Question

Find the rank of the following matrix:
(a)$A=\left[\begin{array}{ccc}1& 3& 4\\ 2& 6& 8\end{array}\right]$.
(b)$A=\left[\begin{array}{ccc}3& -1& 2\\ -6& 2& -4\\ -3& 1& -2\end{array}\right]$.

Answer 1

A is $2\times 3$ matrix and we can have minors of order 1 and 2. Minor of order 2.
$\left|\begin{array}{ccc}1& 3& 4\\ 2& 6& 8\end{array}\right|$= 6-6=0
Other minors of order 2 are
$\left|\begin{array}{cc}3& 4\\ 6& 8\end{array}\right|$= 0 and $\left|\begin{array}{cc}1& 4\\ 2& 8\end{array}\right|$=0
Thus all minors of order 2 are zero, because of . identical lines, Hence, we have only minors of order 1, which will be elements of the given matrix and are not all equal to zero or at least one of them is not equal to zero. Hence, the rank of A is 1.
(b) Proceed as above:. The minors of order 3 and 2 are all zero because of identical lines.
$R_2$ =-2$R_1$,$R_3$ =- $R_1$ $\therefore$ Rank A = 1