Let \(\mathrm{A}=\mathrm{I}_{2}-2 \mathrm{MM}^{\mathrm{T}}\), where M is real matrix of order \(2 \times 1\) such that the relation \(\mathrm{M}^{\mathrm{T}} \mathrm{M}=\mathrm{I}_{1}\) holds. If \(\lambda\) is a real number such that the relation \(\mathrm{AX}=\lambda \mathrm{X}\) holds for some non-zero real matrix X of order \(2 \times 1\), then the sum of squares of all possible values of \(\lambda\) is equal to :

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Matrices

Let

A = I_{2} - 2 MM^{T}

, where M is real matrix of order

2 \times 1

such that the relation

M^{T} M = I_{1}

holds. If

λ

is a real number such that the relation

AX = λ X

holds for some non-zero real matrix X of order

2 \times 1

, then the sum of squares of all possible values of

λ

is equal to :

Answer: 2

M

is a real

2 \times 1

matrix with

M^{T} M = I_{1}

, i.e.

M^{T} M = 1

(a scalar), so

A = I_{2} - 2 M M^{T}

A^{2} = (I - 2 M M^{T}) (I - 2 M M^{T}) = I - 4 M M^{T} + 4 M M^{T} M M^{T} = I - 4 M M^{T} + 4 M (M^{T} M) M^{T} .

Since

M^{T} M = 1

A^{2} = I - 4 M M^{T} + 4 M M^{T} = I .

Given

A X = λ X

for some nonzero

X

, apply

A

again:

A^{2} X = λ A X = λ^{2} X .

But

A^{2} = I

, so

A^{2} X = X

. Hence

λ^{2} X = X

, and since

X \neq = 0

λ^{2} = 1 ⟹ λ = \pm 1.

Sum of squares of all possible values of

λ

1^{2} + (- 1)^{2} = 2

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Integer answer