Can I use back substitution to invert a matrix?

[NewStudent200]

Hello all,

Say I had n equations, with n variables, so that
C1 = C1(f1), C1 is a funtion of f1 only
C2 = C2(f1, f2), C2 is a function of f1 and f2
...
Cn = Cn(f1, f2...fn), Cn is a function of all n vairables

I can calculate the matrix dC/ df where each line is the derivative of Ci with respect to the n vairables, The matrix will look something lile

a 0 0 0... 0
b c 0 0... 0
...
u v w x... z

Now if I was after the matrix df/dC then am I basically just trying to invert this matrix?

Thanks very much for any help

 

[HallsofIvy]

Hello all,

Say I had n equations, with n variables, so that
C1 = C1(f1), C1 is a funtion of f1 only
C2 = C2(f1, f2), C2 is a function of f1 and f2
...
Cn = Cn(f1, f2...fn), Cn is a function of all n vairables

I can calculate the matrix dC/ df where each line is the derivative of Ci with respect to the n vairables, The matrix will look something lile

a 0 0 0... 0
b c 0 0... 0
...
u v w x... z

Now if I was after the matrix df/dC then am I basically just trying to invert this matrix?

Thanks very much for any help

Yes, and, for a triangular matrix, that is fairly simple.

 

[NewStudent200]

Great. Thanks very much. Is there a particular algorithm that is recommended for a triangular matrix?

Regards,

 

[HallsofIvy]

Just "back substitution". If your matrix is, say,
\begin{bmatrix}a & 0 & 0 \\ b & c & 0\\ d & e & f\end{bmatrix}
then its inverse will be a matrix of the form
\begin{bmatrix}u & 0 & 0 \\ v & w & 0\\ x & y & z\end{bmatrix}
such that
\begin{bmatrix}a & 0 & 0 \\ b & c & 0\\ d & e & f\end{bmatrix}\begin{bmatrix}u & 0 & 0 \\ v & w & 0\\ x & y & z\end{bmatrix}= \begin{bmatrix}1 & 0 & 0 \\0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}
You must have au= 1 so u= 1/a. Then bu+ cv= b/a+ cv= 0 so v= -b/ac. cw= 1 so w= 1/c, etc.