Question

Find A⁻¹ using column transformations:

A = ${\displaystyle \left[\begin{array}{cc}5& 3\\ 3& -2\end{array}\right]}$

Answer 1

We know that A⁻¹ = I

∴ ${\displaystyle \left[\begin{array}{cc}5& 3\\ 3& -2\end{array}\right]}$ A⁻¹ = ${\displaystyle \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]}$

Applying C₁ → 2C₂ – C₁, we get

${\displaystyle \left[\begin{array}{cc}1& 3\\ -7& -2\end{array}\right]}$ A⁻¹ = ${\displaystyle \left[\begin{array}{cc}-1& 0\\ 2& 1\end{array}\right]}$

Applying C₂ → C₂ – 3C₁, we get

${\displaystyle \left[\begin{array}{cc}1& 0\\ -7& 19\end{array}\right]}$ A⁻¹ = ${\displaystyle \left[\begin{array}{cc}-1& 3\\ 2& -5\end{array}\right]}$

Applying C₂ → ${\displaystyle \left(\frac{1}{19}\right)}$ C₂, we get

${\displaystyle \left[\begin{array}{cc}1& 0\\ -7& 1\end{array}\right]}$ A⁻¹ = ${\displaystyle \left[\begin{array}{cc}-1& \frac{3}{19}\\ 2& \frac{-5}{19}\end{array}\right]}$

Applying C₁ → C₁ + 7C₂, we get

${\displaystyle \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]}$ A⁻¹ = ${\displaystyle \left[\begin{array}{cc}\frac{2}{19}& \frac{3}{19}\\ \frac{3}{19}& \frac{-5}{19}\end{array}\right]}$

∴ A⁻¹ = ${\displaystyle \frac{1}{19}\left[\begin{array}{cc}2& 3\\ 3& -5\end{array}\right]}$