LA2.2.5

Rank of $A\le 3\,$ since A is of 3rd order.

$|A|=4(-6+6)-2(-12+12)+3(-8+8)=0\,$

Since $|A|=0\,$ rank of $A < 3\,$ i.e $r(A)\le 2\,$

Consider the determinants of 2nd order submatrices

$\begin{vmatrix} 4 & 2 \\ 8 & 4 \end{vmatrix}=0\,$

$\begin{vmatrix} 2 & 3 \\ 4 & 6 \end{vmatrix}=0\,$

$\begin{vmatrix} 4 & 3 \\ 8 & 6 \end{vmatrix}=0\,$

$\begin{vmatrix} 4 & 2 \\ -2 & -1 \end{vmatrix}=0\,$

$\begin{vmatrix} 4 & 3 \\ -1 & -15 \end{vmatrix}=0\,$

$\begin{vmatrix} 4 & 3 \\ -2 & -15 \end{vmatrix}=0\,$

Since all 2nd order submatrices have zero determinants i.e 2nd order minors are all zero.

So $r(A)< 2\,$

Since A is a non-zero matrix $r(A)> 0\,$

Thus the rank of A is 1.