"Solution: \n\\\\We ~ know~ that~ Basis ~of ~vector ~space ~V ~is~ a ~linearly~ independent ~set~ that~ spans~ V.\n\\\\ \\therefore ~dimension~ of ~V = Card(basis ~of~ V)\n\\\\Now ~to ~check ~ linearly ~independent~, we ~need ~to~ write~ given ~vectors~ in ~matrix ~\n\\\\form ~and ~reduce~ it ~to ~row ~echelon ~form.\n\\\\ \\therefore ~Number ~of ~non- zero~ rows ~are~ the~ dimensions ~of~ the ~subspace~ spanned ~\\\\by ~the~ given~ vectors.\n\\\\ \\therefore~ We~ form~ a~ matrix ~whose ~rows~ are~ given ~by ~these ~vectors ~and~ reduce~ ~\\\\it ~to~ echelon~ form\n\\\\ \\therefore \\begin{bmatrix}\n1 & 0 & 2\\\\\n2 & 0 & 1\\\\\n1 & 0 & 1\n\\end{bmatrix}\n\\\\Apply~ R_2 -2R_1 \\rightarrow R_2\n\\\\ ~~~~\\begin{bmatrix}\n1 & 0 & 2\\\\\n0 & 0 & -3\\\\\n1 & 0 & 1\n\\end{bmatrix}\n\\\\ Apply ~R_3 -R_1 \\rightarrow ~R_3 \n\\\\ ~~~~\\begin{bmatrix}\n1 & 0 & 2\\\\\n0 & 0 & -3\\\\\n0 & 0 & -1\n\\end{bmatrix}\n\\\\Apply~(R_3\/-3) \\rightarrow~R_3 \n\\\\~~~~\\begin{bmatrix}\n1 & 0 & 2\\\\\n0 & 0 & 1\\\\\n0 & 0 & -1\n\\end{bmatrix}\n\\\\Apply~R_3+R_2 \\rightarrow R_3\n\\\\~~~~\\begin{bmatrix}\n1 & 0 & 2\\\\\n0 & 0 & 1\\\\\n0 & 0 & 0\n\\end{bmatrix}\n\\\\ Here~non-zero~ vectors ~are~2.\n\\\\ \\therefore 2 ~is ~the~ dimension~ of ~the~ subspace~ spanned~ by ~the~given ~vectors~ in V3(R)."
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