જો શક્ય હોય તો,પ્રારંભિક હાર રૂપાંતરણોનો ઉપયોગ કરીને નીચેના શ્રેણિકનો વ્યસ્ત શોધો:
$\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]$

(N/A) ધારો કે $A = \left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]$. પ્રારંભિક હાર રૂપાંતરણોનો ઉપયોગ કરીને $A^{-1}$ શોધવા માટે,આપણે $A = IA$ નો ઉપયોગ કરીએ છીએ.
$\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right] = \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A$
$R_1 \rightarrow \frac{1}{2}R_1$ લાગુ કરતા:
$\left[\begin{array}{ccc}1 & 0 & -\frac{1}{2} \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right] = \left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A$
$R_2 \rightarrow R_2 - 5R_1$ લાગુ કરતા:
$\left[\begin{array}{ccc}1 & 0 & -\frac{1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 1 & 3\end{array}\right] = \left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ -\frac{5}{2} & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A$
$R_3 \rightarrow R_3 - R_2$ લાગુ કરતા:
$\left[\begin{array}{ccc}1 & 0 & -\frac{1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & \frac{1}{2}\end{array}\right] = \left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ -\frac{5}{2} & 1 & 0 \\ \frac{5}{2} & -1 & 1\end{array}\right] A$
$R_3 \rightarrow 2R_3$ લાગુ કરતા:
$\left[\begin{array}{ccc}1 & 0 & -\frac{1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & 1\end{array}\right] = \left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ -\frac{5}{2} & 1 & 0 \\ 5 & -2 & 2\end{array}\right] A$
$R_1 \rightarrow R_1 + \frac{1}{2}R_3$ અને $R_2 \rightarrow R_2 - \frac{5}{2}R_3$ લાગુ કરતા:
$\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] = \left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right] A$
આમ,$A^{-1} = \left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$.