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Let $U$ be a $3$ by $3$ upper triangular matrix with all diagonal entries non zero. We then know that $U$ is invertible. Show that its inverse is also upper triangular. Also show this fact for a general $n$ by $n$ matrix.

What i tried

To invert the $3$ by $3$ matrix, we need to write the identity matrix in this way and then try to switch sides between the original matrix matrix and the identity matrix.But by doing so we observe that the upper triangular portion of the identity matrix is not affected and hence the inverse remains upper triangular. Could anyone explain this question to me Thanks.$$ A = \left[ {\begin{array}{cc} 1 & 2 & 1 \\ 0 & 1& 0 \\ 0 & 0 & -1 \end{array} } \right]\left[ {\begin{array}{cc} 1 & 0 & 0 \\ 0 & 1& 0 \\ 0 & 0 & 1 \end{array} } \right] $$

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    $\begingroup$ what question? for your example, just try it and apply Gaussian elimination. for the general case, induction. $\endgroup$ – user251257 Sep 10 '15 at 16:26
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What row operations do you need to apply in order to reduce an upper triangular matrix to the identity?

Start at the bottom, scale to reduce to a row with a $1$ in the last column of the row. Use this row to eliminate all non-zero entries in the rows above the last one. Now, move up one row and scale to get a $1$ in the next to the last entry. Use this row to eliminate all non-zero entries in the next to last column from the rows above that. Continue in the same way. What happens to the identity matrix on the right of the augmented matrix as you perform these operations?

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