Question

Find A⁻¹ by adjoint method and by elementary transformations if A = $\left[\begin{array}{ccc}1&2&3\\-1&1&2\\1&2&4\end{array}\right]$

Answer 1

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

Find the inverse:

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

R3 = R3 - 2xR2

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

Swap R3 with R1

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

R1 = R1 ÷ 3

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

R2 = R1 + R1

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

R3 = R3 - R2

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

R3 = R3 - R1

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

R2 = R2 - R3

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

R3 = R3 - R2

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

Swap R3 with R2

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

The inverse is:

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