# Inverse of $\begin{bmatrix}1&2&3\\2&3&4\\3&4&6\end{bmatrix}$ is

$\begin{array}{1 1}(a)\;\begin{bmatrix}-2&0&1\\0&3&2\\1&-2&1\end{bmatrix}&(b)\;\begin{bmatrix}2 &0&-1\\0&-3&2\\-1&2&-1\end{bmatrix}\\(c)\;\begin{bmatrix}1&2&3\\2&3&4\\3&4&6\end{bmatrix}&(d)\;None\;of\;these\end{array}$

Let $A=\begin{bmatrix}1&2&3\\2&3&4\\3&4&6\end{bmatrix}$
Then $adj A=\begin{bmatrix}2 &0&-1\\0&-3&2\\-1&2&-1\end{bmatrix}$
Also $\mid A\mid =\begin{vmatrix}1&2&3\\2&3&4\\3&4&6\end{vmatrix}$
$\Rightarrow 1(2)-2(0)+3(-1)$
$\Rightarrow -1$
$A^{-1}=\large\frac{adj\;A}{\mid A\mid}$
$\qquad=\begin{bmatrix}-2&0&1\\0&3&2\\1&-2&1\end{bmatrix}$
Hence (a) is the correct answer.