Question: Using Julia or Jupyter Language please. 7. Demo 2.3.6 showed solutions of Ax=b, where A=10000110000110001100011,b=0001 Use Function 2.3.5 to solve with =0.1 and =10,100,103,,1012, tabulating

Using Julia or Jupyter Language please.

Ax=b, where A=10000110000110001100011,b=0001 Use Function 2.3.5 to solve with =0.1 and =10,100,103,,1012,

7. Demo 2.3.6 showed solutions of Ax=b, where A=10000110000110001100011,b=0001 Use Function 2.3.5 to solve with =0.1 and =10,100,103,,1012, tabulating the values of and x11. (This kind of behavior is explained in Section 2.8.) Backward substitution to solve an upper triangular linear system

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