- #1

- 351

- 81

- TL;DR Summary
- An upper triangular matrix.

If I have a ##n\times n## matrix

$$

U=

\begin{bmatrix}

u_{11} & u_{12} &u_{13} & \cdots u_{1n} \\

0 & u_{22} & u_{23} & \cdots u_{2n} \\

0&0 &u_{33} &\cdots u_{3n}\\

\vdots & \vdots &\vdots & \cdots \vdots \\

0 & 0 & 0 &\cdots u_{nn}

\end{bmatrix}

$$

Now, I don't want to use the fact that it's determinant is nonzero therefore I can find its inverse (because that's what I'm trying to prove).

I want to know if I can convert that matrix into an identity matrix using Gauss-Jordan elimination method? Well, you see in integrals we usually look for if the integral

$$

U=

\begin{bmatrix}

u_{11} & u_{12} &u_{13} & \cdots u_{1n} \\

0 & u_{22} & u_{23} & \cdots u_{2n} \\

0&0 &u_{33} &\cdots u_{3n}\\

\vdots & \vdots &\vdots & \cdots \vdots \\

0 & 0 & 0 &\cdots u_{nn}

\end{bmatrix}

$$

Now, I don't want to use the fact that it's determinant is nonzero therefore I can find its inverse (because that's what I'm trying to prove).

I want to know if I can convert that matrix into an identity matrix using Gauss-Jordan elimination method? Well, you see in integrals we usually look for if the integral

*can*be found out, in that subject we can prove if something is possible without finding the thing needed; while here how can I ascertain myself that I can convert ##U## into an identity matrix (well if I can do that then it has an inverse) without applying the processes of Gauss-Jordan elimination method.