**Proposition:** A matrix with a row consisting of all zeroes cannot be inverted, nor can one with a column of all zeroes.

*Proof*

Let *A* denote some \(n\times n\) matrix where one of its rows has only zeroes --- let us refer to this as row *r*. Then take another \(n \times n\) matrix *B* and suppose toward contradiction that \(B = A^{-1}\). Then the product *AB* must equal the identity matrix.

From the definition of matrix multiplication, the entry in row r and column j of the product of *AB* is given by:

\[\sum_{k=1}^{n} A_{rk} B_{kj} = A_{r1}B_{1j} + A_{r2}B_{2j} + ... + A_{rn}B_{nj}.\]

Since \(A_{r1} = A_{r2} = ... = A_{rn} = 0\), we have that for every column in \(AB\), the entry in row *r* will be zero. This is a contradiction, as the identity matrix cannot have a row consisting of all zero. Therefore, *A* does not have an inverse.

The same is true of a matrix with a column of zeroes. Let us adjust our assumptions, and say that *A* now has a column *c* of zeroes. If *A* has *B* as its inverse, then \(AB = BA = I\). The following sum gives the entry in row *i* and column *c* of *BA*,

\[\sum_{k=1}^n B_{ik} A_{kc} = B_{i1}A_{1c} + B_{i2}A_{2c} + ... + B_{in}A_{nc}.\]

Similarly, since \(A_{1c} = A_{2c} = A_{nc} = 0\), then column c in *BA* will be all zero. Hence, the product cannot be the identity matrix which again raises our contradiction. Therefore, a matrix with a column containing only zeroes does not have an inverse.

\(\blacksquare\)