Question

Complete the following activity to find inverse of matrix using elementary column transformations and hence verify.

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

C₁ → C₁ + C₃

${\displaystyle \left[\begin{array}{ccc}\text{( )}& 0& -1\\ \text{( )}& 1& 0\\ \text{( )}& 1& 3\end{array}\right]}$ B⁻¹ = ${\displaystyle \left[\begin{array}{ccc}\text{( )}& 0& 0\\ \text{( )}& 1& 0\\ \text{( )}& 0& 1\end{array}\right]}$

C₃ → C₃ + C₁ 

${\displaystyle \left[\begin{array}{ccc}1& 0& 0\\ \text{( )}& 1& \text{( )}\\ 3& 1& \text{( )}\end{array}\right]}$ B⁻¹ = ${\displaystyle \left[\begin{array}{ccc}1& 0& \text{( )}\\ 0& 1& 0\\ \text{( )}& 0& \text{( )}\end{array}\right]}$

C₁ → C₁ – 5C₂, C₃ → C₃ – 5C₂

${\displaystyle \left[\begin{array}{ccc}1& \text{( )}& 0\\ 0& 1& 0\\ \text{( )}& 1& \text{( )}\end{array}\right]}$ B⁻¹ = ${\displaystyle \left[\begin{array}{ccc}1& 0& \text{( )}\\ \text{( )}& 1& -5\\ 1& \text{( )}& 2\end{array}\right]}$

C₁ → C₁ – 2C₃, C₂ → C₂ – C₃ 

${\displaystyle \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]}$ B⁻¹ = ${\displaystyle \left[\begin{array}{ccc}3& -1& \text{( )}\\ \text{( )}& 6& -5\\ 5& \text{( )}& \text{( )}\end{array}\right]}$

B⁻¹ =  ${\displaystyle \left[\begin{array}{ccc}\text{( )}& \text{( )}& \text{( )}\\ \text{( )}& \text{( )}& \text{( )}\\ \text{( )}& \text{( )}& \text{( )}\end{array}\right]}$

${\displaystyle \left[\begin{array}{ccc}2& \text{( )}& -1\\ \text{( )}& 1& 0\\ 0& 1& \text{( )}\end{array}\right]\left[\begin{array}{ccc}3& \text{( )}& \text{( )}\\ \text{( )}& 6& \text{( )}\\ \text{( )}& -2& \text{( )}\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]}$

Answer 1

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

Applying C₁ → C₁ + C₃, we get

${\displaystyle \left[\begin{array}{ccc}1& 0& -1\\ 5& 1& 0\\ 3& 1& 3\end{array}\right]}$ B⁻¹ = ${\displaystyle \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 1& 0& 1\end{array}\right]}$

Applying C₃ → C₃ + C₁, we ge

${\displaystyle \left[\begin{array}{ccc}1& 0& 0\\ 5& 1& 5\\ 3& 1& 6\end{array}\right]}$ B⁻¹ = ${\displaystyle \left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 1& 0& 2\end{array}\right]}$

Applying C₁ → C₁ – 5C₂, C₃ → C₃ – 5C₂, we get

${\displaystyle \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ -2& 1& -1\end{array}\right]}$ B⁻¹ = ${\displaystyle \left[\begin{array}{ccc}1& 0& 1\\ -5& 1& -5\\ 1& 0& 2\end{array}\right]}$

Applying C₁ → C₁ – 2C₃, C₂ → C₂ – C₃, we get

${\displaystyle \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]}$ B⁻¹ = ${\displaystyle \left[\begin{array}{ccc}3& -1& 1\\ -15& 6& -5\\ 5& -2& 2\end{array}\right]}$

B⁻¹ =  ${\displaystyle \left[\begin{array}{ccc}3& -1& 1\\ -15& 6& -5\\ 5& -2& 2\end{array}\right]}$

${\displaystyle \left[\begin{array}{ccc}2& 0& -1\\ 5& 1& 0\\ 0& 1& 3\end{array}\right]\left[\begin{array}{ccc}3& -1& 1\\ -15& 6& -5\\ 5& -2& 2\end{array}\right]=\left[\begin{array}{ccc}6-0-5& -2+0+2& 2-0-2\\ 15-15+0& -5+6-0& 5-5+0\\ 0-15+15& 0+6-6& 0-5+6\end{array}\right]}$

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