### Teorija

Pieņem, ka ir dota trīs lineāru vienādojumu sistēma ar trim nezināmajiem $$x$$, $$y$$ un $$z$$.
$\left\{\begin{array}{c}{a}_{1}x+{b}_{1}y+{c}_{1}z={d}_{1}\\ {a}_{2}x+{b}_{2}y+{c}_{2}z={d}_{2}\\ {a}_{3}x+{b}_{3}y+{c}_{3}z={d}_{3}\end{array}\right\$

Skaitļus ${a}_{i}$, ${b}_{i}$ un ${c}_{i}$ ,ja $$i=1,2,3$$, sauc par koeficientiem.
Skaitļus ${d}_{i}$, ja $$i=1,2,3$$, sauc par brīvajiem locekļiem.

Skaitli $\mathrm{\Delta }$ = $\left|\begin{array}{ccc}{a}_{1}& {b}_{1}& {c}_{1}\\ {a}_{2}& {b}_{2}& {c}_{2}\\ {a}_{3}& {b}_{3}& {c}_{2}\end{array}\right|$ sauc par sistēmas determinantu. Tas sastādīts no koeficientiem pie nezināmajiem $$x$$, $$y$$ un $$z$$.
Skaitļus ${\mathrm{\Delta }}_{x}$ = $\left|\begin{array}{ccc}{d}_{1}& {b}_{1}& {c}_{1}\\ {d}_{2}& {b}_{2}& {c}_{2}\\ {d}_{3}& {b}_{3}& {c}_{3}\end{array}\right|$, ${\mathrm{\Delta }}_{y}$ = $\left|\begin{array}{ccc}{a}_{1}& {d}_{1}& {c}_{1}\\ {a}_{2}& {d}_{2}& {c}_{2}\\ {a}_{3}& {d}_{3}& {c}_{3}\end{array}\right|$, ${\mathrm{\Delta }}_{z}$ = $\left|\begin{array}{ccc}{a}_{1}& {b}_{1}& {d}_{1}\\ {a}_{2}& {b}_{2}& {d}_{2}\\ {a}_{3}& {b}_{3}& {d}_{3}\end{array}\right|$ sauc par nezināmo determinantiem. Nezināmo determinantus iegūst no sistēmas determinanta, aizvietojot tajā koeficientu pie atbilstoša nezināmā ar brīvajiem locekļiem ${d}_{1}$, ${d}_{2}$, ${d}_{3}$.

Sistēmai ir atrisinājums, kuru var aprēķināt pēc Krāmera formulām:
$x=\frac{{\mathrm{\Delta }}_{x}}{\mathrm{\Delta }}$, $y=\frac{{\mathrm{\Delta }}_{y}}{\mathrm{\Delta }}$, $z=\frac{{\mathrm{\Delta }}_{z}}{\mathrm{\Delta }}$.

Piemērs:
$\left\{\begin{array}{c}5x+3y-z=-1\\ x-y+z=0\\ 2x-3y-z=5\end{array}\right\$

$\mathrm{\Delta }=\left|\begin{array}{ccc}5& 3& -1\\ 1& -1& 1\\ 2& -3& -1\end{array}\right|=30$

${\mathrm{\Delta }}_{x}=\left|\begin{array}{ccc}-1& 3& -1\\ 0& -1& 1\\ 5& -3& -1\end{array}\right|=6$

${\mathrm{\Delta }}_{y}=\left|\begin{array}{ccc}5& -1& -1\\ 1& 0& 1\\ 2& 5& -1\end{array}\right|=-33$

${\mathrm{\Delta }}_{z}=\left|\begin{array}{ccc}5& 3& -1\\ 1& -1& 0\\ 2& -3& 5\end{array}\right|=-39$

$x=\frac{{\mathrm{\Delta }}_{x}}{\mathrm{\Delta }}=\frac{6}{30}=0,2$
$y=\frac{{\mathrm{\Delta }}_{y}}{\mathrm{\Delta }}=\frac{-33}{30}=-1,1$
$z=\frac{{\mathrm{\Delta }}_{z}}{\mathrm{\Delta }}=\frac{-39}{30}=-1,3$

Atbilde:
$$x = 0,2$$; $$y = -1,1$$; $$z = -1,3$$