# 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?

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.

• I thought so,that s easy :) – Jack Jan 13 at 18:47

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 $$\left(\begin{array}{rrrrrr} 1 & 0 & 0 & \cdots & 0 & 0\\ -x & 1 & 0 & \cdots & 0 & 0\\ x^2 & -x & 1 & \cdots & 0 & 0 \\ -x^3 & x^2 & -x & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ (-1)^{n-1}x^{n-1} & (-1)^{n-2}x^{n-2} & (-1)^{n-3}x^{n-3} & \cdots & -x & 1 \end{array}\right).$$

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.

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