Systems of Linear Equations

Adding: $3x=9\Rightarrow x=3$. Then $y=x-2=1$. Solution: $(3,1)$.

Cramer: $x_i={\displaystyle \frac{\det (A_i)}{\det (A)}}$.

$x=y$. Then $2y+z=6$ and $y+z=5$, so $y=1$ and $z=4$. Solution: $(1,1,4)$.

$x={\displaystyle \frac{\det (A_1)}{\det (A)}}={\displaystyle \frac{-10}{-5}}=2$.

The second equation is $2\times$ the first, so it provides no new information. The system has infinitely many solutions.

Multiplying the first by $2$: $2x+2y=2\mathrm{\ne }5$. Hence the system is inconsistent — no solution.

Row 2 is a multiple of Row 1, so it doesn't contribute. The non-zero rows after reduction span a 2-dimensional space, so rank $=2$.

Let $a={\displaystyle \frac{1}{x}}$, $b={\displaystyle \frac{1}{y}}$, $c={\displaystyle \frac{1}{z}}$. The system becomes $a+b={\displaystyle \frac{3}{5}}$, $a+c={\displaystyle \frac{2}{3}}$, $b+c={\displaystyle \frac{1}{4}}$.

The question asks for ${\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{y}}=a+b={\displaystyle \frac{3}{5}}$ — directly from the first equation.

R2 − R1: $y+2z=8$. R3 − R1: $3y+8z=30$. From the first: $y=8-2z$. Substituting: $3(8-2z)+8z=30\Rightarrow 24+2z=30\Rightarrow z=3$. Then $y=2$, $x=1$. Solution: $(1,2,3)$.

A square homogeneous system has non-trivial (non-zero) solutions iff $\det (A)=0$.