7.3-bo'lim: Kengaytirilgan matritsalar (augmented matrices)
Ushbu bo'limda biz chiziqli tenglamalar sistemasini yechishning uchinchi asosiy usuli — matritsalar usuli bilan tanishamiz. Ikki noma'lumli ikkita tenglama sistemasi uchun bu usul birinchi bo'limda ko'rib chiqqan algebraik usullarimizdan birmuncha murakkabroq ko'rinishi mumkin. Biroq uchta va undan ko'p noma'lumli kattaroq sistemalar uchun bu usul oldingi bo'limda ko'rgan an'anaviy almashtirish usuliga qaraganda ancha tizimli, qulay va kamroq xatoliklarga olib keladi.
Usulning o'ziga o'tishdan oldin, dastlab zarur ta'riflarni keltirib o'tamiz.
Tenglamalar sistemasi uchun kengaytirilgan matritsa (augmented matrix) — bu sonlardan tashkil topgan to'g'ri to'rtburchakli jadval bo'lib, unda har bir satr bitta tenglamaning barcha koeffitsiyentlari va ozod hadini (tenglikning o'ng tomonidagi sonni) ifodalaydi, har bir ustun esa tegishli bitta o'zgaruvchining koeffitsiyentlariga mos keladi.
Keling, buni aniq misolda ko'rib chiqaylik. Oldingida o'rgangan quyidagi uch noma'lumli tenglamalar sistemasini olaylik:
\[\begin{align*}x - 2y + 3z & = 7\\ 2x + y + z & = 4\\ - 3x + 2y - 2z & = - 10\end{align*}\]Ushbu sistema uchun kengaytirilgan matritsa quyidagi ko'rinishda yoziladi:
\[\left[ {\begin{array}{rrr|r}1 & -2 & 3 & 7 \\ 2 & 1 & 1 & 4 \\ -3 & 2 & -2 & -10\end{array}} \right]\]Birinchi satr birinchi tenglamaning barcha sonlaridan iborat: birinchi ustunda \(x\) ning koeffitsiyenti (1), ikkinchi ustunda \(y\) ning koeffitsiyenti (-2), uchinchi ustunda \(z\) ning koeffitsiyenti (3) va oxirgi ustunda ozod had (7) joylashgan. Xuddi shuningdek, ikkinchi va uchinchi satrlar mos ravishda ikkinchi va uchinchi tenglamalarga to'g'ri keladi. Vertikal chiziq dastlabki tenglamalardagi tenglik belgisini ajratib ko'rsatish uchun xizmat qiladi.
Keyingi navbatda, matritsalar ustidagi satr elementar amallari (elementary row operations)ni o'rganamiz. Elementar amallar jami uch xil bo'lib, quyida ularning har birining matematik belgilanishi va yuqoridagi matritsaga qo'llanilishi keltirilgan:
- Ikkita satr o'rnini almashtirish. Bu amalda \(i\)-satr va \(j\)-satrdagi barcha elementlar o'zaro almashtiriladi. Buning uchun qo'llaniladigan belgilash: \({R_i} \leftrightarrow {R_j}\). Masalan:
\[\left[ {\begin{array}{rrr|r}1 & -2 & 3 & 7 \\ 2 & 1 & 1 & 4 \\ -3 & 2 & -2 & -10\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} \leftrightarrow {R_3}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}-3 & 2 & -2 & -10 \\ 2 & 1 & 1 & 4 \\ 1 & -2 & 3 & 7\end{array}} \right]\]
Ko'rib turganingizdek, uchinchi satrning har bir elementi birinchi satrga o'tadi va birinchi satrning har bir elementi uchinchi satrga tushadi. Satrdagi barcha elementlarni to'liq ko'chirganingizga ishonch hosil qiling.
- Satrni noldan farqli o'zgarmas songa ko'paytirish (yoki bo'lish). Ushbu amalda \(i\)-satrning har bir elementi \(c\) o'zgarmas soniga ko'paytiriladi. Belgilanishi: \(c{R_i}\). Agar satrni songa bo'lmoqchi bo'lsak, \(\frac{1}{c}{R_i}\) kabi yozamiz. Masalan:
\[\left[ {\begin{array}{rrr|r}1 & -2 & 3 & 7 \\ 2 & 1 & 1 & 4 \\ -3 & 2 & -2 & -10\end{array}} \right]\begin{array}{*{20}{c}}{ - 4{R_3}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1 & -2 & 3 & 7 \\ 2 & 1 & 1 & 4 \\ 12 & -8 & 8 & 40\end{array}} \right]\]
Bunda uchinchi satrning har bir soni \(-4\) ga ko'paytirildi: \(-3(-4)=12\), \(2(-4)=-8\), \(-2(-4)=8\), \(-10(-4)=40\). Ishoralarga juda ehtiyot bo'ling.
- Satrga boshqa satrning songa ko'paytirilganini qo'shish. Ushbu amalda \(i\)-satr o'rniga shu \(i\)-satr bilan \(j\)-satrning \(c\) soniga ko'paytirilgan yig'indisi yoziladi: \({R_i} + c{R_j} \to {R_i}\). Masalan, 3-satrga 1-satrning \(-4\) ga ko'paytirilganini qo'shamiz (ya'ni \(R_3 - 4R_1 \to R_3\)):
\[\left[ {\begin{array}{rrr|r}1 & -2 & 3 & 7 \\ 2 & 1 & 1 & 4 \\ -3 & 2 & -2 & -10\end{array}} \right]\begin{array}{*{20}{c}}{{R_3} - 4{R_1} \to {R_3}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1 & -2 & 3 & 7 \\ 2 & 1 & 1 & 4 \\ -7 & 10 & -14 & -38\end{array}} \right]\]
Har bir element bo'yicha hisoblash quyidagicha bajarildi:
\[\begin{align*} - 3 - 4\left( 1 \right) & = - 7\\ 2 - 4\left( { - 2} \right) & = 10\\ - 2 - 4\left( 3 \right) & = - 14\\ - 10 - 4\left( 7 \right) & = - 38\end{align*}\]Bu amal aynan eng ko'p ishlatiladigan asosiy amaldir, shuning uchun ishoralar bilan ishlashda nihoyatda diqqatli bo'lish talab etiladi.
Xo'sh, kengaytirilgan matritsalar va satr amallari yordamida tenglamalar sistemasi qanday yechiladi? Dastlab ikkita noma'lumli ikkita tenglama sistemasini ko'raylik:
\[\begin{align*}ax + by & = p\\ cx + dy & = q\end{align*}\]Dastlab ushbu sistema uchun kengaytirilgan matritsani tuzamiz:
\[\left[ {\begin{array}{rr|r}a & b & p \\ c & d & q\end{array}} \right]\]So'ngra satr elementar amallaridan foydalanib, uni quyidagi soddalashtirilgan ko'rinishga (birlik matritsa holatiga) keltiramiz:
\[\left[ {\begin{array}{rr|r}1 & 0 & h \\ 0 & 1 & k\end{array}} \right]\]Matritsa ushbu ko'rinishga kelgach, masala yechilgan hisoblanadi: sistemaning yagona yechimi to'g'ridan-to'g'ri \(x = h\) va \(y = k\) bo'ladi!
Ushbu kuchli va universal usul Gauss-Jordan usuli (Gauss-Jordan Elimination) deb ataladi.
- \(\begin{align*}3x - 2y & = 14\\ x + 3y & = 1\end{align*}\)
- \(\begin{align*} - 2x + y & = - 3\\ x - 4y & = - 2\end{align*}\)
- \(\begin{align*}3x - 6y & = - 9\\ - 2x - 2y & = 12\end{align*}\)
a \(\begin{align*}3x - 2y & = 14\\ x + 3y & = 1\end{align*}\) Yechim
Dastlab berilgan sistema uchun kengaytirilgan matritsani yozamiz:
\[\left[ {\begin{array}{rr|r} 3 & -2 & 14 \\ 1 & 3 & 1\end{array}} \right]\]Bizning maqsadimiz — chap tomonda 1 va 0 lardan iborat birlik matritsani hosil qilishdir. Buning uchun yuqori chap burchakdagi elementni 1 ga aylantiramiz. Ikkinchi satrning birinchi elementi 1 bo'lgani sababli, 1- va 2-satrlar o'rnini almashtirish eng qulay yo'ldir:
\[\left[ {\begin{array}{rr|r}3 & -2 & 14 \\ 1 & 3 & 1\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} \leftrightarrow {R_2}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1 & 3 & 1 \\ 3 & -2 & 14\end{array}} \right]\]Keyingi qadam — hosil bo'lgan 1 ning ostidagi sonni (ya'ni 3 ni) 0 ga aylantirish. Buning uchun 2-satrdan 1-satrning 3 baravarini ayiramiz (\(R_2 - 3R_1 \to R_2\)):
\[\left[ {\begin{array}{rr|r}1 & 3 & 1 \\ 3 & -2 & 14\end{array}} \right]\begin{array}{*{20}{c}}{{R_2} - 3{R_1} \to {R_2}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1 & 3 & 1 \\ 0 & -11 & 11\end{array}} \right]\]Endi ikkinchi satrning ikkinchi elementini 1 ga aylantiramiz. Buning uchun butun 2-satrni \(-11\) ga bo'lamiz (\(-\frac{1}{11}R_2 \to R_2\)):
\[\left[ {\begin{array}{rr|r}1 & 3 & 1 \\ 0 & -11 & 11\end{array}} \right]\begin{array}{*{20}{c}}{ - \frac{1}{{11}}{R_2}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1 & 3 & 1 \\ 0 & 1 & -1\end{array}} \right]\]Oxirgi qadam — 1-satrdagi 3 ni 0 ga aylantirish. Buning uchun 1-satrdan 2-satrning 3 baravarini ayiramiz (\(R_1 - 3R_2 \to R_1\)):
\[\left[ {\begin{array}{rr|r}1 & 3 & 1 \\ 0 & 1 & -1\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} - 3{R_2} \to {R_1}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1 & 0 & 4 \\ 0 & 1 & -1\end{array}} \right]\]Matritsa talab qilingan shaklga keldi. Demak, sistemaning yagona yechimi: \(x = 4\) va \(y = -1\).
b \(\begin{align*} - 2x + y & = - 3\\ x - 4y & = - 2\end{align*}\) Yechim
Kengaytirilgan matritsani tuzamiz va ketma-ket satr amallarini bajaramiz:
\[\left[ {\begin{array}{rr|r}-2 & 1 & -3 \\ 1 & -4 & -2\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} \leftrightarrow {R_2}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1 & -4 & -2 \\ -2 & 1 & -3\end{array}} \right]\begin{array}{*{20}{c}}{{R_2} + 2{R_1} \to {R_2}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1 & -4 & -2 \\ 0 & -7 & -7\end{array}} \right]\]Ikkinchi satrni \(-7\) ga bo'lamiz va so'ngra birinchi satrdagi \(-4\) ni yo'qotamiz:
\[\left[ {\begin{array}{rr|r}1 & -4 & -2 \\ 0 & -7 & -7\end{array}} \right]\begin{array}{*{20}{c}}{ - \frac{1}{7}{R_2}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1 & -4 & -2 \\ 0 & 1 & 1\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} + 4{R_2} \to {R_1}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1 & 0 & 2 \\ 0 & 1 & 1\end{array}} \right]\]Demak, yechim: \(x = 2\) va \(y = 1\).
c \(\begin{align*}3x - 6y & = - 9\\ - 2x - 2y & = 12\end{align*}\) Yechim
Kengaytirilgan matritsani yozamiz. Birinchi satr barcha elementlari 3 ga bo'lingani uchun, 1-satrni to'g'ridan-to'g'ri 3 ga bo'lib olamiz:
\[\left[ {\begin{array}{rr|r}3 & -6 & -9 \\ -2 & -2 & 12\end{array}} \right]\begin{array}{*{20}{c}}{\frac{1}{3}{R_1}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1 & -2 & -3 \\ -2 & -2 & 12\end{array}} \right]\begin{array}{*{20}{c}}{{R_2} + 2{R_1} \to {R_2}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1 & -2 & -3 \\ 0 & -6 & 6\end{array}} \right]\] \[\left[ {\begin{array}{rr|r}1 & -2 & -3 \\ 0 & -6 & 6\end{array}} \right]\begin{array}{*{20}{c}}{ - \frac{1}{6}{R_2}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1 & -2 & -3 \\ 0 & 1 & -1\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} + 2{R_2} \to {R_1}}\\ \to \end{array}\left[ {\begin{array}{rr|r}1 & 0 & -5 \\ 0 & 1 & -1\end{array}} \right]\]Demak, yechim: \(x = -5\) va \(y = -1\).
Shuni ta'kidlash kerakki, elementar satr amallarini qo'llashda tanlangan yo'l yagona emas. Har kim hisoblash uchun o'ziga eng qulay va oson ko'ringan amallar ketma-ketligini tanlashi mumkin. Qaysi yo'ldan bormang, agar hisoblashlar to'g'ri bajarilsa, yakuniy javob doimo bir xil chiqadi.
Endi uchta noma'lumli uchta tenglamadan iborat kattaroq sistemalarga o'tamiz. Bu holda ham umumiy mantiq to'liq saqlanib qoladi — maqsadimiz matritsani quyidagi ko'rinishga keltirishdir:
\[\left[ {\begin{array}{rrr|r}1 & 0 & 0 & p \\ 0 & 1 & 0 & q \\ 0 & 0 & 1 & r\end{array}} \right]\]Bu holatda sistemaning yagona yechimi darhol ma'lum bo'ladi: \(x = p\), \(y = q\) va \(z = r\).
- \(\begin{align*}3x + y - 2z & = 2\\ x - 2y + z & = 3\\ 2x - y - 3z & = 3\end{align*}\)
- \(\begin{align*}3x + y - 2z & = - 7\\ 2x + 2y + z & = 9\\ - x - y + 3z & = 6\end{align*}\)
a \(\begin{align*}3x + y - 2z & = 2\\ x - 2y + z & = 3\\ 2x - y - 3z & = 3\end{align*}\) Yechim
Kengaytirilgan matritsani tuzamiz va 1- hamda 2-satrlarni almashtirib, yuqori chap burchakka 1 ni chiqaramiz:
\[\left[ {\begin{array}{rrr|r}3 & 1 & -2 & 2 \\ 1 & -2 & 1 & 3 \\ 2 & -1 & -3 & 3\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} \leftrightarrow {R_2}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1 & -2 & 1 & 3 \\ 3 & 1 & -2 & 2 \\ 2 & -1 & -3 & 3\end{array}} \right]\]Birinchi ustundagi 1 ning ostidagi sonlarni (3 va 2 ni) 0 ga aylantirish uchun uchinchi elementar amalni qo'llaymiz:
\[\left[ {\begin{array}{rrr|r}1 & -2 & 1 & 3 \\ 3 & 1 & -2 & 2 \\ 2 & -1 & -3 & 3\end{array}} \right]\begin{array}{*{20}{c}}{{R_2} - 3{R_1} \to {R_2}}\\{{R_3} - 2{R_1} \to {R_3}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1 & -2 & 1 & 3 \\ 0 & 7 & -5 & -7 \\ 0 & 3 & -5 & -3\end{array}} \right]\]Endi ikkinchi satrni 7 ga bo'lamiz:
\[\left[ {\begin{array}{rrr|r}1 & -2 & 1 & 3 \\ 0 & 7 & -5 & -7 \\ 0 & 3 & -5 & -3\end{array}} \right]\begin{array}{*{20}{c}}{\frac{1}{7}{R_2}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1 & -2 & 1 & 3 \\ 0 & 1 & -\frac{5}{7} & -1 \\ 0 & 3 & -5 & -3\end{array}} \right]\]Uchinchi satrdagi 3 ni 0 ga aylantiramiz (\(R_3 - 3R_2 \to R_3\)):
\[\left[ {\begin{array}{rrr|r}1 & -2 & 1 & 3 \\ 0 & 1 & -\frac{5}{7} & -1 \\ 0 & 3 & -5 & -3\end{array}} \right]\begin{array}{*{20}{c}}{{R_3} - 3{R_2} \to {R_3}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1 & -2 & 1 & 3 \\ 0 & 1 & -\frac{5}{7} & -1 \\ 0 & 0 & -\frac{20}{7} & 0\end{array}} \right]\]Uchinchi satrni \(-\frac{7}{20}\) ga ko'paytirib, pastki o'ng burchakdagi 1 ni hosil qilamiz:
\[\left[ {\begin{array}{rrr|r}1 & -2 & 1 & 3 \\ 0 & 1 & -\frac{5}{7} & -1 \\ 0 & 0 & -\frac{20}{7} & 0\end{array}} \right]\begin{array}{*{20}{c}}{ - \frac{7}{{20}}{R_3}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1 & -2 & 1 & 3 \\ 0 & 1 & -\frac{5}{7} & -1 \\ 0 & 0 & 1 & 0\end{array}} \right]\]Uchinchi ustundagi 1 ning ustidagi elementlarni 0 ga aylantiramiz:
\[\left[ {\begin{array}{rrr|r}1 & -2 & 1 & 3 \\ 0 & 1 & -\frac{5}{7} & -1 \\ 0 & 0 & 1 & 0\end{array}} \right]\begin{array}{*{20}{c}}{{R_2} + \frac{5}{7}{R_3} \to {R_2}}\\{{R_1} - {R_3} \to {R_1}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1 & -2 & 0 & 3 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 0\end{array}} \right]\]Nihoyat, 1-satrdagi \(-2\) ni 0 ga aylantiramiz (\(R_1 + 2R_2 \to R_1\)):
\[\left[ {\begin{array}{rrr|r}1 & -2 & 0 & 3 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 0\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} + 2{R_2} \to {R_1}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 0\end{array}} \right]\]Kengaytirilgan matritsa to'liq soddalashtirildi. Demak, yechim:
\[x = 1,\,\,\,y = -1,\,\,\,z = 0\]b \(\begin{align*}3x + y - 2z & = - 7\\ 2x + 2y + z & = 9\\ - x - y + 3z & = 6\end{align*}\) Yechim
Kengaytirilgan matritsani yozamiz. Kasrlardan qochish uchun 1-satrdan 2-satrni ayirib (\(R_1 - R_2 \to R_1\)), yuqorida 1 ni hosil qilamiz:
\[\left[ {\begin{array}{rrr|r}3 & 1 & -2 & -7 \\ 2 & 2 & 1 & 9 \\ -1 & -1 & 3 & 6\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} - {R_2} \to {R_1}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1 & -1 & -3 & -16 \\ 2 & 2 & 1 & 9 \\ -1 & -1 & 3 & 6\end{array}} \right]\]1-ustundagi qolgan sonlarni 0 ga aylantiramiz:
\[\left[ {\begin{array}{rrr|r}1 & -1 & -3 & -16 \\ 2 & 2 & 1 & 9 \\ -1 & -1 & 3 & 6\end{array}} \right]\begin{array}{*{20}{c}}{{R_2} - 2{R_1} \to {R_2}}\\{{R_3} + {R_1} \to {R_3}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1 & -1 & -3 & -16 \\ 0 & 4 & 7 & 41 \\ 0 & -2 & 0 & -10\end{array}} \right]\]Kasrlardan qochish uchun 2- va 3-satrlarni o'zaro almashtiramiz:
\[\left[ {\begin{array}{rrr|r}1 & -1 & -3 & -16 \\ 0 & 4 & 7 & 41 \\ 0 & -2 & 0 & -10\end{array}} \right]\begin{array}{*{20}{c}}{{R_2} \leftrightarrow {R_3}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1 & -1 & -3 & -16 \\ 0 & -2 & 0 & -10 \\ 0 & 4 & 7 & 41\end{array}} \right]\]2-satrni \(-2\) ga bo'lamiz:
\[\left[ {\begin{array}{rrr|r}1 & -1 & -3 & -16 \\ 0 & -2 & 0 & -10 \\ 0 & 4 & 7 & 41\end{array}} \right]\begin{array}{*{20}{c}}{ - \frac{1}{2}{R_2}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1 & -1 & -3 & -16 \\ 0 & 1 & 0 & 5 \\ 0 & 4 & 7 & 41\end{array}} \right]\]E'tibor bering: 2-satrda 3-ustun elementi nol bo'lib chiqdi, bu hisoblashlarimizni sezilarli darajada yengillashtiradi. 3-satrdagi 4 ni yo'qotamiz (\(R_3 - 4R_2 \to R_3\)):
\[\left[ {\begin{array}{rrr|r}1 & -1 & -3 & -16 \\ 0 & 1 & 0 & 5 \\ 0 & 4 & 7 & 41\end{array}} \right]\begin{array}{*{20}{c}}{{R_3} - 4{R_2} \to {R_3}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1 & -1 & -3 & -16 \\ 0 & 1 & 0 & 5 \\ 0 & 0 & 7 & 21\end{array}} \right]\]3-satrni 7 ga bo'lamiz:
\[\left[ {\begin{array}{rrr|r}1 & -1 & -3 & -16 \\ 0 & 1 & 0 & 5 \\ 0 & 0 & 7 & 21\end{array}} \right]\begin{array}{*{20}{c}}{\frac{1}{7}{R_3}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1 & -1 & -3 & -16 \\ 0 & 1 & 0 & 5 \\ 0 & 0 & 1 & 3\end{array}} \right]\]Endi 1-satrdagi \(-3\) ni yo'qotish uchun \(R_1 + 3R_3 \to R_1\) amalini bajaramiz:
\[\left[ {\begin{array}{rrr|r}1 & -1 & -3 & -16 \\ 0 & 1 & 0 & 5 \\ 0 & 0 & 1 & 3\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} + 3{R_3} \to {R_1}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1 & -1 & 0 & -7 \\ 0 & 1 & 0 & 5 \\ 0 & 0 & 1 & 3\end{array}} \right]\]Nihoyat, 1-satrdagi \(-1\) ni yo'qotish uchun \(R_1 + R_2 \to R_1\) amalini bajaramiz:
\[\left[ {\begin{array}{rrr|r}1 & -1 & 0 & -7 \\ 0 & 1 & 0 & 5 \\ 0 & 0 & 1 & 3\end{array}} \right]\begin{array}{*{20}{c}}{{R_1} + {R_2} \to {R_1}}\\ \to \end{array}\left[ {\begin{array}{rrr|r}1 & 0 & 0 & -2 \\ 0 & 1 & 0 & 5 \\ 0 & 0 & 1 & 3\end{array}} \right]\]Kengaytirilgan matritsa to'liq yechildi. Sistemaning yagona yechimi:
\[x = -2,\,\,\,y = 5,\,\,\,z = 3\]Gauss-Jordan usuli yordamida uchta va undan ortiq tenglamali sistemalarni yechish qat'iy algoritmga ega bo'lib, hisoblash jarayonida chalkashliklarni bartaraf etishga xizmat qiladi. Qaysi qulay satr amallarini tanlamang, agar arifmetik amallar to'g'ri bajarilsa, natija doimo aniq va bir xil bo'ladi.