Inverse of an invertible upper triangular matrix of order 3 Prove that the inverse of an invertible upper triangular matrix of order 3 is invertible and upper triangular.
I have checked all the similar questions but I couldn't understand any of them.
I supposed random 3x3 upper triangular matrix and tried to find its inverse, but it came out lower triangular matrix, not the upper triangular. Can anyone please give me a suggestion, how to prove it?
$$
A=\left(\begin{array}{rrr}%
a&b&c\\%
0&d&e\\%
0&0&f\\%
\end{array}\right)%
$$
$$
x11=\left(\begin{array}{rrr}%
d&e\\%
0&f\\%
\end{array}\right)%
=df, x12=-\left(\begin{array}{rrr}%
0&e\\%
0&f\\%
\end{array}\right)%
=0, x13=\left(\begin{array}{rrr}%
0&d\\%
0&0\\%
\end{array}\right)%
=0
$$
$$
x21=-\left(\begin{array}{rrr}%
b&c\\%
0&f\\%
\end{array}\right)%
=-bf, X22=\left(\begin{array}{rrr}%
a&c\\%
0&f\\%
\end{array}\right)%
=af, X23=-\left(\begin{array}{rrr}%
a&b\\%
0&0\\%
\end{array}\right)%
=0
$$
$$
x31=\left(\begin{array}{rrr}%
b&c\\%
d&e\\%
\end{array}\right)%
=bc-cd, x32=-\left(\begin{array}{rrr}%
a&c\\%
0&e\\%
\end{array}\right)%
=ac, x31=\left(\begin{array}{rrr}%
a&b\\%
0&d\\%
\end{array}\right)%
=ad
$$
$$
adjoint A = \left(\begin{array}{rrr}%
df&0&0\\%
-bf&af&0\\%
bc-cd&-ac&ad\\%
\end{array}\right)%
$$
$$
det A = a\left(\begin{array}{rrr}%
d&e\\%
0&f\\%
\end{array}\right)%
=adf
$$
$$
Inverse-A =1/adf  \left(\begin{array}{rrr}%
df&0&0\\%
-bf&af&0\\%
bc-cd&-ac&ad\\%
\end{array}\right)%
$$
It came out lower triangular matrix. Is there any way to make it upper triangular matrix?
 A: As the question was made once more alive I will give for it the answer in a much more  general sense, valid not only  for the upper-triangularity  property of   matrices, but also for other properties if they are present in the described below circumstances.
Now suppose that for some matrices $A,B$ you consider a pattern of entries, say it could be mentioned "upper-triangularity" (UT)   and you have proved that for any   matrices with UT property  the sum $A+B$ and   the product $AB$ preserves UT (what is easy to prove). 
If so then also powers $A^k$ preserve UT.  
Consequently since any inverse can be expressed as polynomial $p(A)$  of $A$  directly calculated from Cayley-Hamilton theorem then also $ A^{-1} $ has the UT property.
A: It is not too difficult to solve directly $$
\left(\begin{array}{rrr}%
a&b&c\\%
0&d&e\\%
0&0&f\\%
\end{array}\right)%
\left(\begin{array}{rrr}%
x&y&z\\%
0&y&v\\%
0&0&w\\%
\end{array}\right)%
=
\left(\begin{array}{rrr}%
1&0&0\\%
0&1&0\\%
0&0&1\\%
\end{array}\right)%
$$
giving
$$ \left(\begin{array}{rrr}%
1/a& -b/(ad)&(be-cd)/(afd)\\%
0&1/d&-e/(fd)\\%
0&0&1/f\\%
\end{array}\right)%
$$ from which we see directly that the matrix is invertible if all $a,d$ and $f$ are different from zero. 
A: Just to be add a simple answer, I will add another one. Note that, by the $QR$ decomposition, for any invertible upper-triangular matrix $U$, there exist an upper-triangular matrix $R$ and an orthogonal matrix $Q$ so that $U^{-1} = Q R.$ Hence, ${(R U)^{-1}} = Q.$ This implies that, inverting both sides, $$R U = Q^{T}.$$ Since the product of two upper-triangular matrices are still upper triangular,  we have found an  orthogonal upper-triangular matrix, which is impossible unless $Q=D$, for some diagonal Matrix $D$, see this link.. Thus the result follows by noting that $$U^{-1}=Q R = DR,$$ which is upper-triangular.
