# Prove that the given determinant is not zero

Question Statement:-

If $|a_1|\gt |a_2|+|a_3|, |b_2|\gt|b_1|+|b_3|$ and $|c_3|\gt |c_1|+|c_2|$, then show that $$\begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} \neq 0$$

Attempt at a solution:-

Consider the following system of equations with the conditions $|a_1|\gt |a_2|+|a_3|, |b_2|\gt|b_1|+|b_3|$ and $|c_3|\gt |c_1|+|c_2|$:- $$a_1x +a_2y+a_3z=0 \tag{1}$$ $$b_1x +b_2y+b_3z=0 \tag{2}$$ $$c_1x +c_2y+c_3z\neq0 \tag{3}$$

From $(1)$ and $(2)$, we have $$\dfrac{x}{\begin{vmatrix} a_2 & a_3\\ b_2 & b_3 \end{vmatrix}}=\dfrac{-y}{\begin{vmatrix} a_1 & a_3\\ b_1 & b_3 \end{vmatrix}}=\dfrac{z}{\begin{vmatrix} a_1 & a_2\\ b_1 & b_2 \end{vmatrix}}$$

Using the above relation in $(3)$, we get $$\begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} \neq 0$$

As you can see, this solution seems like stupidity of the highest kind which is due to my not being able to come up with a good line of thought for the solution. So, please guide me on what should be my approach or line of thought while attempting this problem.

Also it seems that if somehow I could show that the inequation $(3)$ is indeed correct and was arrived upon by using the given conditions in the question then my solution is also correct. So, if you could also help me on that front.

In fact, I'm really having hard time to understand what you did. In particular how do you derive $$\dfrac{x}{\begin{vmatrix} a_2 & a_3\\ b_2 & b_3 \end{vmatrix}}=\dfrac{-y}{\begin{vmatrix} a_1 & a_3\\ b_1 & b_3 \end{vmatrix}}=\dfrac{z}{\begin{vmatrix} a_1 & a_2\\ b_1 & b_2 \end{vmatrix}}?$$ and why do you suppose $c_1x +c_2y+c_3z\neq0$?
Regarding the prove. Suppose as you did that $(x,y,z) \in \mathbb R^3$ is a triple of reals such that $$\left \{ \begin{array}{l c l} a_1x +a_2y+a_3z &=0 \\ b_1x +b_2y+b_3z &=0 \\ c_1x +c_2y+c_3z &= 0 \end{array} \tag{a} \right.$$ Suppose that $(x,y,z) \neq (0,0,0)$ and that $\vert x \vert \ge \vert y \vert \ge \vert z \vert$. Necessarily, $\vert x \vert >0$ and $$\vert a_1 \vert = \left\vert a_2 \frac{y}{x} + a_3 \frac{y}{x}\right\vert \le \vert a_2 \vert \left\vert \frac{y}{x} \right\vert + \vert a_3 \vert \left\vert \frac{z}{x} \right\vert \le \vert a_2 \vert + \vert a_3 \vert$$ in contradiction with your hypothesis $|a_1|\gt |a_2|+|a_3|$. If $x$ is not having the largest absolute value, you can mimic the proof using the other equations. As our hypothesis $(x,y,z) \neq (0,0,0)$ leads to a contradiction, we have to conclude that $(0,0,0)$ is the only solution of the equations $(a)$, which implies as desired that the determinant is non-zero.
The result can be extended to a matrix $A=(A_{i,j})$ with size $n \ge 1$. If for all $j \in \{1, \dots, n\}$ you have $$\vert A_{j,j} \vert > \sum_{i \neq j} \vert A_{j,i} \vert$$ then $\det A \neq 0$.
• I came up with $$\dfrac{x}{\begin{vmatrix} a_2 & a_3\\ b_2 & b_3 \end{vmatrix}}=\dfrac{-y}{\begin{vmatrix} a_1 & a_3\\ b_1 & b_3 \end{vmatrix}}=\dfrac{z}{\begin{vmatrix} a_1 & a_2\\ b_1 & b_2 \end{vmatrix}}$$ using the cross multiplication method for the equations $(1)$ and $(2)$, you can verify it too. As for how I came up with $c_1x +c_2y+c_3z\neq0$, believe me I don't know myself hence I wrote to help me regarding that in the last paragraph of my original post Commented Oct 28, 2016 at 12:08
If the determinant were zero, one of the three rows would be a linear combination of the other two, say $(c_1,c_2,c_3)=x(a_1,a_2,a_3)+y(b_1,b_2,b_3)$. So we have $$|xa_1|\geq |xa_2|+|xa_3|$$ $$|yb_2| \geq |yb_1|+|yb_3|$$ And adding up we get $$|xa_1|+|yb_2| \geq |xa_2|+|xa_3|+|yb_1|+|yb_3|$$ $$|xa_1|-|yb_1|+|yb_2|-|xa_2|\geq|xa_3|+|yb_3|$$ But \begin{align}|xa_3|+|yb_3|&\geq|xa_3+yb_3|\\&=|c_3|\\&>|c_1|+|c_2|\\&= |xa_1+yb_1| + |xa_2+yb_2| \\&\geq|xa_1|-|yb_1|+|yb_2|-|xa_2| \end{align} So $|xa_1|-|yb_1|+|yb_2|-|xa_2|>|xa_1|-|yb_1|+|yb_2|-|xa_2|$, a contradiction. This argument generalizes easily to $n \times n$ matrices.