Question

If A=[1 2 3], [1 1 5], [2 4 7] then find A-1 by elementary column transformations
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Answer 1

Given :

$\left[\begin{array}{ccc}1&2&3\\1&1&5\\2&4&7\end{array}\right]$

To find :

Inverse of A

Solution:

$\left[\begin{array}{ccc}1&2&3\\1&1&5\\2&4&7\end{array}\right] = A. \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$

C₂ => C₂ - 2C₁

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

C₁ => C₁ + C₂

C₂ => -C₂

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

C₃ => C₃ - 2C₂

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

C₁ => C₁ - 2C₃

$\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] =A. \left[\begin{array}{ccc}13&2&-7\\-3&-1&2\\-2&0&1\end{array}\right]$

I = A × A⁻¹

So,

$A^{-1 } =\left[\begin{array}{ccc}13&2&-7\\-3&-1&2\\-2&0&1\end{array}\right]$