# Determine inverse matrix of $\left(\begin{matrix} 0 & 3 \\ 0 & 6 \end{matrix} \right)$ using Gauss-Jordan method

I need to find the inverse of the following matrix with Gauss-Jordan method, but apparently, checking with a calculator, it does not exist:

$$\left(\begin{matrix} 0 & 3 \\ 0 & 6 \end{matrix} \right)$$

How can we apply Gauss-Jordan to the previous matrix, and from that determine if the inverse matrix exists or not?

I think the problem is that we cannot make the upper left $0$ $1$, right?

• The matrix isn't invertible. – Git Gud Mar 11 '15 at 16:23
• @GitGud I already know it, but why? – nbro Mar 11 '15 at 16:24
• Perhaps you meant to type "(...) calculator, it doesn't exist"? Let $X$ be your matrix. By definition, $X$ being invertible means that there exists a $2\times2$ matrix $\color{grey}{Y=}\begin{pmatrix}y_1& y_2\\y_3 & y_4 \end{pmatrix}$ such that $YX=I_2=XY$. Now compute $XY$ explicitly. – Git Gud Mar 11 '15 at 16:27
• Of course it doesn't exist the inverse: there is a completely null column! – Von Neumann Jan 30 '16 at 10:58

$A$ is invertible if it row-reduces to the identity. Your matrix row reduces to $\left(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right)$ and is thus not invertible.

• I was thinking that this is the answer. – nbro Mar 11 '15 at 16:30

A linear map (or matrix) $A$ is invertible $\iff$ bijection.

Ker $A$ contains (1,0), so it can't be.

Matrix A(n*n) is called invertible if $det(A)!=0$ and rank is n.
let $A=\left(\begin{matrix} 0 & 3 \\ 0 & 6 \end{matrix} \right)$
Using formula for determinant matrix 2*2 we get
$det(A)=(a1*a4)-(a2*a3)=(0*6)-(0*3)=0$
If we want to rank of matrix A we just divide second colum by 3 and we get:

$A=\left(\begin{matrix} 0 & 1 \\ 0 & 2 \end{matrix} \right)$ So rank of matrix A is 1. So we can conclude that matrix A is not invertible.

This is a non-invertible matrix. The determinant is $ad-bc=0*6-0*3=0$, so no inverse exists.

• We have not found the concept of determinant yet, that's why I could not calculate it... – nbro Mar 11 '15 at 16:25

Yours is a square matrix with a vanishing determinant, hence it is singular (non-invertible).

From the perspective of the Gauss-Jordan method, you could argue as follows: the rows of the matrix are dependent (and in this case $\text{row}_2=2\times\text{row}_1$) which means that the system of equations we get with the Gauss-Jordan method is also not independent, i.e. not solvable (you have more unknowns than equations). Therefore the matrix is not invertible.

• Ok, but using just Gauss-Jordan, can we determine if the matrix is invertible or not? – nbro Mar 11 '15 at 16:25

what about the fact that $A(1,0)^T=(0,0)$ and if $A$ were invertible you can multiply the last equation by $A^{-1}$ on the left and end up with $(1,0)^T = (0,0)^T.$ that is certainly not true, therefore $A$ could not be invertible.