Let A be a 3 $\times$ 3 real matrix such that

$$A\left( {\matrix{ 1 \cr 1 \cr 0 \cr } } \right) = \left( {\matrix{ 1 \cr 1 \cr 0 \cr } } \right);A\left( {\matrix{ 1 \cr 0 \cr 1 \cr } } \right) = \left( {\matrix{ { - 1} \cr 0 \cr 1 \cr } } \right)$$ and $$A\left( {\matrix{ 0 \cr 0 \cr 1 \cr } } \right) = \left( {\matrix{ 1 \cr 1 \cr 2 \cr } } \right)$$.

If $X = {({x_1},{x_2},{x_3})^T}$ and I is an identity matrix of order 3, then the system $(A - 2I)X = \left( {\matrix{ 4 \cr 1 \cr 1 \cr } } \right)$ has :

<p>Let $$A = \left[ {\matrix{ a & b & c \cr d & e & f \cr g & h & i \cr } } \right]$$</p> <p>$$A = \left[ {\matrix{ 1 \cr 1 \cr 0 \cr } } \right] = \left[ {\matrix{ 1 \cr 1 \cr 0 \cr } } \right] \Rightarrow \left[ {\matrix{ a & b & c \cr d & e & f \cr g & h & i \cr } } \right]\left[ {\matrix{ 1 \cr 1 \cr 0 \cr } } \right] = \left[ {\matrix{ 1 \cr 1 \cr 0 \cr } } \right] \Rightarrow \matrix{ {a + b = 1} \cr {d + e = 1} \cr {g + h = 0} \cr } $$</p> <p>$$A = \left[ {\matrix{ 1 \cr 0 \cr 1 \cr } } \right] = \left[ {\matrix{ { - 1} \cr 0 \cr 1 \cr } } \right] \Rightarrow \left[ {\matrix{ a & b & c \cr d & e & f \cr g & h & i \cr } } \right]\left[ {\matrix{ 1 \cr 0 \cr 1 \cr } } \right] = \left[ {\matrix{ { - 1} \cr 0 \cr 1 \cr } } \right] \Rightarrow \matrix{ {a + c = - 1} \cr {d + f = 1} \cr {g + i = 0} \cr } $$</p> <p>$$A = \left[ {\matrix{ 0 \cr 0 \cr 1 \cr } } \right] = \left[ {\matrix{ 1 \cr 1 \cr 2 \cr } } \right] \Rightarrow \left[ {\matrix{ a & b & c \cr d & e & f \cr g & h & i \cr } } \right]\left[ {\matrix{ 0 \cr 0 \cr 1 \cr } } \right] = \left[ {\matrix{ 1 \cr 1 \cr 2 \cr } } \right] \Rightarrow \matrix{ {c = 1} \cr {f = 1} \cr {i = 2} \cr } $$</p> <p>Solving will get</p> <p>$a = - 2,\,b = 3,\,c = 1,\,d = - 1,\,e = 2,\,f = 1,\,g = - 1,\,h = 1,\,i = 2$</p> <p>$$A = \left[ {\matrix{ { - 2} & 3 & 1 \cr { - 1} & 2 & 1 \cr { - 1} & 1 & 2 \cr } } \right] \Rightarrow A = 2I = \left[ {\matrix{ { - 4} & 3 & 1 \cr { - 1} & 0 & 1 \cr { - 1} & 1 & 0 \cr } } \right]$$</p> <p>$(A - 2I)x = \left[ {\matrix{ 4 \cr 1 \cr 1 \cr } } \right]$</p> <p>$\Rightarrow - 4{x_1} + 3{x_2} + {x_3} = 4$ ..... (i)</p> <p>$- {x_1} + {x_3} = 1$ ...... (ii)</p> <p>$- {x_1} + {x_2} = 1$ ...... (iii)</p> <p>So 3(iii) + (ii) = (i)</p> <p>$\therefore$ Infinite solution</p>