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

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.

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.

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 = \(\begin{bmatrix}t\\-t\end{bmatrix} = \begin{bmatrix}1\\-1\end{bmatrix}\)

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

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

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

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=\(\begin{bmatrix}1\\1\\1\end{bmatrix}\).