7.4-bo'lim : Kengaytirilgan matritsa haqida ko'proq
2. Quyidagi tenglamalar sistemasi uchun tizimni kengaytirilgan matritsaga aylantiring va tizimning yechimini aniqlash yoki tizimning nomuvofiq yoki bog'liqligini aniqlash uchun to'ldirilgan matritsa usullaridan foydalaning.
\[\begin{align*}7x - 8y & = - 12\\ - 4x + 2y & = 3\end{align*}\]Yechim
Birinchi qadam tenglamalar sistemasi uchun kengaytirilgan matritsani yozishdir.
\[\left[ {\begin{array}{rr|r}7&{ - 8}&{ - 12}\\{ - 4}&2&3\end{array}} \right]\]2-qadam
Yuqori chap burchakdagi raqamni bitta qilishimiz kerak. Buning uchun bir necha usullar mavjud. Buning bir usuli - \(\frac{1}{7}{R_{\,1}}\) elementar qator operatsiyasidan foydalanish. Biroq, bu birinchi qatordagi boshqa ikkita yozuvga kasrlarni qo'yadi va ulardan qochish yaxshi bo'lishi mumkin.
Shunday qilib, o'rniga quyidagi elementar qator amalini bajaramiz.
\[\left[ {\begin{array}{rr|r}7&{ - 8}&{ - 12}\\{ - 4}&2&3\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,1}} + 2{R_{\,2}} \to {R_{\,1}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rr|r}{ - 1}&{ - 4}&{ - 6}\\{ - 4}&2&3\end{array}} \right]\]Endi bu biz xohlagan narsa emas, chunki yuqori chap tomondagi raqam ijobiy emas, minus. Biroq, biz buni birinchi qatorni -1 ga ko'paytirish orqali osongina tuzatishimiz mumkin.
\[\left[ {\begin{array}{rr|r}{ - 1}&{ - 4}&{ - 6}\\{ - 4}&2&3\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{ - {R_{\,1}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rr|r}1&4&6\\{ - 4}&2&3\end{array}} \right]\]E'tibor bering, ushbu bosqichda ko'rsatilgandek, ushbu muammolarni hal qilishning bir nechta turli yo'llari mavjud. Ba'zilari oraliq bosqichlarni "tartibsizroq" qiladi, ammo biz hal qilish jarayonida qanday yo'l tanlaganimizdan qat'i nazar, biz erishgan yechim bir xil bo'ladi.
3-qadam
Keyinchalik, 1 dan pastdagi -4 ni nolga aylantirishimiz kerak va biz buni quyidagi elementar qator operatsiyasi bilan amalga oshirishimiz mumkin.
\[\left[ {\begin{array}{rr|r}1&4&6\\{ - 4}&2&3\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,2}} + 4{R_{\,1}} \to {R_{\,2}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rr|r}1&4&6\\0&{18}&{27}\end{array}} \right]\]4-qadam
Keyingi qadam, ikkinchi ustunning pastki qismidagi raqamni (bu holda 18) bittaga aylantirishdir. Quyidagi elementar qator operatsiyasi biz uchun buni amalga oshiradi.
\[\left[ {\begin{array}{rr|r}1&4&6\\0&{18}&{27}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{\frac{1}{{18}}{R_{\,2}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{r}}1&4&6\\0&1&{\frac{{27}}{{18}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}1&4&6\\0&1&{\frac{3}{2}}\end{array}} \right]\]Birinchi bosqichda biz kasrlarni kengaytirilgan matritsaga qo'yadigan qadamdan qochishni tanladik, lekin ba'zida, bu bosqichda bo'lgani kabi, ulardan qochib bo'lmaydi.
5-qadam
Nihoyat, biz 4-bosqichda olingan raqamdan yuqoridagi raqamni nolga aylantirishimiz kerak. Buning uchun quyidagi elementar qator amalidan foydalanishimiz mumkin.
\[\left[ {\begin{array}{rr|r}1&4&6\\0&1&{\frac{3}{2}}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,1}} - 4{R_{\,2}} \to {R_{\,1}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rr|r}1&0&0\\0&1&{\frac{3}{2}}\end{array}} \right]\]6-qadam
Biz 5-bosqichda topilgan yakuniy kengaytirilgan matritsadan tizimning yechimini olamiz: \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = 0,\,\,y = \frac{3}{2}}}\).