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$$A=\begin{bmatrix} 1&0&0&....&0&0\\ x&1&0&....&0&0\\ 0&x&1&....&0&0\\ \vdots & \vdots & \ddots & \ddots & \vdots & \vdots\\ \vdots & \vdots & \vdots & \ddots & \ddots & \vdots\\ 0&0&0&... & x&1\\ \end{bmatrix}$$ 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?

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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.

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  • $\begingroup$ I thought so,that s easy :) $\endgroup$ – Jack Jan 13 at 18:47
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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 $$\left(\begin{array}{rrrrrr} 1 & 0 & 0 & \cdots & 0 & 0\\ -x & 1 & 0 & \cdots & 0 & 0\\ x^2 & -x & 1 & \cdots & 0 & 0 \\ -x^3 & x^2 & -x & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ (-1)^{n-1}x^{n-1} & (-1)^{n-2}x^{n-2} & (-1)^{n-3}x^{n-3} & \cdots & -x & 1 \end{array}\right).$$

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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.

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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 $$A^{-1}=\begin{bmatrix} 1&0&0&....&0&0\\ -x&1&0&....&0&0\\ x^2&-x&1&....&0&0\\ \vdots & \vdots & \ddots & \ddots & \vdots & \vdots\\ \vdots & \vdots & \vdots & \ddots & \ddots & \vdots\\ (-1)^{n-1}x^{n-1}&(-1)^{n-2}x^{n-2}&(-1)^{n-3}x^{n-3}&... & -x&1\\ \end{bmatrix}$$

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