## Elementary Linear Algebra 7th Edition

$A$ has no inverse, or is non-invertible (or singular).
To find $A^{-1}$, we have $$\left[ A \ \ I \right]= \left[ \begin {array}{cccccc} 1&0&0&1&0&0\\ 3&0&0&0 &1&0\\ 2&5&5&0&0&1\end {array} \right] .$$ Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows $$\left[ \begin {array}{cccccc} 1&0&0&0&\frac{1}{3}&0\\ 0&1&1 &0&-\frac{2}{15}&\frac{1}{5}\\ 0&0&0&1&-\frac{1}{3}&0\end {array} \right] .$$ Note that the portion of the matrix $A$ has a row of zeros. So it is not possible to rewrite the matrix $\left[ A \ \ I \right]$ in the form $\left[I \ \ A^{-1} \right]$. This means that has no inverse, or is non-invertible (or singular).