250+ TOP MCQs on Eigenvalues and Vectors of a Matrix and Answers

Linear Algebra Multiple Choice Questions on “Eigenvalues and Vectors of a Matrix”.

1. Find the Eigen values for the following 2×2 matrix.
A=(begin{bmatrix}1&8\2&1end{bmatrix}).
a) -3
b) 2
c) 6
d) 4
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&1end{bmatrix} -lambda begin{bmatrix}1&0\0&1end{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.

2. Find the Eigenvalue for the given matrix.
A=(begin{bmatrix}4&1&3\1&3&1\2&0&5end{bmatrix}).
a) 13
b) -3
c) 7.1
d) 8.3
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&5end{bmatrix}-lambda begin{bmatrix}1&0&0\0&1&0\0&0&1end{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&1end{bmatrix}).
a) (begin{bmatrix}0\-1end{bmatrix})
b) (begin{bmatrix}1\-1end{bmatrix})
c) (begin{bmatrix}-1\-1end{bmatrix})
d) (begin{bmatrix}1\0end{bmatrix})
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&1end{bmatrix}-(-2)begin{bmatrix}1&0\0&1end{bmatrix})
[A-λI]=(begin{bmatrix}3+2&5\3&1+2end{bmatrix})
[A-λI]=(begin{bmatrix}5&5\3&3end{bmatrix})
Since, [A-λI]X=0
(begin{bmatrix}5&5\3&3end{bmatrix}
begin{bmatrix}x\yend{bmatrix} = begin{bmatrix}0\0end{bmatrix})
Thus,
5x+5y=0 and 3x+3y=0
Let x=t,
Then, y=-t
X = (begin{bmatrix}t\-tend{bmatrix} = begin{bmatrix}1\-1end{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&7end{bmatrix}).
a) (begin{bmatrix}-1\-1\2end{bmatrix})
b) (begin{bmatrix}-1\1\2end{bmatrix})
c) (begin{bmatrix}-1\-1\-2end{bmatrix})
d) (begin{bmatrix}-1\-2\2end{bmatrix})
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&7end{bmatrix}-(3)begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})
[A-λI]=(begin{bmatrix}3-3&10&5\-2&-3-3&-4\3&5&7-3end{bmatrix})
[A-λI]=(begin{bmatrix}0&10&5\-2&-6&-4\3&5&4end{bmatrix})
Since, [A-λI]X=0
(begin{bmatrix}0&10&5\-2&-6&-4\3&5&4end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}0\0\0end{bmatrix})
Using Row transformation

(1) Interchanging R1 and R2/2
(begin{bmatrix}-1&-3&-2\0&10&5\3&5&4end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}0\0\0end{bmatrix})
(2) R3=R3+3R1
(begin{bmatrix}0&10&5\0&10&5\0&-4&-2end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}0\0\0end{bmatrix})
(3) R2=R2/5 and R3=R3+2R2
(begin{bmatrix}-1&-3&-2\0&2&1\0&0&0end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}0\0\0end{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\2end{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&4end{bmatrix}).
a) 3, (begin{bmatrix}1\1\1end{bmatrix})
b) 9, (begin{bmatrix}1\1\1end{bmatrix})
c) 9, (begin{bmatrix}1\0\1end{bmatrix})
d) 2, (begin{bmatrix}1\0\1end{bmatrix})
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&4end{bmatrix}-lambda begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})
The characteristic polynomial is given by-
λ3-(Sum of diagonal elements) λ2+(Sum of minor of diagonal element)λ-|A|=0

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&4end{bmatrix}-9begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})
[A-λI]=(begin{bmatrix}3-9&4&2\1&6-9&2\1&4&4-9end{bmatrix})
[A-λI]=(begin{bmatrix}-6&4&2\1&-3&2\1&4&-5end{bmatrix})
[A-λI]X=0
(begin{bmatrix}-6&4&2\1&-3&2\1&4&-5end{bmatrix}
begin{bmatrix}x\y\zend{bmatrix} = begin{bmatrix}0\0\0end{bmatrix})

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

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

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

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

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