JEE MAIN - Mathematics (2022 - 25th June Morning Shift - No. 3)
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 :
Explanation
Let $$A = \left[ {\matrix{ a & b & c \cr d & e & f \cr g & h & i \cr } } \right]$$
$$A = \left[ {\matrix{ 1 \cr 1 \cr 0 \cr } } \right] = \left[ {\matrix{ 1 \cr 1 \cr 0 \cr } } \right] \Rightarrow \left[ {\matrix{ a & b & c \cr d & e & f \cr g & h & i \cr } } \right]\left[ {\matrix{ 1 \cr 1 \cr 0 \cr } } \right] = \left[ {\matrix{ 1 \cr 1 \cr 0 \cr } } \right] \Rightarrow \matrix{ {a + b = 1} \cr {d + e = 1} \cr {g + h = 0} \cr } $$
$$A = \left[ {\matrix{ 1 \cr 0 \cr 1 \cr } } \right] = \left[ {\matrix{ { - 1} \cr 0 \cr 1 \cr } } \right] \Rightarrow \left[ {\matrix{ a & b & c \cr d & e & f \cr g & h & i \cr } } \right]\left[ {\matrix{ 1 \cr 0 \cr 1 \cr } } \right] = \left[ {\matrix{ { - 1} \cr 0 \cr 1 \cr } } \right] \Rightarrow \matrix{ {a + c = - 1} \cr {d + f = 1} \cr {g + i = 0} \cr } $$
$$A = \left[ {\matrix{ 0 \cr 0 \cr 1 \cr } } \right] = \left[ {\matrix{ 1 \cr 1 \cr 2 \cr } } \right] \Rightarrow \left[ {\matrix{ a & b & c \cr d & e & f \cr g & h & i \cr } } \right]\left[ {\matrix{ 0 \cr 0 \cr 1 \cr } } \right] = \left[ {\matrix{ 1 \cr 1 \cr 2 \cr } } \right] \Rightarrow \matrix{ {c = 1} \cr {f = 1} \cr {i = 2} \cr } $$
Solving will get
$$a = - 2,\,b = 3,\,c = 1,\,d = - 1,\,e = 2,\,f = 1,\,g = - 1,\,h = 1,\,i = 2$$
$$A = \left[ {\matrix{ { - 2} & 3 & 1 \cr { - 1} & 2 & 1 \cr { - 1} & 1 & 2 \cr } } \right] \Rightarrow A = 2I = \left[ {\matrix{ { - 4} & 3 & 1 \cr { - 1} & 0 & 1 \cr { - 1} & 1 & 0 \cr } } \right]$$
$$(A - 2I)x = \left[ {\matrix{ 4 \cr 1 \cr 1 \cr } } \right]$$
$$ \Rightarrow - 4{x_1} + 3{x_2} + {x_3} = 4$$ ..... (i)
$$ - {x_1} + {x_3} = 1$$ ...... (ii)
$$ - {x_1} + {x_2} = 1$$ ...... (iii)
So 3(iii) + (ii) = (i)
$$\therefore$$ Infinite solution
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