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All entries left blank in the determinant below are zeros.
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$A$ is invertible if and only if RREF of $A$ is $I_3$.
Step | the row operation | the elementary matrix | the inverse of ele mat |
---|---|---|---|
1st | The third row is replaced by the the sum of the first row and the third row | $E_1 = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right]$ | $E_1^{-1} = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right]$ |
2nd | The third row and the second row are interchanged | $E_2 = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right]$ | $E_2^{-1} = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right]$ |
3rd | The third row is replaced by the the sum of the third row and the second row multiplied by $(-2)$ | $E_3 = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -2 & 1 \end{array}\right]$ | $E_3^{-1} = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{array}\right]$ |
4th | The first row is replaced by the the sum of the first row and the third row | $E_4 = \left[\!\begin{array}{rrr} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$ | $E_4^{-1} = \left[\!\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$ |
A nonzero matrix is said to be in row echelon form if its first row is a nonzero row and each nonzero row bellow the first has strictly more leading zeros then the previous row.
A nonzero matrix which is in row echelon form is said to be in reduced row echelon form if the leading entries off all nonzero rows are equal to 1 and this 1 is the only nonzero entry in its column.