1

I am given the following $3 \times 3$ matrix and is told that there is no inverse. $$ \begin{pmatrix} 1&1&-3\\2&1&-3\\1&2&-6 \end{pmatrix} $$

I was asked to apply Gauss-Jordan elimination on this matrix, and so far, I got this:

$$ \begin{pmatrix} 1&0&0 &|&-1&1&0\\ 0&1&-3& | & 2 & -1 & 0\\ 0&2&-6 & | & 1&-1&1 \end{pmatrix} $$

Does it mean that a matrix with no inverse cannot be reduced (on the LHS) to the identity matrix?

bryan.blackbee
  • 4,161
  • 2
  • 29
  • 50
  • Any matrix can be reduced to the so-called row-echelon form (see, eg, http://en.wikipedia.org/wiki/Row_echelon_form). The identity matrix is a special case. If the row-echelon form is the identity matrix, your matrix is invertible. In your case, you have a form with the third row $0$ (after one more step), so the matrix is not invertible (as you already knew). – gnometorule Feb 15 '13 at 06:38

1 Answers1

2

Yes it does mean that a matrix with no inverse cant be reduced to identity matrix in the elimination.

You can easily check it out subtract two times the 2nd row from the third row you will end up in a all zero third row.

The reason behind this is that let a non invertible matrix be reduced to the identity matrix by a series of row operation. We know each row operation can be thought of as a multiplication of elementary matrices from left,so we have after the row operations,

$$E_k.E_{k-1}\dots E_1.A=I\dots (1)$$

Here A is the non invertible matrix and $E_i$s are the elementary matrices,then we have

$$A=E_1^{-1}....E_K^{-1}.I....(1)$$

On the right side we have all the matrices invertible which implies that A must be invertible.This is a contradiction.

Abhra Abir Kundu
  • 8,150