Youin normal form of a 4-4 matrix can i do like this first make A11=1 then make all elements of the first row except a11=0 likewise then make all elements of first column except A11 0 then i make A22=1 then all other elements i.e A32 and A42=0 then make A33=1 then make A43 and A23=0 then make A44=1 if possible and at the end make A34 and A24 =0
Question
Youin normal form of a 4-4 matrix can i do like this first make A11=1 then make all elements of the first row except a11=0 likewise then make all elements of first column except A11 0 then i make A22=1 then all other elements i.e A32 and A42=0 then make A33=1 then make A43 and A23=0 then make A44=1 if possible and at the end make A34 and A24 =0
Solution
Yes, you are correct. The process you described is essentially the process of transforming a matrix into its Reduced Row Echelon Form (RREF). Here are the steps you mentioned, clarified:
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Make A11 = 1. This is done by dividing the entire row by the value of A11.
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Make all other elements in the first row and the first column zero, except for A11. This is done by subtracting suitable multiples of the first row from the other rows, and suitable multiples of the first column from the other columns.
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Make A22 = 1, by dividing the second row by the value of A22.
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Make all other elements in the second row and the second column zero, except for A22. This is done by subtracting suitable multiples of the second row from the other rows, and suitable multiples of the second column from the other columns.
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Repeat this process for A33 and A44.
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Finally, make sure that A34 and A24 are zero. This can be done by subtracting suitable multiples of the fourth row from the second and third rows.
This process will give you a matrix in Reduced Row Echelon Form, where the leading coefficient (also known as the pivot) of each row is 1, and all elements below and above the pivot are zero.
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