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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., $\det\left(\begin{array}{ccc} 1 & -1 & -1\\ 0 & 1-\lambda & 0\\ 0 & 0 & 1-\lambda \end{array}\right)<\det\left(\begin{array}{ccc} 3-\lambda & -1 & -1\\ -1 & 1-\lambda & 0\\ -1 & 0 & 1-\lambda \end{array}\right)$.

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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

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  • $\begingroup$ Thanks! But is there a way to bypass the direct calculation of determinant, because this is a special case of my problem, basically the size of matrix is general $N$, could we look at the problem from the structure of the matrix in the determinant and use the property of $0 < \lambda <1$? $\endgroup$ – user96212 Sep 9 '14 at 20:30

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