Using Julia or Jupyter Language please. 7. Demo 2.3.6 showed solutions of Ax=b, where A=?10000??11000?0?1100????0?110??00?11??,b=??0001?? Use Function 2.3.5 to solve with ?=0.1 and ?=10,100,103,,1012, tabulating the values of ? and ?x1??1?. (This kind of behavior is explained in Section 2.8.) Backward substitution to solve an upper triang

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Using Julia or Jupyter Language please.       7. Demo 2.3.6 showed solutions of Ax=b, where A=?10000??11000?0?1100????0?110??00?11??,b=??0001?? Use Function 2.3.5 to solve with ?=0.1 and ?=10,100,103,,1012, tabulating the values of ? and ?x1??1?. (This kind of behavior is explained in Section 2.8.)  Backward substitution to solve an upper triangular linear system

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