Systems of Linear Equations

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

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

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

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

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

Multiplying the first by $2$: $2x + 2y = 2 \ne 5$. 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 $a=\dfrac{1}{x}$, $b=\dfrac{1}{y}$, $c=\dfrac{1}{z}$. The system becomes $a+b=\dfrac{3}{5}$, $a+c=\dfrac{2}{3}$, $b+c=\dfrac{1}{4}$.

The question asks for $\dfrac{1}{x}+\dfrac{1}{y}=a+b=\dfrac{3}{5}$ — 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 $\det(A) = 0$.