Let A be a \(3 \times 3\) real matrix such that \(\mathrm{A}\left(\begin{array}{l}1 \\ 1 \\ 0\end{array}\right)=\left(\begin{array}{l}1 \\ 1 \\ 0\end{array}\right) ; \mathrm{A}\left(\begin{array}{l}1 \\ 0 \\ 1\end{array}\right)=\left(\begin{array}{c}-1 \\ 0 \\ 1\end{array}\right)\) and \(\mathrm{A}\left(\begin{array}{l}0 \\ 0 \\ 1\end{array}\right)=\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right)\). If \(\mathrm{X}=\left(\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}\right)^{\mathrm{T}}\) and I is an identity matrix of order 3 , then the system \((A-2 I) X=\left(\begin{array}{l}4 \\ 1 \\ 1\end{array}\right)\) has

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Matrices

Let A be a

3 \times 3

real matrix such that

A 110 = 110; A 101 = - 1 01

and

A 001 = 112

.
If

X = (x_{1}, x_{2}, x_{3})^{T}

and I is an identity matrix of order 3 , then the system

(A - 2 I) X = 411

has

Answer: (B)

From the given conditions, we can determine

A = - 2 - 1 - 1 321112

. Then

A - 2 I = - 4 - 1 - 1 301110

. Computing

∣ A - 2 I ∣ = 0

, so the matrix is singular. The system becomes:

- 4 x_{1} + 3 x_{2} + x_{3} = 4

- x_{1} + x_{3} = 1

- x_{1} + x_{2} = 1

. Checking consistency: equation (1) equals

- [(2) + 3 (3)]

, which gives

0 = 0

, confirming the system is consistent with infinitely many solutions.

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Single correct option