7.4-bo'lim : Kengaytirilgan matritsa haqida ko'proq
6. 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.
\[\begin{align*}2x + 3y & = 20\\ 7x + 2y & = 53\end{align*}\]Yechim
Birinchi qadam tenglamalar sistemasi uchun kengaytirilgan matritsani yozishdir.
\[\left[ {\begin{array}{rr|r}2&3&{20}\\7&2&{53}\end{array}} \right]\]2-qadam
Yuqori chap burchakdagi raqamni bitta qilishimiz kerak. Buning uchun bir necha usullar mavjud. Buning bir usuli - \(\frac{1}{2}{R_{\,1}}\) elementar qator operatsiyasidan foydalanish. Biroq, bu birinchi qatordagi boshqa ikkita yozuvga kasrlarni qo'yadi va ulardan qochish yaxshi bo'lishi mumkin.
Bu foydasizdek tuyulishi mumkin bo'lsa-da, keling, quyidagi elementar qator operatsiyasini ko'rib chiqaylik.
\[\left[ {\begin{array}{rr|r}2&3&{20}\\7&2&{53}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,2}} - 3{R_{\,1}} \to {R_{\,2}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rr|r}2&3&{20}\\1&{ - 7}&{ - 7}\end{array}} \right]\]Ushbu operatsiya biz ishlashimiz kerak bo'lgan birinchi qator o'rniga ikkinchi qatorda ishladi. E'tibor bering, biz birinchi ustunning pastki raqamiga 1 qo'yganmiz. Bizga birinchi ustunning yuqori raqamida 1 kerak va biz buni oddiy qatorlarni quyidagi tarzda almashtirish orqali amalga oshirishimiz mumkin,
\[\left[ {\begin{array}{rr|r}2&3&{20}\\1&{ - 7}&{ - 7}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,1}} \leftrightarrow {R_{\,2}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rr|r}1&{ - 7}&{ - 7}\\2&3&{20}\end{array}} \right]\]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 2 ni nolga aylantirishimiz kerak va biz buni quyidagi elementar qator operatsiyasi bilan amalga oshirishimiz mumkin.
\[\left[ {\begin{array}{rr|r}1&{ - 7}&{ - 7}\\2&3&{20}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,2}} - 2{R_{\,1}} \to {R_{\,2}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rr|r}1&{ - 7}&{ - 7}\\0&{17}&{34}\end{array}} \right]\]4-qadam
Keyingi qadam, ikkinchi ustunning pastki qismidagi raqamni (bu holda 17) bittaga aylantirishdir. Quyidagi elementar qator operatsiyasi biz uchun buni amalga oshiradi.
\[\left[ {\begin{array}{rr|r}1&{ - 7}&{ - 7}\\0&{17}&{34}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{\frac{1}{{17}}{R_{\,2}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rr|r}1&{ - 7}&{ - 7}\\0&1&2\end{array}} \right]\]5-qadam
Nihoyat, biz 4-bosqichda olingan raqamdan yuqoridagi raqamni nolga aylantirishimiz kerak. Buning uchun quyidagi elementar qator amalidan foydalanishimiz mumkin.
\[\left[ {\begin{array}{rr|r}1&{ - 7}&{ - 7}\\0&1&2\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,2}} + 7{R_{\,1}} \to {R_{\,2}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rr|r}1&0&7\\0&1&2\end{array}} \right]\]6-qadam
Biz 5-bosqichda topilgan yakuniy kengaytirilgan matritsadan tizimning yechimini olamiz: \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = 7,\,\,y = 2}}\).