Solving triangular systems of equations: Backwards substitution example

Consider the triangular system

    \[ R x=\left(\begin{array}{ccc} 3 & 8 & 3 \\ 0 & 6 & -1 \\ 0 & 0 & -3 \end{array}\right)\left(\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right)=\left(\begin{array}{c} 1 \\ 2 \\ -3 \end{array}\right) \]

We solve for the last variable x_3 first, obtaining (from the last equation) x_3=1. We plug this value of x_3 into the first and second equation, obtaining a new triangular system in two variables x_1,x_2:

    \[ \left(\begin{array}{ll} 3 & 8 \\ 0 & 6 \end{array}\right)\left(\begin{array}{l} x_1 \\ x_2 \end{array}\right)=\left(\begin{array}{c} 1-3 x_3 \\ 2+x_3 \end{array}\right)=\left(\begin{array}{c} -2 \\ 3 \end{array}\right) . \]

We proceed by solving for the last variable x_2. The last equation yields x_2=3 / 6=1 / 2. Plugging this value into the first equation gives

    \[ 3 x_1=-2-8 x_2=-6. \]

We can apply the idea to find the inverse of the square upper triangular matrix R, by solving

    \[ R x^{(1)}=\left(\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right), \quad R x^{(2)}=\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right), \quad R x^{(3)}=\left(\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right) . \]

The matrix \left[x^{(1)}, x^{(2)}, x^{(3)}\right] is then the inverse of R. We find

    \[ R^{-1} = \left(\begin{array}{c|c|c} x^{(1)} & x^{(2)} & x^{(3)} \end{array}\right) = \left(\begin{array}{c|c|c} 1 / 3 & -4 / 9 & 13 / 27 \\ 0 & 1 / 6 & -1 / 18 \\ 0 & 0 & -1 / 3 \end{array}\right) . \]

As illustrated above, the inverse of a triangular matrix is triangular.

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