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Example 1.4.8 Let F 27. Find a basis for the eigenspace of A 3 0 3 corresponding to the eigenvalue X-3. We have to find a basis for the null space of A 31 34 3, so we Fields and Matrix Algebra 17 ut A- 31 in row-reduced echelon form: 3 4 30 4 (3-4-)0 1 0 5 (2-5-)4 0 0 (4 40 We find that x3 is a free variable, and x1--6x3- X3, x2--5x3 -2x3, leading to the basis 2Hello. Can someone explain to me how they got the first matrix show above as

632 5 132 ? This is my only question. Please show process/work. Thanks!

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