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## Main Question or Discussion Point

**How to write the algorithm? I figured out a method to find the inverse.**

The assignment is making use of the property of triangular matrices to find the inverse of a matrix [itex]\displaystyle A[/itex].

The inverse of a triangular matrix(Upper/ Lower) is also triangular(Upper/ Lower) and is easy to find.

[itex]\displaystyle \begin{bmatrix} a & b & c\\ 0 & d & e\\ 0 & 0& f \end{bmatrix} \begin{bmatrix}\frac{1}{a} & x & z\\ 0 & \frac{1}{d} & y \\ 0 & 0 &\frac{1}{f} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}[/itex]

By equating the Upper 0s, [itex]\displaystyle x, y, z[/itex] are readily to be found.

[itex]\displaystyle x = -a^{-1} b d^{-1}[/itex], [itex]\displaystyle y = -d^{-1} e f^{-1}[/itex], [itex]\displaystyle z = a^{-1} b d^{-1} c f^{-1} - a^{-1} c f^{-1}[/itex]

In higher order matrices:

[itex]\displaystyle a d x= - \begin{vmatrix} b \end{vmatrix}[/itex]

[itex]\displaystyle d f y = - \begin{vmatrix} e \end{vmatrix}[/itex]

[itex]\displaystyle a d f z= + \begin{vmatrix} b & c \\ d & e \end{vmatrix}[/itex]

[itex]\displaystyle \pm [/itex] signs follow the plan:

[itex]\displaystyle \begin{bmatrix} + & - & + \\ - & + & -\\ + & -& + \end{bmatrix} \sim \begin{bmatrix} + & x & z \\ - & + & y\\ + & - & + \end{bmatrix}[/itex]

If [itex]\displaystyle A[/itex] is invertible, find [itex]\displaystyle A^{-1}[/itex] by changing [itex]\displaystyle A[/itex] triangular...

[itex]\displaystyle AX=I[/itex]

[itex]\displaystyle E_3E_2E_1AX=E_3E_2E_1I[/itex]

[itex]\displaystyle E_jAX=E_jI[/itex]

Take Row operation and Column operation on [itex]\displaystyle E_jA[/itex] to become a triangular matrix

[itex]\displaystyle R_jE_jAC_j = U[/itex] is triangular.

[itex]\displaystyle (R_jE_jAC_j)^{-1} = U^{-1}[/itex] can be found by the method above.

[itex]\displaystyle (R_jE_jAC_j)^{-1}=C_j^{-1}A^{-1}E_j^{-1}R_j^{-1} = U^{-1}[/itex]

[itex]\displaystyle A^{-1}=C_j(C_j^{-1}A^{-1}E_j^{-1}R_j^{-1})R_jE_j = C_j(U^{-1})R_jE_j[/itex]

[itex]\displaystyle E_j = E_jAX = E_jI[/itex] is computed in the beginning.

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