Página — Matematica 9° Secundaria (pág. 52)

\begin{aligned}&|A_{y}|=3[(1)(1)-(-3)(-2)]-2[(-6)(-1)-(-3)(-1)]+1[(-6)(-2)(1)(-1)]\\ &\\ &|A_{y}|=3[1-6]-2[-6-3]+1[12+1]\\ &\\ &|A_{y}|=3(-5)-2(-9)+1(13)\\ &\\ &|A_{y}|=-15+18+13\\ &\\ &|A_{y}|=16\\ \end{aligned}

Por último, reemplazamos de acuerdo a \left|\mathsf{A}_{z}\right|=\left|\begin{matrix}a_{_1}&b_{_1}&d_{_1}\\ a_{_2}&b_{_2}&d_{_2}\\ a_{_3}&b_{_3}&d_{_3}\end{matrix}\right| y desarrollamos \left|\mathrm{A}_{z}\right|=\left|\begin{array}{c c c|c}3&-2&-6\\2&3&1\\1&-4&-3\end{array}\right|=3\left|\begin{array}{c c|c}3&1&-2&-6\\-4&-3&-2&-4&-3\end{array}\right|+1\left|\begin{array}{c c|c}-2&-6\\-2&-6\\3&1\end{array}\right|

\left|\mathsf{A}_{z}\right|=\left|\begin{matrix}a_{_1}&b_{_1}&d_{_1}\\ a_{_2}&b_{_2}&d_{_2}\\ a_{_3}&b_{_3}&d_{_3}\end{matrix}\right|\left|\mathrm{A}_{z}\right|=\left|\begin{array}{c c c|c}3&-2&-6\\2&3&1\\1&-4&-3\end{array}\right|=3\left|\begin{array}{c c|c}3&1&-2&-6\\-4&-3&-2&-4&-3\end{array}\right|+1\left|\begin{array}{c c|c}-2&-6\\-2&-6\\3&1\end{array}\right|

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

\left|\mathrm{A}_{-}\right|=3[-9+4]-2[6-24]+1[-2+18]

|A_{z}|=3(-5)-2(-18)+1(16)

|A_{z}|=-15+36+16

|A_{z}|=37

De esta manera tenemos que:\left|\mathrm{A}\right|=4,\left|\mathrm{A}_{x}\right|=15,\left|\mathrm{A}_{y}\right|=16,\left|\mathrm{A}_{z}\right|=37

\mathrm{Pro}\quad\mathrm{lo}\quad\mathrm{t}=\frac{\left|A_{x}\right|}{\left|A\right|}=\frac{15}{4};\quad y=\frac{\left|A_{y}\right|}{\left|A\right|}=\frac{16}{4}=4;\quad z=\frac{\left|A_{z}\right|}{\left|A\right|}=\frac{37}{4}

La solución del sistema es:\left(\frac{15}{4},4,\frac{37}{4}\right)

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