Question

Find the inverse of matrix A = ${\displaystyle \left[\begin{array}{ccc}1& 0& 1\\ 0& 2& 3\\ 1& 2& 1\end{array}\right]}$ by using elementary row transformations 

Answer 1

A = ${\displaystyle \left[\begin{array}{ccc}1& 0& 1\\ 0& 2& 3\\ 1& 2& 1\end{array}\right]}$ 

∴ |A| = ${\displaystyle \left|\begin{array}{ccc}1& 0& 1\\ 0& 2& 3\\ 1& 2& 1\end{array}\right|}$ 

= 1(2 – 6) – 0 + 1(0 – 2)

= 1(– 4) + 1(– 2)

= – 4 – 2

= – 6 ≠ 0

∴ A^–1 exists.

Consider AA^–1 = I

∴ ${\displaystyle \left[\begin{array}{ccc}1& 0& 1\\ 0& 2& 3\\ 1& 2& 1\end{array}\right]}$ A^–1 = ${\displaystyle \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]}$

Applying R₃ → R₃ – R₁, we get

${\displaystyle \left[\begin{array}{ccc}1& 0& 1\\ 0& 2& 3\\ 0& 2& 0\end{array}\right]}$ A^–1 = ${\displaystyle \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ -1& 0& 1\end{array}\right]}$

Applying R₂ → ${\displaystyle \left(\frac{1}{2}\right)}$ R₂, we get

${\displaystyle \left[\begin{array}{ccc}1& 0& 1\\ 0& 1& \frac{3}{2}\\ 0& 2& 0\end{array}\right]}$ A^–1 = ${\displaystyle \left[\begin{array}{ccc}1& 0& 0\\ 0& \frac{1}{2}& 0\\ -1& 0& 1\end{array}\right]}$

Applying R₃ → R₃ – 2R₂, we get

${\displaystyle \left[\begin{array}{ccc}1& 0& 1\\ 0& 1& \frac{3}{2}\\ 0& 0& -3\end{array}\right]}$ A^–1 = ${\displaystyle \left[\begin{array}{ccc}1& 0& 0\\ 0& \frac{1}{2}& 0\\ -1& -1& 1\end{array}\right]}$

Applying R₃ → ${\displaystyle (-\frac{1}{3})}$ R₃, we get

${\displaystyle \left[\begin{array}{ccc}1& 0& 1\\ 0& 1& \frac{3}{2}\\ 0& 0& 1\end{array}\right]}$ A^–1 = ${\displaystyle \left[\begin{array}{ccc}1& 0& 0\\ 0& \frac{1}{2}& 0\\ \frac{1}{3}& \frac{1}{3}& -\frac{1}{3}\end{array}\right]}$

Applying R₁ → R₁ – R₃, R₂ → R₂ – ${\displaystyle \left(\frac{3}{2}\right)}$ R₃, we get

${\displaystyle \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]}$ A^–1 = ${\displaystyle \left[\begin{array}{ccc}\frac{2}{3}& -\frac{1}{3}& \frac{1}{3}\\ -\frac{1}{2}& 0& \frac{1}{2}\\ \frac{1}{3}& \frac{1}{3}& -\frac{1}{3}\end{array}\right]}$

∴ A^–1 = ${\displaystyle \left[\begin{array}{ccc}\frac{2}{3}& -\frac{1}{3}& \frac{1}{3}\\ -\frac{1}{2}& 0& \frac{1}{2}\\ \frac{1}{3}& \frac{1}{3}& -\frac{1}{3}\end{array}\right]}$