Question

Find the inverse of $\left[\begin{array}{ccc}1&0&1\\0&2&3\\1&2&1\end{array}\right]$ by adjoint method.

Answer 1

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

Find the determinant:

$\left|\begin{array}{ccc}1&0&1\\0&2&3\\1&2&1\end{array}\right| = 1(2 - 6) - 0(1 - 3) + (0 - 2) = -6 \neq 0$

Find the co-factors:

$C_{11} = \left|\begin{array}{cc}2&3\\2&1\end{array}\right| = (2 \times 1) - (2 \times 3) = -4$

$C_{21} = - \left|\begin{array}{cc}0&1\\2&1\end{array}\right| = - [(0 \times 1) - (2 \times 1)] = 2$

$C_{31} = \left|\begin{array}{cc}0&1\\2&3\end{array}\right| = (0 \times 3) - (1 \times 2) = -2$

$C_{12} = - \left|\begin{array}{cc}0&3\\1&1\end{array}\right| = -[(0 \times 1) - (3 \times 1)] = 3$

$C_{22} = \left|\begin{array}{cc}1&1\\1&1\end{array}\right| = (1 \times 1) - (1 \times 1) = 0$

$C_{32} = - \left|\begin{array}{cc}1&1\\0&3\end{array}\right| = - [(1 \times 3) - (0 \times 1)] = -3$

$C_{13} = \left|\begin{array}{cc}0&2\\1&3\end{array}\right| = (0 \times 3) - (1 \times 2) = -2$

$C_{23} = - \left|\begin{array}{cc}1&0\\1&2\end{array}\right| = - [ (1 \times 2) - (1 \times 0)] = -2$

$C_{33} = \left|\begin{array}{cc}1&0\\1&2\end{array}\right| = (1 \times 2) - (1 \times 0) = 2$

Adjoint Matrix:

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

Find the inverse:

$-\dfrac{1}{6} \left[\begin{array}{ccc}-4&2&-2\\3&0&-3\\-2&-2&2\end{array}\right] = \left[\begin{array}{ccc}2/3&-1/3&1/3\\-1/2&0&1/2\\1/3&1/3&-1/3\end{array}\right]$