Problem 2

Consider the system

a_{11}x_1+a_{12}x_2+a_{13}x_3=0

a_{21}x_1+a_{22}x_2+a_{23}x_3=0

a_{31}x_1+a_{32}x_2+a_{33}x_3=0

. The diagonal coefficients

a_{11},a_{22},a_{33}

are positive, the other coefficients are negative, and the sum of the coefficients in each equation is positive. Prove that the system has only the solution

x_1=x_2=x_3=0

Step 1 of 4: Convert the row-sum condition

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In plain words

The diagonal term is stronger than the two opposing terms combined.

a_{ii}>0,\\quad a_{ij}<0\\ (i\ne j),\\quad a_{ii}+a_{ij}+a_{ik}>0\\Longrightarrow a_{ii}>|a_{ij}|+|a_{ik}|.

Detailed analysis

In row $i$ , the two off-diagonal entries are negative. Thus positivity of the row sum is exactly the strict diagonal-dominance inequality $a_{ii}>|a_{ij}|+|a_{ik}|$ .