Let A be a \(3 \times 3\) real matrix such that \(\mathrm{A}\left(\begin{array}{l}1 \\ 0 \\ 1\end{array}\right)=2\left(\begin{array}{l}1 \\ 0 \\ 1\end{array}\right), \mathrm{A}\left(\begin{array}{l}-1 \\ 0 \\ 1\end{array}\right)=4\left(\begin{array}{l}-1 \\ 0 \\ 1\end{array}\right), \mathrm{A}\left(\begin{array}{l}0 \\ 1 \\ 0\end{array}\right)=2\left(\begin{array}{l}0 \\ 1 \\ 0\end{array}\right)\). Then, the system \((A-3 I)\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)\) has

JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Matrices

Let A be a

3 \times 3

real matrix such that

A 101 = 2101, A - 1 01 = 4 - 1 01, A 010 = 2010

.
Then, the system

(A - 3 I) x y z = 123

has

Answer: (A)

The three given vectors

(1, 0, 1)^{T}, (- 1, 0, 1)^{T}, (0, 1, 0)^{T}

are eigenvectors of

A

with eigenvalues

2, 4, 2

respectively, and they form a basis of

R^{3}

(they are linearly independent). So

A

is diagonalisable with eigenvalues

2, 4, 2

.

Then

A - 3 I

has eigenvalues

2 - 3 = - 1, 4 - 3 = 1, 2 - 3 = - 1

— all nonzero.

Since none of the eigenvalues of

A - 3 I

is zero,

det (A - 3 I) = (- 1) (1) (- 1) = 1 \neq = 0

, so

A - 3 I

is invertible. An invertible coefficient matrix means the system

(A - 3 I) x y z = 123

has a unique solution, for any right-hand side.

JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Single correct option