Chapter 2 - Matrix Algebra - 2.2 Exercises - Page 112: 33

See Solution

$\begin{bmatrix} 1 & 0 & 0\\ 1 & 1 & 0\\ 1 & 1 & 1\\ \end{bmatrix}^{-1}=$$\begin{bmatrix} 1 & 0 & 0\\ -1 & 1 & 0\\ 0 & -1 & 1\\ \end{bmatrix}$ $\begin{bmatrix} 1 & 0 & 0&0\\ 1 & 1 & 0&0\\ 1 & 1 & 1&0\\ 1&1&1&1 \end{bmatrix}^{-1}=$$\begin{bmatrix} 1 & 0 & 0&0\\ -1 & 1 & 0&0\\ 0 & -1 & 1&0\\ 0&0 & -1 & 1\\ \end{bmatrix}$ The inverse of a matrix in this form will be 1s on the major diagonal and -1 on the diagonal below the major diagonal $a_{xx}$ where x is an integer will always be in the form $1*n-1*(n-1)$ because the number of 1s in one column is always one more than the number of 1s in the next column and the first column is multiplied by 1 and the next column is multilpied by -1 for a sum of products of 1.