$$A=\begin{bmatrix}
1&0&0&….&0&0\\
x&1&0&….&0&0\\
0&x&1&….&0&0\\
\vdots & \vdots & \ddots & \ddots & \vdots & \vdots\\
\vdots & \vdots & \vdots & \ddots & \ddots & \vdots\\
0&0&0&… & x&1\\
\end{bmatrix}$$
I tried to find its determinant through recursive methods but I couldn't find the adjoint in order to get the inverse. How can I solve it?
[Math] find the inverse of the following matrix
inverselinear algebramatricestoeplitz-matrices
Best Answer
The determinant of a lower-triangular matrix is the product of the diagonal elements, hence (in your case) it's $1$.
As for finding the inverse...a direct approach might make some sense. If you want $$ AM = I $$ then the first column of $M$ must start with a $1$, right, because the dot product of the first row of $A$ and the first col of $M$ must be $1$. Now look at the dot product of the second row of $A$ with the first column of $M$...that tells you want the SECOND entry of the first row of $M$ must be. Repeat until you're done.