Página — Matematica 9° Secundaria (pág. 50)

\left|\mathsf{A}_{z}\right|=\left|\begin{array}{ccc}1&4&-6\\2&-8&12\\3&4&-3\\\end{array}\right|=1\left|\begin{array}{cc}-8&12\\4&-3\\\end{array}\right|-2\left|\begin{array}{cc}4&-6\\4&-3\\\end{array}\right|+3\left|\begin{array}{cc}4&-6\\-8&12\\\end{array}\right|

\left|\mathrm{A}_{-}\right|=1\left[(-8)(-3)-(4)(12)\right]-2\left[(4)(-3)-(4)(-6)\right]+3\left[(4)(12)-(-8)(-6)\right]

|A_{z}|=1[24-48]-2[-12+24]+3[48-48]

|A_{z}|=1(-24)-2(12)-3(0)

|A_{z}|=-24-24+0

|A_{z}|=-48

De esta manera tenemos que:|A|=-24,\left|A_{x}\right|=-12,\left|A_{y}\right|=3,\left|A_{z}\right|=-48

\mathrm{Pro}\quad\mathrm{lo}\quad\mathrm{t}=\frac{\left|A_{x}\right|}{\left|A\right|}=\frac{-12}{-24}=\frac{1}{2};\quad y=\frac{\left|A_{y}\right|}{\left|A\right|}=-\frac{3}{24}=-\frac{1}{8};\quad z=\frac{\left|A_{z}\right|}{\left|A\right|}=-\frac{-48}{-24}=2.

La solución del sistema es:\left(\frac{1}{2},-\frac{1}{8},2\right)

Ejemplo 2

Resolver el siguiente sistema de ecuaciones utilizando determinantes.

Resolvamos

\begin{cases}3x-2y-z=-6\\2x+3y-2z=1\\\quad x-4y+z=-3\end{cases}

Identificamos las constantes para poder armar la matriz de 3\times3

\begin{array}{l}a_{1}=3\ b_{1}=-2\ c_{1}=-1\ d_{1}=-6\\a_{2}=2\ b_{2}=3\ c_{2}=-2\ d_{2}=1\\a_{3}=1\ b_{3}=-4\ c_{3}=1\ d_{3}=-3\end{array}

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