Suppose we have a linear equation with parameter $0 <\lambda <1$ as
$\left(\begin{array}{ccc} 3-\lambda & -1 & -1\\ -1 & 1-\lambda & 0\\ -1 & 0 & 1-\lambda \end{array}\right)\left(\begin{array}{c} v_{1}\\ v_{2}\\ v_{4} \end{array}\right)=\left(\begin{array}{c} 1\\ 0\\ 0 \end{array}\right)v_{3}$
According to Cramer's rule, we have
$$ v_{1}=\frac{\det\left(\begin{array}{ccc} 1 & -1 & -1\\ 0 & 1-\lambda & 0\\ 0 & 0 & 1-\lambda \end{array}\right)}{\det\left(\begin{array}{ccc} 3-\lambda & -1 & -1\\ -1 & 1-\lambda & 0\\ -1 & 0 & 1-\lambda \end{array}\right)} v_3 $$
Now, how to show $|v_{1}|>|v_{3}|$, i.e., $\det\left(\begin{array}{ccc} 1 & -1 & -1\\ 0 & 1-\lambda & 0\\ 0 & 0 & 1-\lambda \end{array}\right)<\det\left(\begin{array}{ccc} 3-\lambda & -1 & -1\\ -1 & 1-\lambda & 0\\ -1 & 0 & 1-\lambda \end{array}\right)$.
We explicitly compute the two determinants (i.e. using Sarrus' rule): $\det\begin{pmatrix} 1&-1&-1\\ 0&1-\lambda&0\\ 0&0&1-\lambda \end{pmatrix}=1\cdot(1-\lambda)^2=\lambda^2-2\lambda+1$
$\det\begin{pmatrix} 3-\lambda&-1&-1\\ -1&1-\lambda&0\\ -1&0&1-\lambda \end{pmatrix} = (3-\lambda)(1-\lambda)^2-(1-\lambda)-(1-\lambda)=\lambda^3+\lambda^2-5\lambda+1$
$\lambda^2-2\lambda+1<\lambda^3+\lambda^2-5\lambda+1$ follows immediately:
$\lambda^3-3\lambda>0$
$\lambda(\lambda^2-3)>0$
And since $0<\lambda<1$, this always holds