Review : Systems of Equations — Question 3

Question 3 – Solution

Strategy. Choose an equation with a nonzero coefficient for elimination. An inconvenient ordering does not change the information in a system.

Step 1: Eliminate without dividing by zero. The second equation gives $x=3+y-2z$. Substituting into the third gives $3y-3z=\delta$, or $y-z=\delta/3$. Together with $2y-z=1$, this determines $y,z$, then $x$. For $\delta=0$ the result is $\boxed{(x,y,z)=(2,1,1)}$. Swapping the first two equations would also provide an immediate nonzero $x$ coefficient; neither approach requires a zero division.

Step 2: Solve for the perturbation exactly. Subtract $y-z=\delta/3$ from $2y-z=1$ to obtain $y=1-\delta/3$. It follows that $$\boxed{x=2+\delta,\qquad y=1-\frac\delta 3,\qquad z=1-\frac{2\delta}3.}$$ Only the nonzero constants $1$ and $3$ were used as divisors, so this calculation applies to every real $\delta$.

Step 3: Check the original measurements and uniqueness. Substitution gives $$\begin{aligned} 2y-z&=2-2\delta/3-1+2\delta/3=1,\\ x-y+2z&=2+\delta-1+\delta/3+2-4\delta/3=3,\\ 2x+y+z&=4+2\delta+1-\delta/3+1-2\delta/3=6+\delta. \end{aligned}$$ Every solution was forced by reversible substitutions and nonzero divisions. Thus exactly one solution exists for every real $\delta$.

Step 4: Identify a blind measurement. The combination $x+3y=2+\delta+3-\delta=\boxed{5}$ is independent of $\delta$. In contrast, $z$ determines $\boxed{\delta=\tfrac 32(1-z)}$ uniquely. The first two equations fix certain combinations even when the last datum changes; a measurement that stays constant along that change cannot reveal it.