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

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no solution

infinitely many solutions

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exactly two solutions

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Detailed Solution

$A = [\begin{array}{l} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}] \Rightarrow A [\begin{array}{l} 0 \\ 0 \\ 1 \end{array}] = [\begin{array}{l} c_{1} \\ c_{2} \\ c_{3} \end{array}] = [\begin{array}{l} 1 \\ 1 \\ 2 \end{array}] \Rightarrow c_{1} = 1, c_{2} = 1, c_{3} = 2 A [\begin{array}{l} 1 \\ 0 \\ 1 \end{array}] = [\begin{array}{l} c_{1} + a_{1} \\ c_{2} + a_{2} \\ c_{3} + a_{3} \end{array}] = [\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix}] \Rightarrow a_{1} = - 2, a_{2} = - 1, a_{3} = - 1 A [\begin{array}{l} 1 \\ 1 \\ 0 \end{array}] = [\begin{array}{l} a_{1} + b_{1} \\ a_{2} + b_{2} \\ a_{3} + b_{3} \end{array}] = [\begin{array}{l} 1 \\ 1 \\ 0 \end{array}] \Rightarrow b_{1} = 3, b_{2} = 2, b_{3} = 1 \Rightarrow A = [\begin{array}{l} - 2 & 3 & 1 \\ - 1 & 2 & 1 \\ - 1 & 1 & 2 \end{array}] \Rightarrow A - 2 I = [\begin{array}{l} - 4 & 3 & 1 \\ - 1 & 0 & 1 \\ - 1 & 1 & 0 \end{array}] | A - 2 I | = 0 Now, [\begin{array}{l} - 4 & 3 & 1 \\ - 1 & 0 & 1 \\ - 1 & 1 & 0 \end{array}] [\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}] = [\begin{array}{l} 4 \\ 1 \\ 1 \end{array}] - 4 x_{1} + 3 x_{2} + x_{3} = 4 (1) - x_{1} + x_{3} = 1 (2) - x_{1} + x_{2} = 1 (3) (1) - [(2) + 3 (3)] 0 = 0 \Rightarrow infinite solutions$