Discover whether
\[ A = \left[ \begin{array}{c c c c} 1 & 2 & 3 & 4 \\ 0 & 2 & 3 & 4 \\ 0 & 0 & 3 & 4 \\ 0 & 0 & 0 & 4 \end{array} \right] \]
is invertible, and find \(A^{-1}\) if it exists.
Solution
All diagonal entries are nonzero, so \(A\) is invertible. Direct multiplication verifies that \[ A^{-1}= \begin{bmatrix} 1&-1&0&0\\ 0&\frac12&-\frac12&0\\ 0&0&\frac13&-\frac13\\ 0&0&0&\frac14 \end{bmatrix}. \] Indeed, multiplying \(A\) by this matrix gives \(I_4\): each diagonal product is \(1\), while the successive superdiagonal cancellations are \[ -1+2\cdot\frac12=0, \qquad 2\left(-\frac12\right)+3\left(\frac13\right)=0, \qquad 3\left(-\frac13\right)+4\left(\frac14\right)=0, \] and the remaining upper entries cancel similarly. Hence the displayed matrix is \(A^{-1}\).