7.4-bo'lim : Kengaytirilgan matritsa haqida ko'proq
8. 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.
\[\begin{align*}3x - 9z & = 33\\ 7x - 4y - z & = - 15\\ 4x + 6y + 5z & = - 6\end{align*}\]Yechim
Birinchi qadam tenglamalar sistemasi uchun kengaytirilgan matritsani yozishdir.
\[\left[ {\begin{array}{rrr|r}3&0&{ - 9}&{33}\\7&{ - 4}&{ - 1}&{ - 15}\\4&6&5&{ - 6}\end{array}} \right]\]Birinchi qatorning ikkinchi ustunidagi nolga e'tibor bering. Eslatib o'tamiz, ikkinchi ustun har bir tenglamadagi \(y\) koeffitsientlariga to'g'ri keladi va birinchi tenglamada \(y\) yo'qligi sababli koeffitsient nolga teng bo'lishi kerak.
2-qadam
Yuqori chap burchakdagi raqamni bitta qilishimiz kerak. Biz buni quyidagi elementar qator operatsiyasi bilan osonlikcha bajarishimiz mumkin.
\[\left[ {\begin{array}{rrr|r}3&0&{ - 9}&{33}\\7&{ - 4}&{ - 1}&{ - 15}\\4&6&5&{ - 6}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{\frac{1}{3}{R_{\,1}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}1&0&{ - 3}&{11}\\7&{ - 4}&{ - 1}&{ - 15}\\4&6&5&{ - 6}\end{array}} \right]\]3-qadam
Keyinchalik, birinchi ustundagi 1 ostidagi 7 va 4 raqamlarini nolga aylantirishimiz kerak va biz buni quyidagi elementar qator operatsiyalari bilan amalga oshirishimiz mumkin.
\[\left[ {\begin{array}{rrr|r}1&0&{ - 3}&{11}\\7&{ - 4}&{ - 1}&{ - 15}\\4&6&5&{ - 6}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,2}} - 7{R_{\,1}} \to {R_{\,2}}}\\{{R_{\,3}} - 4{R_{\,1}} \to {R_{\,3}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}1&0&{ - 3}&{11}\\0&{ - 4}&{20}&{ - 92}\\0&6&{17}&{ - 50}\end{array}} \right]\]4-qadam
Endi biz ikkinchi qatordagi -4 ni bittaga aylantirishimiz kerak va buni quyidagi elementar qator operatsiyasi bilan bajarish mumkin.
\[\left[ {\begin{array}{rrr|r}1&0&{ - 3}&{11}\\0&{ - 4}&{20}&{ - 92}\\0&6&{17}&{ - 50}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{ - \frac{1}{4}{R_{\,2}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}1&0&{ - 3}&{11}\\0&1&{ - 5}&{23}\\0&6&{17}&{ - 50}\end{array}} \right]\]5-qadam
Keyinchalik, uchinchi qatordagi 6 ni nolga aylantirishimiz kerak. Quyidagi elementar qator operatsiyasi biz uchun buni amalga oshiradi.
\[\left[ {\begin{array}{rrr|r}1&0&{ - 3}&{11}\\0&1&{ - 5}&{23}\\0&6&{17}&{ - 50}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,3}} - 6{R_{\,2}} \to {R_{\,3}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}1&0&{ - 3}&{11}\\0&1&{ - 5}&{23}\\0&0&{47}&{ - 188}\end{array}} \right]\]6-qadam
OK, uchinchi qatordagi 47 ni bittaga aylantirishimiz kerak va buni quyidagicha qilishimiz mumkin:
\[\left[ {\begin{array}{rrr|r}1&0&{ - 3}&{11}\\0&1&{ - 5}&{23}\\0&0&{47}&{ - 188}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{\frac{1}{{47}}{R_{\,3}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}1&0&{ - 3}&{11}\\0&1&{ - 5}&{23}\\0&0&1&{ - 4}\end{array}} \right]\]7-qadam
Keyinchalik, uchinchi ustundagi -5 va -3 ni nolga aylantirishimiz kerak. Quyidagi elementar qator operatsiyalari biz uchun buni amalga oshiradi.
\[\left[ {\begin{array}{rrr|r}1&0&{ - 3}&{11}\\0&1&{ - 5}&{23}\\0&0&1&{ - 4}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,1}} + 3{R_{\,3}} \to {R_{\,1}}}\\{{R_{\,2}} + 5{R_{\,3}} \to {R_{\,2}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}1&0&0&{ - 1}\\0&1&0&3\\0&0&1&{ - 4}\end{array}} \right]\]8-qadam
Odatda biz yana bir qadam qilishimiz kerak edi. Birinchi qator va ikkinchi ustundagi raqamni nolga aylantirishimiz kerak. Biroq, bu holatda u erda allaqachon nol bor va shuning uchun bu bosqichda hech qanday ish yo'q.
Keyin kengaytirilgan matritsaning yakuniy shakli:
\[\left[ {\begin{array}{rrr|r}1&0&0&{ - 1}\\0&1&0&3\\0&0&1&{ - 4}\end{array}} \right]\]Ushbu qadam ko'rsatganidek, biz vaqti-vaqti bilan "bepul" raqam olamiz. Boshqacha qilib aytadigan bo'lsak, biz oraliq bosqichga qo'ygan ishimiz nafaqat bu bosqichda biz qidirgan raqamni beradi, balki keyingi bosqichda bizga kerak bo'lgan raqamni ham qo'yadi. Yoki, bu holatda bo'lgani kabi, bizga kerak bo'lgan raqam boshidanoq mavjud edi.
9-qadam
Biz 8-bosqichda topilgan yakuniy kengaytirilgan matritsadan tizimning yechimini olamiz: \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = - 1,\,\,y = 3,\,\,z = - 4}}\).