There is a theorem that an $n \times n$ matrix $A$ is invertible if and only if $A$ is row equivalent to $I_n$, and in this case, any sequence of elementary row operations that reduces $A$ to $I_n$ also transforms $I_n$ to $A^{-1}$.
Can anyone shows how the proof is?
Thanks a lot.
$\endgroup$ 32 Answers
$\begingroup$To show that this is true, use can use the idea of Elementary Matrices.
If you want to perform one of the three elementary row operations on some matrix $M$ (row switching, row adding, or row scaling), this is equivalent to pre-multiplying $M$ by one of three special types of matrices, which I generically denote with $E$.
The result of a series of $n$ row operations on $M$ has the form $E_n E_{n-1} ... E_ 3 E_2 E_1 M$.
Each elementary matrix $E_i$ is invertible, so if $M$ is row equivalent to the identity matrix $I$,
$$I = E_n E_{n-1} ... E_ 3 E_2 E_1 M$$
then the inverse of $M$ has the form,
$$M^{-1} = E_n E_{n-1} ... E_ 3 E_2 E_1$$
So a matrix being invertible and a matrix being row-equivalent to the identity are the same thing.
Examples of elementary matrices:
Row-switching matrices are just the identity with the appropriate rows swapped. This matrix swaps rows 2 and 3:
$$\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right)$$
Row-scaling matrices are just the identity with the appropriate row scaled. This matrix scales the second row by 5:
$$\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 1 \end{array} \right)$$
Row-adding matrices are just the identity with an extra off-diagonal non-zero element $E_{i,j}$ which adds that number times row $j$ to row $i$. This matrix adds -2 times row 3 to row 1:
$$\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -2 & 0 & 1 \end{array} \right)$$
It isn't hard to show that each of these matrices is invertible.
$\endgroup$ 6 $\begingroup$This is the Gauss-Jordan elimination method and it is a simple consequence of row-echelon form reduction method to solve a linear system.. For a proof you can see: Proof of inverse matrices, with method of Gauss / Jordan
$\endgroup$ 0
