Question

Solve for minor

(a) $A=\left[\begin{array}{ccc}1& 2& 3\\ 2& 4& 7\\ 3& 6& 10\end{array}\right]$
(b) $A=\left[\begin{array}{ccc}1& 2& 3\\ 2& 3& 4\\ 4& 10& 18\end{array}\right]$

Answer 1

(a) $A$ is $3\times 3$ matrix and we can have minors of order 1, 2, 3.
Minor of order 3.
$\left|\begin{array}{ccc}1& 2& 3\\ 2& 4& 7\\ 3& 6& 10\end{array}\right|$ Apply $R_3-(R_1+R_2)$
$=\left|\begin{array}{ccc}1& 2& 3\\ 2& 4& 7\\ 0& 0& 1\end{array}\right|=0$
Minor of order 2.
$\left|\begin{array}{cc}1& 2\\ 2& 4\end{array}\right|=0,\left|\begin{array}{cc}2& 4\\ 3& 6\end{array}\right|=0$
$\left|\begin{array}{cc}2& 3\\ 4& 7\end{array}\right|=14-12=2\ne 0$
Hence, there is at least one minor of order 2 which is not zero whereas all the minors of order $2+1$ i.e. 3 are zero. Hence $\rho \left(A\right)=2$.
(b) In this case minors of orders $3=-2\ne 0$
$\therefore \rho \left(A\right)=3$