Question

Using elementary transformation, find the inverse of the matrix given​

Answer 1

$\sf \: Given \: A = \left[\begin{array}{ccc} \sf1 & \sf 3 & \sf - 2 \\ \sf - 3& \sf0 & \sf - 5 \\\sf2 & \sf5 & \sf0\end{array}\right]$

we know that, A = IA

$\sf \left[\begin{array}{ccc} \sf1 & \sf 3 & \sf - 2 \\ \sf - 3& \sf0 & \sf - 5 \\\sf2 & \sf5 & \sf0\end{array}\right] = \left[\begin{array}{ccc} \sf1 & \sf 0 & \sf 0 \\ \sf 0& \sf1 & \sf 0 \\\sf0 & \sf0 & \sf1\end{array}\right]A$

Applying R₂ → R₂ + 3R₁

$\sf \left[\begin{array}{ccc} \sf1 & \sf 3 & \sf - 2 \\ \sf 0& \sf9 & \sf - 11 \\\sf2 & \sf5 & \sf0\end{array}\right] = \left[\begin{array}{ccc} \sf1 & \sf 0 & \sf 0 \\ \sf 3& \sf1 & \sf 0 \\\sf0 & \sf0 & \sf1\end{array}\right]A$

Applying R₃ → R₃ + 2R₁

$\sf \left[\begin{array}{ccc} \sf1 & \sf 3 & \sf - 2 \\ \sf 0& \sf9 & \sf - 11 \\\sf0 & \sf - 1 & \sf4\end{array}\right] = \left[\begin{array}{ccc} \sf1 & \sf 0 & \sf 0 \\ \sf 3& \sf1 & \sf 0 \\\sf - 2 & \sf0 & \sf1\end{array}\right]A$

Applying R₂ → 1/9R₂

$\sf \left[\begin{array}{ccc} \sf1 & \sf 3 & \sf - 2 \\ \\ \sf 0& \sf1 & \sf \dfrac{ - 11}{9} \\ \\\sf0 & \sf - 1 & \sf4\end{array}\right] = \left[\begin{array}{ccc} \sf1 & \sf 0 & \sf 0 \\ \\ \sf \dfrac{1}{3} & \sf \dfrac{1}{9} & \sf 0 \\ \\\sf - 2 & \sf0 & \sf1\end{array}\right]A$

Applying R₁ → R₁ - 3R₂

$\sf \left[\begin{array}{ccc} \sf1 & \sf 0 & \sf \dfrac{5}{3} \\ \\ \sf 0& \sf1 & \sf \dfrac{ - 11}{9} \\ \\\sf0 & \sf - 1 & \sf4\end{array}\right] = \left[\begin{array}{ccc} \sf0& \sf \dfrac{ - 1}{3} & \sf 0 \\ \\ \sf \dfrac{1}{3} & \sf \dfrac{1}{9} & \sf 0 \\ \\\sf - 2 & \sf0 & \sf1\end{array}\right]A$

Applying R₃ → R₃ + R₂

$\sf \left[\begin{array}{ccc} \sf1 & \sf 0 & \sf \dfrac{5}{3} \\ \\ \sf 0& \sf1 & \sf \dfrac{ - 11}{9} \\ \\\sf0 & \sf 0 & \sf \dfrac{25}{9} \end{array}\right] = \left[\begin{array}{ccc} \sf0& \sf \dfrac{ - 1}{3} & \sf 0 \\ \\ \sf \dfrac{1}{3} & \sf \dfrac{1}{9} & \sf 0 \\ \\\sf \dfrac{ - 5}{3} & \sf \dfrac{1}{9} & \sf1\end{array}\right]A$

Applying R₃ → 9/25R₃

$\sf \left[\begin{array}{ccc} \sf1 & \sf 0 & \sf \dfrac{5}{3} \\ \\ \sf 0& \sf1 & \sf \dfrac{ - 11}{9} \\ \\\sf0 & \sf 0 & \sf1 \end{array}\right] = \left[\begin{array}{ccc} \sf0& \sf \dfrac{ - 1}{3} & \sf 0 \\ \\ \sf \dfrac{1}{3} & \sf \dfrac{1}{9} & \sf 0 \\ \\ \sf \dfrac{ - 3}{5} & \sf \dfrac{1}{25} & \sf \dfrac{9}{25} \end{array}\right]A$

Applying R₁ → R₁ - 5/3R₃

$\sf \left[\begin{array}{ccc} \sf1 & \sf 0 & \sf 0 \\ \\ \sf 0& \sf1 & \sf \dfrac{ - 11}{9} \\ \\\sf0 & \sf 0 & \sf1 \end{array}\right] = \left[\begin{array}{ccc} \sf0& \sf \dfrac{ - 2}{5} & \sf \dfrac{ - 3}{5} \\ \\ \sf \dfrac{1}{3} & \sf \dfrac{1}{9} & \sf 0 \\ \\ \sf \dfrac{ - 3}{5} & \sf \dfrac{1}{25} & \sf \dfrac{9}{25} \end{array}\right]A$

Applying R₂ → R₂ + 11/9R₃

$\sf \left[\begin{array}{ccc} \sf1 & \sf 0 & \sf 0 \\ \sf 0& \sf1 & \sf 0 \\\sf0 & \sf 0 & \sf1 \end{array}\right] = \left[\begin{array}{ccc} \sf1& \sf \dfrac{ - 2}{5} & \sf \dfrac{ - 3}{5} \\ \\ \sf - \dfrac{2}{5} & \sf \dfrac{4}{25} & \sf \dfrac{11}{25} \\ \\ \sf \dfrac{ - 3}{5} & \sf \dfrac{1}{25} & \sf \dfrac{9}{25} \end{array}\right]A$

we know, I = A¯¹A

$\sf Thus, A^{-1} = \left[\begin{array}{ccc} \sf1& \sf \dfrac{ - 2}{5} & \sf \dfrac{ - 3}{5} \\ \\ \sf - \dfrac{2}{5} & \sf \dfrac{4}{25} & \sf \dfrac{11}{25} \\ \\ \sf \dfrac{ - 3}{5} & \sf \dfrac{1}{25} & \sf \dfrac{9}{25} \end{array}\right]A$