Question

If $|a_1|>|a_2|+|a_3|,|b_2|>|b_1|+|b_3|$ and $|c_3|>|c_1|+|c_2|$, then show $\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|=0$

Answer 1

Let the given determinant be equal to zero. Then there exists $x,y,$ and $z,$ not all zero, such that

$a_{1}x+a_{2}y+a_{3}z=0$

$b_{1}x+b_{2}y+b_{3}z=0$

and $c_{1}x+c_{2}y+c_{3}z=0$

assume that $|x|\ge |y|\ge |z|$ and $x\ne 0$. Then from $a_{1}x=(-a_{2}y)+(-a_{3}z)$

$\therefore |a_{1}x|=|-a_{2}y-a_{3}z|\le |a_{2}y|+|a_{3}z|$

$\Rightarrow |a_1||x|\le |a_2||y|+|a_3||z|$

But $x\ne 0$

i.e., $|a_1|\le |a_2|+|a_3|$

similarly $|b_2|\le |b_1|+|b_3|$

$|c_3|\le |c_1|+|c_2|$

which is contradiction. Hence the assumption that the determinant is zero must be wrong.