Why can’t i convert a matrix order n with a last row all zeros into an identity matrix order n

matrices

I saw a phrase in my book that says " it's impossible to convert a matrix with a last row all zeros into an identity matrix ".

But I was curious if there is any combination of elementary operations to convert it or a proof for the above statement.

So can someone help me?

Best Answer

An identity matrix is a square matrix with $1$'s all along the diagonal. Thus, an identity matrix has a $1$ in every column (and every row).

There are three elementary row operations. None of them are capable of taking a matrix that has all zeros in a given column and transforming it into a matrix with a non-zero element in that column.

Swap Rows

If you swap two rows that both have zeros in the $n$th column, those rows will continue to have zeros in the $n$th column.

Scalar Multiplication

It is impossible to multiply a row with a zero in the $n$th column by a scalar to cause it to have a non-zero element in the $n$th column, since there is no scalar by which you can multiply $0$ to get anything other than $0$.

Row Addition

If multiple rows have $0$'s in the $n$th column, it's impossible to add them together and get a row that has a non-zero element in the $n$th column. That would require adding zero to itself to get something other than $0$, which is impossible.

Since it's impossible to create a non-zero element in the $n$th row of a matrix if the $n$th row contains all zeros, it is impossible to convert said matrix into an identity matrix, which requires a non-zero element ($1$) in every column.