Consider the matrix A = 1.2969 0.8648 0.2161 0.1441 Using MATLAB, compute the following and discuss the results: (a) Find the condition number (in any norm), determinant, and inverse of A. (b) Let B be the matrix obtained from A by rounding off to three decimal places, i.e., 1.2969 -> 1.297. Find the determinant and inverse of B. Can you explain why A−1 and B−1 differ? (c) Set b1 = (1.2969, 0.2161)T and and do x = A\b1. You will find the solution x = (1, 0)T. Repeat the process but with a vector b2 obtained from b1 by rounding off to three decimal places. Explain what happens. Why do the two answers differ by so much?
Consider the matrix A = 1.2969 0.8648 0.2161 0.1441 Using MATLAB, compute the following and discuss the results: (a) Find the condition number (in any norm), determinant, and inverse of A. (b) Let B be the matrix obtained from A by rounding off to three decimal places, i.e., 1.2969 -> 1.297. Find the determinant and inverse of B. Can you explain why A−1 and B−1 differ? (c) Set b1 = (1.2969, 0.2161)T and and do x = A\b1. You will find the solution x = (1, 0)T. Repeat the process but with a vector b2 obtained from b1 by rounding off to three decimal places. Explain what happens. Why do the two answers differ by so much?
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Your Question:
Consider the matrix
A =
1.2969 0.8648
0.2161 0.1441
Using MATLAB, compute the following and discuss the results:
(a) Find the condition number (in any norm), determinant, and inverse of A.
(b) Let B be the matrix obtained from A by rounding off to three decimal places,
i.e., 1.2969 -> 1.297. Find the determinant and inverse of B. Can you explain
why A−1 and B−1 differ?
(c) Set b1 = (1.2969, 0.2161)T and and do x = A\b1. You will find the solution
x = (1, 0)T. Repeat the process but with a vector b2 obtained from b1 by rounding off to three decimal places. Explain what happens. Why do the two answers differ by so much?
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