I want to find the inverse of the following matrix $$A=\begin{bmatrix}
1&0&0&....&0&0\\
x&1&0&....&0&0\\
0&x&1&....&0&0\\
 \vdots & \vdots & \ddots & \ddots & \vdots & \vdots\\
 \vdots & \vdots & \vdots & \ddots & \ddots & \vdots\\
0&0&0&... & x&1\\
\end{bmatrix}$$
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?
 A: The inverse can easily be obtained by Gaussian Elimination. Sutract $x$-times the first row from the second row; then subtract $x$ times the second row from the third row; etc. You get
$$\left(\begin{array}{rrrrrr}
1 & 0 & 0 & \cdots & 0 & 0\\
-x & 1 & 0 & \cdots & 0 & 0\\
x^2 & -x & 1 & \cdots & 0 & 0 \\
-x^3 & x^2 & -x & \cdots & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
(-1)^{n-1}x^{n-1} & (-1)^{n-2}x^{n-2} & (-1)^{n-3}x^{n-3} & \cdots & -x & 1
\end{array}\right).$$ 
A: As John Hughes already noted in his answer, the determinant of a triangular matrix is the product of its diagonal elements.  
As for the inverse, here’s a slightly tricky method. If $A$ is $n\times n$, then its characteristic polynomial is $(\lambda-1)^n$. By the Cayley-Hamilton theorem, $(A-I)^n = 0$. In other words, $N=A-I$ is nilpotent of index at most $n$. Since $I$ and $N$ commute, expand $A^{-1} = (I+N)^{-1}$ as the formal power series$$(I+N)^{-1} = I-N+N^2-N^3+\cdots.$$ This series is truncated after at most $n$ terms, therefore $$A^{-1} = \sum_{k=0}^{n-1} (A-I)^k.$$ (We’ll define the $0$th power of any matrix as the identity so that this works for $x=0$, too.) The powers of $N$ have a particularly simple pattern, which I’ll leave for you to discover. Hint: write $N$ as $xJ$ and then think about what the product $J\mathbf v$ is in terms of the components of the vector $\mathbf v$.  
An expression for $A^{-1}$ can also be obtained directly from $(A-I)^n=0$ by expanding and rearranging, but I think this obscures the simple and important underlying structure that you’ll encounter many times in the future.
A: You can use Gauss Jordan Method to find the inverse of the matrix.
$$A=\begin{bmatrix}
1&0&0&....&0&0\\
x&1&0&....&0&0\\
0&x&1&....&0&0\\
 \vdots & \vdots & \ddots & \ddots & \vdots & \vdots\\
 \vdots & \vdots & \vdots & \ddots & \ddots & \vdots\\
0&0&0&... & x&1\\
\end{bmatrix}$$
$$A:I=\begin{bmatrix}
1&0&0&....&0&0&:&1&0&0&....&0&0 \\
x&1&0&....&0&0&:&0&1&0&....&0&0 \\
0&x&1&....&0&0 &:&0&0&1&....&0&0\\
 \vdots & \vdots & \ddots & \ddots & \vdots & \vdots&: & \vdots & \vdots & \ddots & \ddots & \vdots & \vdots \\
 \vdots & \vdots & \vdots & \ddots & \ddots & \vdots &:&\vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\
0&0&0&... & x&1 &:&0&0&0&....&0&1\\
\end{bmatrix}$$
 You can use row elementary operations to remove x from each row. For 2nd row , it would be $$R_2-x*R_1$$
For 3rd row, it would be $$R_3-x*R_2$$ and so on.
Similar operation on Identity matrix will give you the inverse
It would be
$$A^{-1}=\begin{bmatrix}
1&0&0&....&0&0\\
-x&1&0&....&0&0\\
x^2&-x&1&....&0&0\\
 \vdots & \vdots & \ddots & \ddots & \vdots & \vdots\\
 \vdots & \vdots & \vdots & \ddots & \ddots & \vdots\\
(-1)^{n-1}x^{n-1}&(-1)^{n-2}x^{n-2}&(-1)^{n-3}x^{n-3}&... & -x&1\\
\end{bmatrix}$$
A: 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. 
