1
Question
The matrix A = ⎡⎢
⎢
⎢⎣320120−1012032⎤⎥
⎥
⎥⎦ has three distinct eigen values and one of its eigen vectors is ⎡⎢⎣101⎤⎥⎦
which one of the following can be anooter eigen vector of A?
The matrix A = ⎡⎢
⎢
⎢⎣320120−1012032⎤⎥
⎥
⎥⎦ has three distinct eigen values and one of its eigen vectors is ⎡⎢⎣101⎤⎥⎦
which one of the following can be anooter eigen vector of A?
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Solution
The correct option is C ⎡⎢⎣10−1⎤⎥⎦
A = ⎡⎢
⎢
⎢⎣320120−1012032⎤⎥
⎥
⎥⎦
To find the eigen values of A, det(A−λI) = 0 i.e.,
∣∣
∣
∣∣32−λ 0120−1−λ012032−λ ∣∣
∣
∣∣ = 0
(32−λ)[(−1−λ)(32−λ)]+12[0−12(−1−λ)] = 0
(−1−λ)[(32−λ)2−14] = 0
−(1+λ)[λ2−3λ+2] = 0
or −(1+λ)(λ−1)(λ−2) = 0
Hence eigen values of A are -1, 1 and 2.
Now, let X be an eigen vector of A associated to λ, then
AX = λX
So, ⎡⎢
⎢
⎢⎣320120−1012032⎤⎥
⎥
⎥⎦ ⎡⎢⎣101⎤⎥⎦ = λ⎡⎢⎣101⎤⎥⎦ On solving it λ = 2 so given Eigen vector corresponds to λ = 2
Thus for λ = +1 by taking X = ⎡⎢⎣xyz⎤⎥⎦, we have
AX = λX
(A-I)X = 0
⎡⎢
⎢⎣120120−2012012⎤⎥
⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣000⎤⎥⎦
⇒ 12x +12z = 0
& -2y = 0
So y = 0 & x = -z. Let z = k then x = -k
So Eigen Vector for λ = 1,
X = ⎡⎢⎣−K0K⎤⎥⎦∼⎡⎢⎣−101⎤⎥⎦∼⎡⎢⎣10−1⎤⎥⎦
Hence option (c) satisfies.
A = ⎡⎢ ⎢ ⎢⎣320120−1012032⎤⎥ ⎥ ⎥⎦
To find the eigen values of A, det(A−λI) = 0 i.e.,
∣∣ ∣ ∣∣32−λ 0120−1−λ012032−λ ∣∣ ∣ ∣∣ = 0
(32−λ)[(−1−λ)(32−λ)]+12[0−12(−1−λ)] = 0
(−1−λ)[(32−λ)2−14] = 0
−(1+λ)[λ2−3λ+2] = 0
or −(1+λ)(λ−1)(λ−2) = 0
Hence eigen values of A are -1, 1 and 2.
Now, let X be an eigen vector of A associated to λ, then
AX = λX
So, ⎡⎢ ⎢ ⎢⎣320120−1012032⎤⎥ ⎥ ⎥⎦ ⎡⎢⎣101⎤⎥⎦ = λ⎡⎢⎣101⎤⎥⎦ On solving it λ = 2 so given Eigen vector corresponds to λ = 2
Thus for λ = +1 by taking X = ⎡⎢⎣xyz⎤⎥⎦, we have
AX = λX
(A-I)X = 0
⎡⎢ ⎢⎣120120−2012012⎤⎥ ⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣000⎤⎥⎦
⇒ 12x +12z = 0
& -2y = 0
So y = 0 & x = -z. Let z = k then x = -k
So Eigen Vector for λ = 1,
X = ⎡⎢⎣−K0K⎤⎥⎦∼⎡⎢⎣−101⎤⎥⎦∼⎡⎢⎣10−1⎤⎥⎦
Hence option (c) satisfies.
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