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?
On
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.
On
A linear map (or matrix) $A$ is invertible $\iff$ bijection.
Ker $ A$ contains (1,0), so it can't be.
On
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.
On
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.
$A$ 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.