JEE Main 202431 Jan, Evening shiftMathematicsHardMatrices and Determinants245× since 2002Let AAA be a 3×33 \times 33×3 real matrix such that A(101)=2(101),A(−101)=4(−101),A(010)=2(010). A\left(\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right)=2\left(\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right), A\left(\begin{array}{l} -1 \\ 0 \\ 1 \end{array}\right)=4\left(\begin{array}{l} -1 \\ 0 \\ 1 \end{array}\right), A\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right)=2\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right) \text {. }A101=2101,A−101=4−101,A010=2010. Then, the system (A−3I)(xyz)=(123)(A-3 I)\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)(A−3I)xyz=123 has :AOption A. exactly two solutionsBOption B. infinitely many solutionsCOption C. unique solutionDOption D. no solutionCheckSkip · show solution
