Solving triangular systems of equations: Backwards substitution example
Consider the triangular system
![\[ R x=\left(\begin{array}{ccc} 3 & 8 & 3 \\ 0 & 6 & -1 \\ 0 & 0 & -3 \end{array}\right)\left(\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right)=\left(\begin{array}{c} 1 \\ 2 \\ -3 \end{array}\right) \]](https://pressbooks.pub/app/uploads/quicklatex/quicklatex.com-fee8a57274a4068a1684d2e0d14ce66e_l3.png "Rendered by QuickLaTeX.com")
We solve for the last variable 
first, obtaining (from the last equation) 
. We plug this value of into the first and second equation, obtaining a new triangular system in two variables 
:
![\[ \left(\begin{array}{ll} 3 & 8 \\ 0 & 6 \end{array}\right)\left(\begin{array}{l} x_1 \\ x_2 \end{array}\right)=\left(\begin{array}{c} 1-3 x_3 \\ 2+x_3 \end{array}\right)=\left(\begin{array}{c} -2 \\ 3 \end{array}\right) . \]](https://pressbooks.pub/app/uploads/quicklatex/quicklatex.com-affca6f11338fed9a0c65ba7d5c1f431_l3.png "Rendered by QuickLaTeX.com")
We proceed by solving for the last variable 
. The last equation yields 
. Plugging this value into the first equation gives
![\[ 3 x_1=-2-8 x_2=-6. \]](https://pressbooks.pub/app/uploads/quicklatex/quicklatex.com-04a7c49bc2951a7923920385363f3cba_l3.png "Rendered by QuickLaTeX.com")
We can apply the idea to find the inverse of the square upper triangular matrix 
, by solving
![\[ R x^{(1)}=\left(\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right), \quad R x^{(2)}=\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right), \quad R x^{(3)}=\left(\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right) . \]](https://pressbooks.pub/app/uploads/quicklatex/quicklatex.com-c4a233390379f2b0995f4b26d8bec974_l3.png "Rendered by QuickLaTeX.com")
The matrix ![\left[x^{(1)}, x^{(2)}, x^{(3)}\right]](https://pressbooks.pub/app/uploads/quicklatex/quicklatex.com-05b0c8a473ad866881470f65dbb1458c_l3.png "Rendered by QuickLaTeX.com")
is then the inverse of . We find
![\[ R^{-1} = \left(\begin{array}{c|c|c} x^{(1)} & x^{(2)} & x^{(3)} \end{array}\right) = \left(\begin{array}{c|c|c} 1 / 3 & -4 / 9 & 13 / 27 \\ 0 & 1 / 6 & -1 / 18 \\ 0 & 0 & -1 / 3 \end{array}\right) . \]](https://pressbooks.pub/app/uploads/quicklatex/quicklatex.com-1029418680ca2bbdc26c31e2f25ba973_l3.png "Rendered by QuickLaTeX.com")
As illustrated above, the inverse of a triangular matrix is triangular.
