Question

Find the inverse of the following matrix by using row elementary transformation ( 231 341 372)

Answer 1

Given :    A = $\left[\begin{array}{ccc}2&3&1\\3&4&1\\3&7&2\end{array}\right]$  

To Find : A⁻¹   using row elementary transformation

Solution:

A = A I

$\left[\begin{array}{ccc}2&3&1\\3&4&1\\3&7&2\end{array}\right]$   =  A  $\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$  

R₃  → R₃ - R₂

$\left[\begin{array}{ccc}2&3&1\\3&4&1\\0&3&1\end{array}\right]$   =  A  $\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&-1&1\end{array}\right]$  

R₁  → R₁ - R₃

$\left[\begin{array}{ccc}2&0&0\\3&4&1\\0&3&1\end{array}\right]$   =  A  $\left[\begin{array}{ccc}1&1&-1\\0&1&0\\0&-1&1\end{array}\right]$  

R₁  → R₁ /2

$\left[\begin{array}{ccc}1&0&0\\3&4&1\\0&3&1\end{array}\right]$   =  A  $\left[\begin{array}{ccc}1/2&1/2&-1/2\\0&1&0\\0&-1&1\end{array}\right]$  

R₂ →  R₂  - R₃

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

R₂ →  R₂  - 3R₁

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

R₃  → R₃ - 3R₂

$\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$     =  A    $\left[\begin{array}{ccc}1/2&1/2&-1/2\\-3/2&1/2&1/2\\9/2&-5/2&-1/2\end{array}\right]$  

I = AA⁻¹

A⁻¹  =   $\left[\begin{array}{ccc}1/2&1/2&-1/2\\-3/2&1/2&1/2\\9/2&-5/2&-1/2\end{array}\right]$  

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