Yuqori chap burchakdagi raqamni bitta qilishimiz kerak. Oldingi muammolar kabi (i.e. ikkita o'zgaruvchiga ega tizimlarni echish) biz buni \(\frac{1}{2}{R_{\,1}}\) elementar qator operatsiyasi bilan tezda amalga oshirishimiz mumkin, lekin bu kasrlarni kengaytirilgan matritsaga qo'yadi va ular, ehtimol, bir necha qadamlar atrofida bo'lishi mumkin va kengaytirilgan matritsa bu o'lchamga ega bo'lganda, imkon qadar uzoq vaqt davomida ulardan qochish juda yaxshi bo'lar edi.

Shunday qilib, keling, quyidagi elementar qator operatsiyasidan boshlaylik.

\[\left[ {\begin{array}{rrr|r}2&5&2&{ - 38}\\3&{ - 2}&4&{17}\\{ - 6}&1&{ - 7}&{ - 12}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,1}} - {R_{\,2}} \to {R_{\,1}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}{ - 1}&7&{ - 2}&{ - 55}\\3&{ - 2}&4&{17}\\{ - 6}&1&{ - 7}&{ - 12}\end{array}} \right]\]

Ushbu operatsiyani bajarish orqali biz plyus kerak bo'lgan joyda salbiyni oldik, ammo keyingi elementar qator operatsiyasi bilan buni osongina tuzatishimiz mumkin.

\[\left[ {\begin{array}{rrr|r}{ - 1}&7&{ - 2}&{ - 55}\\3&{ - 2}&4&{17}\\{ - 6}&1&{ - 7}&{ - 12}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{rrr|r}{ - {R_{\,1}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{r}}1&{ - 7}&2&{55}\\3&{ - 2}&4&{17}\\{ - 6}&1&{ - 7}&{ - 12}\end{array}} \right]\]

Keling, ushbu muammoning qolgan qismiga kirishdan oldin qisqacha eslatma. Uch o'zgaruvchiga ega tizimlarni hal qilish uchun kengaytirilgan matritsalardan foydalanish juda zerikarli jarayon bo'lishi mumkin va hal qilish jarayonida ko'plab mumkin bo'lgan yo'llar mavjud, shuning uchun sizning yechimingiz siz bosib o'tgan yo'lga qarab bu yechimdan farq qilishi mumkin. Yakuniy javoblar, agar biz barcha arifmetikani to'g'ri bajargan bo'lsak, qanday yo'l tutishimizdan qat'i nazar, bir xil bo'ladi.

Keyinchalik, birinchi ustundagi 1 ostidagi 3 va -6 raqamlarini nolga aylantirishimiz kerak va biz buni quyidagi elementar qator operatsiyalari bilan amalga oshirishimiz mumkin.

\[\left[ {\begin{array}{rrr|r}1&{ - 7}&2&{55}\\3&{ - 2}&4&{17}\\{ - 6}&1&{ - 7}&{ - 12}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{rrr|r}{{R_{\,2}} - 3{R_{\,1}} \to {R_{\,2}}}\\{{R_{\,3}} + 6{R_{\,1}} \to {R_{\,3}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{r}}1&{ - 7}&2&{55}\\0&{19}&{ - 2}&{ - 148}\\0&{ - 41}&5&{318}\end{array}} \right]\]

Endi biz ikkinchi qatordagi 19 ni bittaga aylantirishimiz kerak va buni qilishning yagona oson yo'li quyidagi elementar qator operatsiyasi kabi ko'rinadi.

\[\left[ {\begin{array}{rrr|r}1&{ - 7}&2&{55}\\0&{19}&{ - 2}&{ - 148}\\0&{ - 41}&5&{318}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{\frac{1}{{19}}{R_{\,2}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}1&{ - 7}&2&{55}\\0&1&{ - \frac{2}{{19}}}&{ - \frac{{148}}{{19}}}\\0&{ - 41}&5&{318}\end{array}} \right]\]

Birinchi bosqichda biz kasrlarni kengaytirilgan matritsaga qo'yadigan qadamdan qochishni tanladik, lekin ba'zida, bu bosqichda bo'lgani kabi, ulardan qochib bo'lmaydi. Uch o'zgaruvchiga ega tizimlar uchun kengaytirilgan matritsalar bilan fraktsiyalar deyarli muqarrar ravishda paydo bo'ladi va ular ko'pincha "tartibsiz" bo'ladi.

Bu faqat ushbu tizimlarni hal qilishda hal qilishimiz kerak bo'lgan narsa. Biz ularni iloji boricha uzoqroq tutishga harakat qilamiz, lekin ular paydo bo'lganda va hal qilish jarayonini davom ettirganda bundan mustasno.

Keyinchalik, uchinchi qatordagi -41 ni nolga aylantirishimiz kerak. Quyidagi elementar qator operatsiyasi biz uchun buni amalga oshiradi.

\[\left[ {\begin{array}{rrr|r}1&{ - 7}&2&{55}\\0&1&{ - \frac{2}{{19}}}&{ - \frac{{148}}{{19}}}\\0&{ - 41}&5&{318}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,3}} + 41{R_{\,2}} \to {R_{\,3}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}1&{ - 7}&2&{55}\\0&1&{ - \frac{2}{{19}}}&{ - \frac{{148}}{{19}}}\\0&0&{\frac{{13}}{{19}}}&{ - \frac{{26}}{{19}}}\end{array}} \right]\]

Yana, biz kengaytirilgan matritsaga ko'proq kasr qo'yishimiz kerak edi. Bunday muammolar bilan bu hayot haqiqati. Biroq, keyingi bosqichda ko'rib turganimizdek, ular ham ko'pincha yo'qoladi.

OK, uchinchi qatordagi \(\frac{{13}}{{19}}\) ni bittaga aylantirishimiz kerak va buni quyidagicha qilishimiz mumkin:

\[\left[ {\begin{array}{rrr|r}1&{ - 7}&2&{55}\\0&1&{ - \frac{2}{{19}}}&{ - \frac{{148}}{{19}}}\\0&0&{\frac{{13}}{{19}}}&{ - \frac{{26}}{{19}}}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{\frac{{19}}{{13}}{R_{\,3}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}1&{ - 7}&2&{55}\\0&1&{ - \frac{2}{{19}}}&{ - \frac{{148}}{{19}}}\\0&0&1&{ - 2}\end{array}} \right]\]

Keyinchalik, \( - \frac{2}{{19}}\) va uchinchi ustundagi 2 ni nolga aylantirishimiz kerak. Quyidagi elementar qator operatsiyalari biz uchun buni amalga oshiradi.

\[\left[ {\begin{array}{rrr|r}1&{ - 7}&2&{55}\\0&1&{ - \frac{2}{{19}}}&{ - \frac{{148}}{{19}}}\\0&0&1&{ - 2}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,1}} - 2{R_{\,3}} \to {R_{\,1}}}\\{{R_{\,2}} + \frac{2}{{19}}{R_{\,3}} \to {R_{\,2}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}1&{ - 7}&0&{59}\\0&1&0&{ - 8}\\0&0&1&{ - 2}\end{array}} \right]\]

E'tibor bering, kasrlar endi butunlay yo'qoldi! Bu har doim ham sodir bo'lmaydi, lekin tez-tez sodir bo'ladi, fraktsiyalar oraliq bosqichlarda kiritiladi va keyingi bosqichlarda yo'qoladi.

Yakuniy operatsiya uchun biz ikkinchi ustundagi -7 ni nolga aylantirishimiz kerak va buni quyidagicha qilishimiz mumkin:

\[\left[ {\begin{array}{rrr|r}1&{ - 7}&0&{59}\\0&1&0&{ - 8}\\0&0&1&{ - 2}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,1}} + 7{R_{\,2}} \to {R_{\,1}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}1&0&0&3\\0&1&0&{ - 8}\\0&0&1&{ - 2}\end{array}} \right]\]

Biz 8-bosqichda topilgan yakuniy kengaytirilgan matritsadan tizimning yechimini olamiz: \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = 3,\,\,y = - 8,\,\,z = - 2}}\).