Question

Find the inverse of $\left[\begin{array}{ccc}1&2&3\\1&1&5\\2&4&7\end{array}\right]$ by adjoint method.

Answer 1

Matrix:

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

Find the determinant:

$\left|\begin{array}{ccc}1&2&3\\1&1&5\\2&4&7\end{array}\right| = 1(1 \times 7 - 4 \times 5) - 2(1 \times 7 - 2 \times 5) + 3(1 \times 4 - 1 \times 2) = -1 \neq 0$

Find the cofactor matrix:

$C_{11} = \left|\begin{array}{ccc}1&5\\4&7\end{array}\right| = (1 \times 7)- (5 \times 4) = -13$

$C_{12} = - \left|\begin{array}{ccc}2&3\\4&7\end{array}\right| =-[ (2 \times 7)- (3 \times 4) ]= -2$

$C_{13} = \left|\begin{array}{ccc}2&3\\1&5\end{array}\right| = (2 \times 5)- (3 \times 1) =7$

$C_{21} = - \left|\begin{array}{ccc}1&5\\2&7\end{array}\right| = -[(1 \times 7)- (2 \times 5) ]=3$

$C_{22} = \left|\begin{array}{ccc}1&5\\2&7\end{array}\right| = (1 \times 7)- (2 \times 5) =1$

$C_{23} = - \left|\begin{array}{ccc}1&3\\1&5\end{array}\right| = -[ (1 \times 5)- (1 \times 3) ] =-2$

$C_{31} = \left|\begin{array}{ccc}1&1\\2&4\end{array}\right| = (1 \times 4)- (1 \times 2) =2$

$C_{32} = -\left|\begin{array}{ccc}1&2\\2&4\end{array}\right| =-[ (1 \times 4)- (2 \times 2) ] =0$

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

Adjoint matrix:

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

Inverse matrix:

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