Question

If A = [ x 0 0 0 y 0 0 0 z ] is a nonsingular matrix then find A^-1 by elementary row transformations, Hence, find the inverse of [ 2 0 0 0 1 0 0 0 -1 ]

Answer 1

Given : $A = \left[\begin{array}{ccc}x&0&0\\0&y&0\\0&0&z\end{array}\right]$    

To find : A⁻¹ by elementary row transformations

Solution:

A = AI

$A = \left[\begin{array}{ccc}x&0&0\\0&y&0\\0&0&z\end{array}\right]$  

= > $\left[\begin{array}{ccc}x&0&0\\0&y&0\\0&0&z\end{array}\right]$    = A $\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$  

R₁ →  R₁ /x  , R₂ →  R₂ /y  ,    R₃ →  R₃ /z

=> $\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$    = A $\left[\begin{array}{ccc}1/x&0&0\\0&1/y&0\\0&0&1/z\end{array}\right]$  

I = AA⁻¹

=>  A⁻¹ = $\left[\begin{array}{ccc}1/x&0&0\\0&1/y&0\\0&0&1/z\end{array}\right]$  

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

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