}1&-3&2\\0&-14&14\\0&0&0\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}\)

-14y+14z=0
x-3y+2z=0
Let z=t
Then, y=t and x=t
X=\(\begin{bmatrix}1\\1\\1\end{bmatrix}\).

Sanfoundry Global Education & Learning Series – Linear Algebra.

To practice all areas of Linear Algebra, here is complete set of 1000+ Multiple Choice Questions and Answers.

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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