MHB 5.3 Show that a square matrix with a zero row is not invertible.

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SUMMARY

A square matrix with a zero row is not invertible due to its determinant being zero. For example, the determinant of the matrix $$\begin{pmatrix}0&0&0\\ 0&1&0\\ 3&0&0\end{pmatrix}$$ is zero, indicating that it does not have an inverse. In contrast, the matrix $$\begin{pmatrix}1&0&0\\ 0&1&0\\ 3&0&1\end{pmatrix}$$ has a determinant of one, confirming its invertibility. The discussion emphasizes that any square matrix with a row of zeros will result in a determinant of zero, thus confirming its non-invertibility.

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karush
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Show that a square matrix with a zero row is not invertible.
first a matrix has to be a square to be invertable
if
$$\det \begin{pmatrix}1&0&0\\ 0&1&0\\ 3&0&1\end{pmatrix}=1$$
then $$\begin{pmatrix}
1&0&0\\ 0&1&0\\ 3&0&1\end{pmatrix}^{-1}
=\begin{pmatrix}1&0&0\\ 0&1&0\\ -3&0&1
\end{pmatrix}$$
but if $r_1$ is all zeros
$$\det \begin{pmatrix}
0&0&0\\ 0&1&0\\ 3&0&0
\end{pmatrix}=0$$
then
$$\begin{pmatrix}
0&0&0\\ 0&1&0\\ 3&0&0
\end{pmatrix}^{-1} DNE$$ok I,m not real sure formally why this is ...

I could only do so with an example.:confused:
 
Last edited:
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karush said:
if
$$\det \begin{pmatrix}1&0&0\\ 0&1&0\\ 3&0&1\end{pmatrix}=1$$
then $$\begin{pmatrix}
1&0&0\\ 0&1&0\\ 3&0&1\end{pmatrix}^{-1}
=\begin{pmatrix}1&0&0\\ 0&1&0\\ -3&0&1
\end{pmatrix}$$
This reminds me a joke. A tourist asks a local resident: "If I go down this street, will there be a railway station?" The local replies, "The station will be there even if you don't go down that street".

Suppose the $i$th row of a square matrix $A$ consists entirely of zeros. What is the $i$th row in the matrix $AB$ for any size-compatible matrix $B$? Can $AB$ be the identity matrix?
 
so what would AB look like in your example
 
One cannot say much about about $AB$ in general without knowing more about $A$ and $B$, but we can know the $i$th row of $AB$. I suggest you use the definition of matrix multiplication to find what that row is. If you want, you can consider an example where $A$ is the matrix from your original post where the first row has all zeros and $B$ is an arbitrary 3x3 matrix. However, it is pretty obvious what the $i$th row of $AB$ is in general if the $i$th row of $A$ consists of zeros.
 
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

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