5.2-bo'lim: Ko'rib chiqish: matritsalar va vektorlar
Ushbu bo'lim differensial tenglamalar sistemalari bilan ishlashda vaqti-vaqti bilan qo'llaniladigan ko'plab asosiy tushunchalarni qamrab olish uchun mo'ljallangan. Ushbu bo'limda juda ko'p tafsilotlar bo'lmaydi va biz ko'p sonli misollar bilan ishlamaymiz. Bundan tashqari, ko'p hollarda biz umumiy holatni ko'rib chiqmaymiz, chunki biz differentsial tenglamalar ishida umumiy holatlarga muhtoj emasmiz.
Keling, matritsalar uchun asosiy yozuvlardan boshlaylik. \(n \times m\) (bu ko'pincha hajmi or o'lcham matritsaning) matritsasi \(n\) satrlari va \(m\) ustunlari bo'lgan matritsa bo'lib, \(i^{\text{th}}\) qatori va \(j^{\text{th}}\) ustunidagi yozuv \(a_{ij}\) bilan belgilanadi. Umumiy \(n \times m\) matritsasini yozishning qisqacha usuli quyidagicha.
\[A = {\left( {\begin{array}{*{20}{r}}{{a_{11}}}&{{a_{12}}}& \cdots &{{a_{1m}}}\\{{a_{21}}}&{{a_{22}}}& \cdots &{{a_{2m}}}\\ \vdots & \vdots &{}& \vdots \\{{a_{n1}}}&{{a_{n2}}}& \cdots &{{a_{nm}}}\end{array}} \right)_{n \times m}} = {\left( {{a_{ij}}} \right)_{n \times m}}\]Agar kerak bo'lsa, matritsaning o'lchami yoki o'lchami ko'rsatilgandek yoziladi. Agar bu talab qilinmasa yoki muammo aniq bo'lmasa, obuna o'lchami ko'pincha matritsadan chiqariladi.
Maxsus matritsalar
Vaqti-vaqti bilan foydalanishimiz mumkin bo'lgan bir nechta "maxsus" matritsalar mavjud. Birinchi maxsus matritsa bu kvadrat matritsa. Kvadrat matritsa - bu o'lchami (yoki o'lchami) \(n \times n\) bo'lgan har qanday matritsa. Boshqacha qilib aytganda, u ustunlar bilan bir xil qatorlarga ega. Kvadrat matritsada yuqori chapdan boshlanib, pastki o'ngda tugaydigan diagonal ko'pincha deyiladi. asosiy diagonali.
Biz ko'rib chiqmoqchi bo'lgan keyingi ikkita maxsus matritsalar nol matritsa va identifikatsiya matritsasidir. nol matritsa, \(0_{n \times m}\) bilan belgilangan, barcha yozuvlari nolga teng bo'lgan matritsadir. identifikatsiya matritsasi bu kvadrat \(n \times n\) matritsasi bo'lib, \(I_{n}\) bilan belgilanadi, uning asosiy diagonallari hammasi 1 ga, qolgan barcha elementlari esa nolga teng. Bu erda umumiy nol va identifikatsiya matritsalari.
\[{0_{n \times m}} = {\left( {\begin{array}{*{20}{r}}0&0& \cdots &0\\ \vdots & \vdots &{}& \vdots \\0&0& \cdots &0\end{array}} \right)_{n \times m}}\hspace{0.25in}\hspace{0.25in}{I_n} = {\left( {\begin{array}{*{20}{r}}1&0& \cdots &0\\0&1& \cdots &0\\ \vdots & \vdots & \ddots & \vdots \\0&0& \cdots &1\end{array}} \right)_{n \times n}}\]Matritsa arifmetikasida bu ikki matritsa matritsa ishida haqiqiy sanoq sistemasida nol va bitta akt kabi ishlaydi.
Biz bu erda ko'rib chiqadigan oxirgi ikkita maxsus matritsalar ustun matritsasi va qator matritsasi. Bular bitta ustun yoki bitta satrdan iborat matritsalardir. Umuman olganda, ular,
\[x = {\left( {\begin{array}{*{20}{r}}{{x_1}}\\{{x_2}}\\ \vdots \\{{x_n}}\end{array}} \right)_{n \times 1}}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}y = {\left( {\begin{array}{*{20}{r}}{{y_1}}&{{y_2}}& \cdots &{{y_m}}\end{array}} \right)_{1 \times m}}\]Biz ularni ko'pincha deb nomlaymiz vektorlar.
Arifmetika
Keyinchalik matritsalar bilan bog'liq arifmetikani ko'rib chiqishimiz kerak. bilan boshlaymiz qo'shimcha va ayirish ikkita matritsadan iborat. Deylik, bizda ikkita \(n \times m\) matritsasi bor, \(A\) va \(B\). Bu ikki matritsaning yig'indisi (yoki farqi) u holda,
\[{A_{n \times m}} \pm {B_{n \times m}} = {\left( {{a_{ij}}} \right)_{n \times m}} \pm {\left( {{b_{ij}}} \right)_{n \times m}} = {\left( {{a_{ij}} \pm {b_{ij}}} \right)_{n \times m}}\]Bir xil o'lchamdagi ikkita matritsaning yig'indisi yoki farqi bir xil o'lchamdagi yangi matritsa bo'lib, yozuvlari dastlabki ikkita matritsadan mos keladigan yozuvlarning yig'indisi yoki farqi hisoblanadi. E'tibor bering, biz turli o'lchamdagi yozuvlarni qo'sha olmaymiz yoki ayira olmaymiz.
Keyinchalik, ko'rib chiqaylik skalyar ko'paytirish. Skayar ko'paytirishda biz \(A\) matritsasini doimiy (ba'zan skaler deb ataladi) \(\alpha \) ga ko'paytiramiz. Bu holda biz barcha yozuvlari \(\alpha \) doimiysiga ko'paytirilgan yangi matritsani olamiz.
\[\alpha {A_{n \times m}} = \alpha {\left( {{a_{ij}}} \right)_{n \times m}} = {\left( {\alpha \,{a_{ij}}} \right)_{n \times m}}\]\(A-5B\) ni hisoblang.
Yechim
Bu erda ishdan boshqa qiladigan ko'p narsa yo'q.
\[\begin{align*}A - 5B & = \left( {\begin{array}{*{20}{r}}3&{ - 2}\\{ - 9}&1\end{array}} \right) - 5\left( {\begin{array}{*{20}{r}}{ - 4}&1\\0&{ - 5}\end{array}} \right)\\ & = \left( {\begin{array}{*{20}{r}}3&{ - 2}\\{ - 9}&1\end{array}} \right) - \left( {\begin{array}{*{20}{r}}{ - 20}&5\\0&{ - 25}\end{array}} \right)\\ & = \left( {\begin{array}{*{20}{r}}{23}&{ - 7}\\{ - 9}&{26}\end{array}} \right)\end{align*}\]Biz avval \(B\) ning barcha yozuvlarini 5 ga ko'paytirdik, so'ngra yangi matritsadagi yozuvlarni olish uchun tegishli yozuvlarni ayirdik.
Biz ko'rib chiqadigan yakuniy matritsa operatsiyasi matritsalarni ko'paytirish. Bu erda biz ikkita matritsadan boshlaymiz, \(A_{n \times p}\) va \(B_{p \times m}\). E'tibor bering, \(A\) da \(B\) qatorlar soni bir xil bo'lishi kerak. Agar bu to'g'ri bo'lmasa, biz ko'paytirishni amalga oshira olmaymiz. Agar bu to'g'ri bo'lsa, biz quyidagi ko'paytirishni bajarishimiz mumkin.
\[{A_{n \times p}}{B_{p \times m}} = {\left( {{c_{ij}}} \right)_{n \times m}}\]Yangi matritsa \(n \times m\) o'lchamiga ega bo'ladi va \(i^{\text{th}}\) qatori va \(j^{\text{th}}\) ustunidagi yozuv \(c_{ij}\) \(A\) matritsasining \(i\) qatorini \(j\) matritsasining \(j\) ustuniga ko'paytirish orqali topiladi. Bu har doim ham so'zlarda mantiqiy emas, shuning uchun misolni ko'rib chiqaylik.
\(AB\) ni hisoblang.
Yechim
Yangi matritsa \(2 \times 4\) o'lchamiga ega bo'ladi. Yangi matritsaning 1-satri va 1-ustunidagi yozuv \(A\) ning 1-qatorini \(B\) ning 1-ustuniga koʻpaytirish orqali topiladi. Bu shuni anglatadiki, biz \(A\) qatori va \(B\) ustunidan mos yozuvlarni ko'paytiramiz va keyin natijalarni qo'shamiz. Bu erda barcha yo'l bilan hisoblangan bir nechta yozuvlar mavjud.
\[\begin{align*}{c_{11}} & = \left( 2 \right)\left( 1 \right) + \left( { - 1} \right)\left( { - 4} \right) + \left( 0 \right)\left( 0 \right) = 6\\ {c_{13}} & = \left( 2 \right)\left( { - 1} \right) + \left( { - 1} \right)\left( 1 \right) + \left( 0 \right)\left( 0 \right) = - 3\\ {c_{24}} & = \left( { - 3} \right)\left( 2 \right) + \left( 6 \right)\left( 0 \right) + \left( 1 \right)\left( { - 2} \right) = - 8\end{align*}\]Mana to'liq yechim.
\[C = \left( {\begin{array}{*{20}{r}}6&{ - 3}&{ - 3}&4\\{ - 27}&{21}&9&{ - 8}\end{array}} \right)\]Ushbu oxirgi misolda biz mahsulotni qila olmasligimizga e'tibor bering BA chunki \(B\) ustunlari soni \(A\) qatorlari soniga mos kelmaydi. Shuni ta'kidlash kerakki, biz \(AB\) ni hisoblashimiz \(BA\) ni hisoblashimiz mumkin degani emas. Xuddi shunday, biz \(AB\) va \(BA\) ni ham hisoblay olsak ham, ular bir xil matritsa bo'lishi mumkin yoki bo'lmasligi mumkin.
Aniqlovchi
Biz ko'rib chiqishimiz kerak bo'lgan keyingi mavzu aniqlovchi matritsadan. Determinant aslida kvadrat matritsani oladigan va uni raqamga aylantiruvchi funktsiyadir. Funktsiyaning haqiqiy formulasi biroz murakkab va bu ko'rib chiqish doirasidan tashqarida.
Har qanday kvadrat matritsaning determinantlarini hisoblashning asosiy usuli kofaktorlar usuli deb ataladi. Biz deyarli faqat \(2 \times 2\) matritsalari va vaqti-vaqti bilan \(3 \times 3\) matritsalari bilan shug'ullanmoqchi bo'lganimiz uchun bu erda usulga kirmaymiz. Ushbu holatlarning har biri uchun oddiy formulalar berishimiz mumkin. \(A\) matritsasi determinantining standart yozuvi.
\[\det \left( A \right) = \left| A \right|\]Mana \(2 \times 2\) va \(3 \times 3\) matritsalarining determinanti uchun formulalar.
\[\left| {\begin{array}{*{20}{r}}a&c\\b&d\end{array}} \right| = ad - cb\] \[\left| {\begin{array}{*{20}{r}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{array}} \right| = {a_{11}}\left| {\begin{array}{*{20}{r}}{{a_{22}}}&{{a_{23}}}\\{{a_{32}}}&{{a_{33}}}\end{array}} \right| - {a_{12}}\left| {\begin{array}{*{20}{r}}{{a_{21}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{33}}}\end{array}} \right| + {a_{13}}\left| {\begin{array}{*{20}{r}}{{a_{21}}}&{{a_{22}}}\\{{a_{31}}}&{{a_{32}}}\end{array}} \right|\]Yechim
\(2 \times 2\) uchun uni formulaga ulashdan boshqa qiladigan ish yo'q.
\[\det \left( A \right) = \left| {\begin{array}{*{20}{r}}{ - 9}&{ - 18}\\2&4\end{array}} \right| = \left( { - 9} \right)\left( 4 \right) - \left( { - 18} \right)\left( 2 \right) = 0\]\(3 \times 3\) uchun biz uni formulaga ulashimiz mumkin, ammo \(2 \times 2\) holatidan farqli o'laroq, bu formulani eslab qolish oson emas. Xuddi shu natijaga erishishning osonroq yo'li mavjud. Xuddi shu natijaga erishishning tezroq yo'li quyidagi amallarni bajarishdir. Avval matritsani yozing va birinchi ikkita ustunning nusxasini quyidagi tarzda oxiriga yopishtiring.
\[\det \left( B \right) = \left| {\begin{array}{*{20}{r}}2&3&1\\{ - 1}&{ - 6}&7\\4&5&{ - 1}\end{array}} \right|\,\,\,\,\begin{array}{*{20}{r}}2&3\\{ - 1}&{ - 6}\\4&5\end{array}\]Endi e'tibor bering, chapdan o'ngga uchta diagonal va o'ngdan chapga uchta diagonal bor. Biz nima qilamiz, har bir diagonaldagi yozuvlarni yuqoriga ko'paytiramiz va agar diagonal chapdan o'ngga bo'lsa, biz ularni qo'shamiz va agar diagonal o'ngdan chapga bo'lsa, biz ularni ayitamiz.
Mana bu matritsa uchun ish.
\[\begin{align*}\det \left( B \right) & = \left| {\begin{array}{*{20}{r}}2&3&1\\{ - 1}&{ - 6}&7\\4&5&{ - 1}\end{array}} \right|\,\,\,\,\begin{array}{*{20}{r}}2&3\\{ - 1}&{ - 6}\\4&5\end{array}\\ & = \left( 2 \right)\left( { - 6} \right)\left( { - 1} \right) + \left( 3 \right)\left( 7 \right)\left( 4 \right) + \left( 1 \right)\left( { - 1} \right)\left( 5 \right) - \\ & \hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\left( 3 \right)\left( { - 1} \right)\left( { - 1} \right) - \left( 2 \right)\left( 7 \right)\left( 5 \right) - \left( 1 \right)\left( { - 6} \right)\left( 4 \right)\\ & = 42\end{align*}\]\(3 \times 3\) determinantini olish uchun formuladan yoki qisqa yoʻldan foydalanishingiz mumkin.
Agar matritsaning determinanti nolga teng bo'lsa, biz uni matritsa deb ataymiz birlik va agar matritsaning determinanti nolga teng bo'lmasa, biz matritsani chaqiramiz yagona bo'lmagan. Yuqoridagi misoldagi \(2 \times 2\) matritsasi birlik, \(3 \times 3\) matritsasi esa yagona emas.
Matritsa teskari
Keyinchalik, biz ko'rib chiqishimiz kerak teskari matritsadan. O'lchamdagi \(A\) kvadrat matritsasi berilgan n x \(n\), agar biz bir xil o'lchamdagi boshqa matritsani topsak, \(B\) shundayki,
\[AB = BA = {I_n}\]keyin biz \(B\) deb ataymiz teskari \(A\) va uni \(B=A^{-1}\) bilan belgilang.
\(A\) matritsasining teskarisini hisoblash juda oddiy. Birinchidan, biz yangi matritsa hosil qilamiz,
\[\left( {A\,\,\,{I_n}} \right)\]va keyin oldingi qator amallaridan foydalaning bo'limi va bu matritsani shaklga aylantirishga harakat qiling,
\[\left( {{I_n}\,\,\,B} \right)\]Agar biz buni qila olsak, \(B\) \(A\) ning teskarisidir. Agar buni qila olmasak, \(A\) matritsasining teskarisi yo'q.
Yechim
Biz yangi matritsani birinchi navbatda \(3 \times 3\) identifikatsiya matritsasini ushbu matritsaga yopishtirish orqali hosil qilamiz. Bu
\[\left( {\begin{array}{*{20}{r}}2&1&1\\{ - 5}&{ - 3}&0\\1&1&{ - 1}\end{array}\quad \begin{array}{*{20}{r}}1&0&0\\0&1&0\\0&0&1\end{array}} \right)\]Endi biz birinchi uchta ustunni \(3 \times 3\) identifikatoriga aylantirish uchun qator operatsiyalaridan foydalanamiz. Boshqacha qilib aytadigan bo'lsak, biz yuqori chap burchakdan boshlanadigan diagonalda 1 va dastlabki uchta ustundagi barcha boshqa yozuvlarda nol bo'lishini xohlaymiz.
Agar siz o'ylab ko'rsangiz, bu jarayon biz oxirgi foydalangan jarayonga juda o'xshaydi bo'limi tizimlarini hal qilish uchun, u faqat bir oz uzoqroq ketadi. Mana bu muammo uchun ish.
\[\left( {\begin{array}{*{20}{r}}2&1&1\\{ - 5}&{ - 3}&0\\1&1&{ - 1}\end{array}\quad \begin{array}{*{20}{r}}1&0&0\\0&1&0\\0&0&1\end{array}} \right)\begin{array}{*{20}{c}}{{R_1} \leftrightarrow {R_3}}\\ \Rightarrow \end{array}\left( {\begin{array}{*{20}{r}}1&1&{ - 1}\\{ - 5}&{ - 3}&0\\2&1&1\end{array}\quad \begin{array}{*{20}{r}}0&0&1\\0&1&0\\1&0&0\end{array}} \right)\begin{array}{*{20}{c}}{{R_2} + 5{R_1}}\\{{R_3} - 2{R_1}}\\ \Rightarrow \end{array}\] \[\left( {\begin{array}{*{20}{r}}1&1&{ - 1}\\0&2&{ - 5}\\0&{ - 1}&3\end{array}\quad \begin{array}{*{20}{r}}0&0&1\\0&1&5\\1&0&{ - 2}\end{array}} \right)\begin{array}{*{20}{c}}{\frac{1}{2}{R_2}}\\ \Rightarrow \end{array}\left( {\begin{array}{*{20}{r}}1&1&{ - 1}\\0&1&{\frac{{ - 5}}{2}}\\0&{ - 1}&3\end{array}\quad \begin{array}{*{20}{r}}0&0&1\\0&{\frac{1}{2}}&{\frac{5}{2}}\\1&0&{ - 2}\end{array}} \right)\begin{array}{*{20}{c}}{{R_3} + {R_2}}\\ \Rightarrow \end{array}\] \[\left( {\begin{array}{*{20}{r}}1&1&{ - 1}\\0&1&{\frac{{ - 5}}{2}}\\0&0&{\frac{1}{2}}\end{array}\quad \begin{array}{*{20}{r}}0&0&1\\0&{\frac{1}{2}}&{\frac{5}{2}}\\1&{\frac{1}{2}}&{\frac{1}{2}}\end{array}} \right)\begin{array}{*{20}{c}}{2{R_3}}\\ \Rightarrow \end{array}\left( {\begin{array}{*{20}{r}}1&1&{ - 1}\\0&1&{\frac{{ - 5}}{2}}\\0&0&1\end{array}\quad \begin{array}{*{20}{r}}0&0&1\\0&{\frac{1}{2}}&{\frac{5}{2}}\\2&1&1\end{array}} \right)\begin{array}{*{20}{c}}{{R_2} + \frac{5}{2}{R_3}}\\{{R_1} + {R_3}}\\ \Rightarrow \end{array}\] \[\left( {\begin{array}{*{20}{r}}1&1&0\\0&1&0\\0&0&1\end{array}\quad \begin{array}{*{20}{r}}2&1&2\\5&3&5\\2&1&1\end{array}} \right)\begin{array}{*{20}{c}}{{R_1} - {R_2}}\\ \Rightarrow \end{array}\left( {\begin{array}{*{20}{r}}1&0&0\\0&1&0\\0&0&1\end{array}\quad \begin{array}{*{20}{r}}{ - 3}&{ - 2}&{ - 3}\\5&3&5\\2&1&1\end{array}} \right)\]Shunday qilib, biz dastlabki uchta ustunni \(3 \times 3\) identifikatsiya matritsasiga aylantira oldik, shuning uchun teskari mavjud va shunday bo'ladi,
\[{A^{ - 1}} = \left( {\begin{array}{*{20}{r}}{ - 3}&{ - 2}&{ - 3}\\5&3&5\\2&1&1\end{array}} \right)\]Shunday qilib, teskari mavjud bo'lgan misol bor edi. Keling, teskari mavjud bo'lmagan misolni ko'rib chiqaylik.
Yechim
Bunday holda biz yangi matritsani olish uchun \(2 \times 2\) identifikatoriga murojaat qilamiz va keyin birinchi ikkita ustunni \(2 \times 2\) identifikator matritsasiga aylantirishga harakat qilamiz.
\[\left( {\begin{array}{*{20}{r}}1&{ - 3}&1&0\\{ - 2}&6&0&1\end{array}} \right)\,\,\,\begin{array}{*{20}{c}}{2{R_1} + {R_2}}\\ \Rightarrow \end{array}\,\,\left( {\begin{array}{*{20}{r}}1&{ - 3}&1&0\\0&0&2&1\end{array}} \right)\,\,\]Va biz uzoqqa borishimiz shart emas. \(2 \times 2\) identifikatori birinchi ikkita ustunda bo'lishi uchun biz ikkinchi ustunning ikkinchi yozuvida 1 va birinchi ustunning ikkinchi yozuvida 0 bo'lishimiz kerak. Biroq, ikkinchi ustunning ikkinchi yozuvida 1 ni olishning hech qanday usuli yo'q, bu birinchi ustundagi ikkinchi yozuvda 0 ni saqlaydi. Shuning uchun biz birinchi ikkita ustunda \(2 \times 2\) identifikatorini ololmaymiz va shuning uchun \(B\) ning teskarisi mavjud emas.
Biz bu teskari munozarani quyidagi fakt bilan qoldiramiz.
Teskari mavjud yoki yo'qligini aniqlash
\(A\) kvadrat matritsasi berilgan.
- Agar \(A\) yagona bo'lmasa, \(A^{-1}\) mavjud bo'ladi.
- Agar \(A\) birlik bo'lsa, \(A^{-1}\) mavjud bo'lmaydi.
Oldingi ikkita misol uchun bu haqiqatni tekshirishni sizga qoldiraman.
Tenglamalar sistemalari qayta ko'rib chiqildi
Biz tenglamalar sistemasini tezda qayta ko'rib chiqishimiz kerak. Keling, umumiy tenglamalar sistemasidan boshlaylik.
\[\begin{equation}\begin{aligned}{a_{11}}{x_1} + {a_{12}}{x_2} + \cdots + {a_{1n}}{x_n} & = {b_1}\\ {a_{21}}{x_1} + {a_{22}}{x_2} + \cdots + {a_{2n}}{x_n} & = {b_2}\\ \vdots \hspace{0.8in} & \\ {a_{n1}}{x_1} + {a_{n2}}{x_2} + \cdots + {a_{nn}}{x_n} & = {b_n}\end{aligned}\label{eq:eq1}\end{equation}\]Endi olish uchun har bir tomonni vektorga aylantiring,
\[\left( {\begin{array}{*{20}{r}}{{a_{11}}{x_1} + {a_{12}}{x_2} + \cdots + {a_{1n}}{x_n}}\\{{a_{21}}{x_1} + {a_{22}}{x_2} + \cdots + {a_{2n}}{x_n}}\\ \vdots \\{{a_{n1}}{x_1} + {a_{n2}}{x_2} + \cdots + {a_{nn}}{x_n}}\end{array}} \right) = \left( {\begin{array}{*{20}{r}}{{b_1}}\\{{b_2}}\\ \vdots \\{{b_n}}\end{array}} \right)\]Ushbu tenglamaning chap tomonini matritsani ko'paytirish deb hisoblash mumkin.
\[\left( {\begin{array}{*{20}{r}}{{a_{11}}}&{{a_{12}}}& \cdots &{{a_{1n}}}\\{{a_{21}}}&{{a_{22}}}& \cdots &{{a_{2n}}}\\ \vdots & \vdots & \ddots & \vdots \\{{a_{n1}}}&{{a_{n2}}}& \cdots &{{a_{nn}}}\end{array}} \right)\left( {\begin{array}{*{20}{r}}{{x_1}}\\{{x_2}}\\ \vdots \\{{x_n}}\end{array}} \right) = \left( {\begin{array}{*{20}{r}}{{b_1}}\\{{b_2}}\\ \vdots \\{{b_n}}\end{array}} \right)\]Belgini biroz soddalashtirish,
\[\begin{equation}A\vec x = \vec b \label{eq:eq2}\end{equation}\]Bu yerda, \(\vec x\) - komponentlari dastlabki tenglamalar sistemasidagi noma'lumlar bo'lgan vektor. Biz \(\eqref{eq:eq2}\) ni \(\eqref{eq:eq1}\) tenglamalar sistemasining matritsa shakli deb ataymiz va \(\eqref{eq:eq2}\) ni yechish \(\eqref{eq:eq1}\) ni yechishga teng. Yechish jarayoni bir xil. \(\eqref{eq:eq2}\) uchun kengaytirilgan matritsa
\[\left( {A\,\,\,\vec b} \right)\]Kengaytirilgan matritsaga ega bo'lgach, biz matritsa shaklida yozilmagan tizimda bo'lgani kabi davom etamiz.
\(\eqref{eq:eq2}\) yechimlari haqida bizda quyidagi fakt ham bor.
Yechimlarning mumkin bo'lgan raqamlari
\(\eqref{eq:eq2}\) tenglama tizimini hisobga olsak, biz yechimlar uchun quyidagi uchta imkoniyatdan biriga egamiz.
- Hech qanday yechim bo'lmaydi.
- Aynan bitta yechim bo'ladi.
- Cheksiz ko'p echimlar bo'ladi.
Aslida, biz hozir biroz uzoqroqqa borishimiz mumkin. Biz noma'lumlar bilan bir xil miqdordagi tenglamalarga egamiz deb taxmin qilganimiz uchun \(\eqref{eq:eq2}\)-dagi \(A\) matritsasi kvadrat matritsadir va shuning uchun biz uning determinantini hisoblashimiz mumkin. Bu quyidagi faktni beradi.
\(A\) yechimlari va yagonaligi
\(\eqref{eq:eq2}\) da tenglamalar sistemasini hisobga olsak, bizda quyidagilar mavjud.
- Agar \(A\) yagona bo'lmasa, tizim uchun aynan bitta yechim bo'ladi.
- Agar \(A\) singular bo'lsa, u holda tizimga yechim bo'lmaydi yoki cheksiz ko'p echimlar bo'ladi.
Bir jinsli tizimning matritsa shakli
\[\begin{equation}A\vec x = \vec 0 \label{eq:eq3}\end{equation}\]bu erda \(\vec 0\) barcha nollarning vektori. Bir hil tizimda biz \(\vec x = \vec 0\) yechimiga ega bo'lishimiz kafolatlanadi. Bir hil tizimlar uchun yuqoridagi fakt shundan iboratki,
Bir jinsli sistemaning yechimlari
\(\eqref{eq:eq3}\) bir jinsli tizimini hisobga olgan holda bizda quyidagilar mavjud.
- Agar \(A\) yagona bo'lmasa, yagona yechim \(\vec x = \vec 0\) bo'ladi.
- Agar \(A\) yagona bo'lsa, tizim uchun nolga teng bo'lmagan cheksiz ko'p echimlar bo'ladi.
Chiziqli mustaqillik / chiziqli bog'liqlik
Bu mavzuni birinchi marta ko'rib turganimiz yo'q. Biz chiziqli mustaqillik va chiziqli bog'liqlikni ham ko'rdik orqaga biz ikkinchi tartibli differentsial tenglamalarni ko'rib chiqayotganimizda. Ushbu bo'limda biz funktsiyalar bilan shug'ullangan edik, ammo bu erda tushuncha aslida bir xil. Agar biz \(n\) vektorlaridan boshlasak,
\[{\vec x_1},\,\,{\vec x_2},\,\, \ldots ,\,\,{\vec x_n}\]Agar biz doimiylarni topa olsak, \(c_{1}\), \(c_{2}\), …, \(c_{n}\) kamida ikkita nolga teng
\[\begin{equation}{c_1}{\vec x_1} + {c_2}{\vec x_2} + \, \ldots + {c_n}{\vec x_n} = \vec 0 \label{eq:eq4}\end{equation}\]u holda vektorlarni chiziqli bog'liq deb ataymiz. Agar \(\eqref{eq:eq4}\) da ishlaydigan yagona konstantalar \(c_{1}=0\), \(c_{2}\)=0, …, \(c_{n}=0\) bo‘lsa, vektorlarni chiziqli mustaqil deb ataymiz.
Agar biz \(n\) vektorlarining har biri \(n\) komponentlariga ega deb taxmin qilsak, i.e. vektorlarning har biri shunday ko'rinadi,
\[\vec x = \left( {\begin{array}{*{20}{r}}{{x_1}}\\{{x_2}}\\ \vdots \\{{x_n}}\end{array}} \right)\]chiziqli mustaqillik va chiziqli bog'liqlik uchun juda oddiy testni olishimiz mumkin. E'tibor bering, bunday bo'lishi shart emas, lekin biz barcha ishimizda \(n\) vektorlari bilan ishlaymiz, ularning har birida \(n\) komponentlari mavjud.
\(n\) vektorlarining chiziqli mustaqilligi uchun test
\(n\) vektorlarining har biri \(n\) komponentlarini hisobga olgan holda,
\[{\vec x_1},\,\,{\vec x_2},\,\, \ldots ,\,\,{\vec x_n}\]matritsa hosil qilish,
\[X = \left( {\begin{array}{*{20}{r}}{{{\vec x}_1}}&{{{\vec x}_2}}& \cdots &{{{\vec x}_n}}\end{array}} \right)\]Shunday qilib, \(X\) matritsasi \(i^{\text{th}}\) ustuni \(i^{\text{th}}\) vektori \({\vec x_i}\) bo'lgan matritsadir. Keyin,
- Agar \(X\) birlik bo'lmasa (i.e. \(\det(X)\) nolga teng emas) u holda \(n\) vektorlari chiziqli mustaqil va
- agar \(X\) birlik bo'lsa (i.e. \(\det(X)=0\)) keyin \(n\) vektorlari chiziqli bog'liq bo'lib, \(\eqref{eq:eq4}\) ni haqiqatga aylantiruvchi konstantalarni tizimni yechish orqali topish mumkin.
\[X\,\vec c = \vec 0\]
bu erda \(\vec c\) - \(\eqref{eq:eq4}\) konstantalarini o'z ichiga olgan vektor.
Yechim
Shunday qilib, birinchi navbatda \(X\) ni hosil qilish va uning determinantini hisoblash kerak.
\[X = \left( {\begin{array}{*{20}{r}}1&{ - 2}&6\\{ - 3}&1&{ - 2}\\5&4&1\end{array}} \right)\quad \quad \Rightarrow \hspace{0.25in}\hspace{0.25in}\det \left( X \right) = - 79\]Bu matritsa yagona emas va shuning uchun vektorlar chiziqli mustaqildir.
Yechim
Oxirgi misolda bo'lgani kabi, avval \(X\) ni hosil qiling va uning determinantini hisoblang.
\[X = \left( {\begin{array}{*{20}{r}}1&{ - 4}&2\\{ - 1}&1&{ - 1}\\3&{ - 6}&4\end{array}} \right)\quad \quad \hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}\det \left( X \right) = 0\]Demak, bu vektorlar chiziqli bog'liqdir. Endi vektorlar orasidagi munosabatni topishimiz kerak. Bu shuni anglatadiki, biz \(\eqref{eq:eq4}\) ni haqiqatga aylantiradigan doimiylarni topishimiz kerak.
Shunday qilib, biz tizimni hal qilishimiz kerak
\[X\,\vec c = \vec 0\]Mana bu tizim uchun kengaytirilgan matritsa va yechim ishlaydi.
\[\left( {\begin{array}{*{20}{r}}1&{ - 4}&2\\{ - 1}&1&{ - 1}\\3&{ - 6}&4\end{array}\quad \begin{array}{*{20}{r}}0\\0\\0\end{array}} \right)\begin{array}{*{20}{c}}{{R_2} + {R_1}}\\{{R_3} - 3{R_1}}\\ \Rightarrow \end{array}\left( {\begin{array}{*{20}{r}}1&{ - 4}&2\\0&{ - 3}&1\\0&6&{ - 2}\end{array}\quad \begin{array}{*{20}{r}}0\\0\\0\end{array}} \right)\begin{array}{*{20}{c}}{{R_3} + 2{R_2}}\\ \Rightarrow \end{array}\left( {\begin{array}{*{20}{r}}1&{ - 4}&2\\0&{ - 3}&1\\0&0&0\end{array}\quad \begin{array}{*{20}{r}}0\\0\\0\end{array}} \right)\begin{array}{*{20}{c}}{ - \frac{1}{3}{R_2}}\\ \Rightarrow \end{array}\] \[\left( {\begin{array}{*{20}{r}}1&{ - 4}&2\\0&1&{ - \frac{1}{3}}\\0&0&0\end{array}\quad \begin{array}{*{20}{r}}0\\0\\0\end{array}} \right)\begin{array}{*{20}{c}}{{R_1} + 4{R_2}}\\ \Rightarrow \end{array}\left( {\begin{array}{*{20}{r}}1&0&{\frac{2}{3}}\\0&1&{ - \frac{1}{3}}\\0&0&0\end{array}\quad \begin{array}{*{20}{r}}0\\0\\0\end{array}} \right)\quad \Rightarrow \quad \begin{array}{*{20}{r}}{{c_1} + \frac{2}{3}{c_3} = 0}\\{{c_2} - \frac{1}{3}{c_3} = 0}\\{0 = 0}\end{array}\quad \Rightarrow \quad \begin{array}{*{20}{l}}{{c_1} = - \frac{2}{3}{c_3}}\\{{c_2} = \frac{1}{3}{c_3}}\\{}\end{array}\]Endi biz doimiylar uchun haqiqiy qiymatlarni xohlaymiz, shuning uchun \({c_3} = 3\) dan foydalansak, \({c_1} = - 2\), \({c_2} = 1\) va \({c_3} = 3\) quyidagi yechimni olamiz. O'zaro munosabatlar shundan keyin.
\[ - 2{\vec x^{(1)}} + {\vec x^{(2)}} + 3{\vec x^{(3)}} = \left( {\begin{array}{*{20}{r}}0\\0\\0\end{array}} \right)\]Matritsalar bilan hisoblash
Haqiqatan ham, matritsalar bilan hisob-kitoblar bilan shug'ullanishimiz mumkinligiga ishonch hosil qilishdan boshqa ko'p narsa yo'q.
Birinchidan, shu paytgacha biz faqat raqamlari bo'lgan matritsalarni yozuvlar sifatida ko'rib chiqdik, ammo matritsadagi yozuvlar ham funktsiyalar bo'lishi mumkin. Shunday qilib, biz matritsalarni quyidagi shaklda ko'rishimiz mumkin:
\[A\left( t \right) = \left( {\begin{array}{*{20}{r}}{{a_{11}}\left( t \right)}&{{a_{12}}\left( t \right)}& \cdots &{{a_{1n}}\left( t \right)}\\{{a_{21}}\left( t \right)}&{{a_{22}}\left( t \right)}& \cdots &{{a_{2n}}\left( t \right)}\\ \vdots & \vdots &{}& \vdots \\{{a_{m1}}\left( t \right)}&{{a_{m2}}\left( t \right)}& \cdots &{{a_{mn}}\left( t \right)}\end{array}} \right)\]Endi bu shakldagi matritsani differensiallash va integrasiyalash haqida gapirish mumkin. Ushbu shakldagi matritsani farqlash yoki integrasiya qilish uchun biz qiladigan yagona narsa individual yozuvlarni farqlash yoki birlashtirishdir.
\[A'\left( t \right) = \left( {\begin{array}{*{20}{r}}{{{a'}_{11}}\left( t \right)}&{{{a'}_{12}}\left( t \right)}& \cdots &{{{a'}_{1n}}\left( t \right)}\\{{{a'}_{21}}\left( t \right)}&{{{a'}_{22}}\left( t \right)}& \cdots &{{{a'}_{2n}}\left( t \right)}\\ \vdots & \vdots &{}& \vdots \\{{{a'}_{m1}}\left( t \right)}&{{{a'}_{m2}}\left( t \right)}& \cdots &{{{a'}_{mn}}\left( t \right)}\end{array}} \right)\] \[\int{{A\left( t \right)\,dt}} = \left( {\begin{array}{*{20}{r}}{\int{{{a_{11}}\left( t \right)\,dt}}}&{\int{{{a_{12}}\left( t \right)\,dt}}}& \cdots &{\int{{{a_{1n}}\left( t \right)\,dt}}}\\{\int{{{a_{21}}\left( t \right)\,dt}}}&{\int{{{a_{22}}\left( t \right)\,dt}}}& \cdots &{\int{{{a_{2n}}\left( t \right)\,dt}}}\\ \vdots & \vdots &{}& \vdots \\{\int{{{a_{m1}}\left( t \right)\,dt}}}&{\int{{{a_{m2}}\left( t \right)\,dt}}}& \cdots &{\int{{{a_{mn}}\left( t \right)\,dt}}}\end{array}} \right)\]Shunday qilib, biz bunday narsaga duch kelganimizda, bundan hayajonlanmang. Biz odatdagidek farqlang yoki birlashtiring.
Ushbu bo'limda biz chiziqli algebradan juda ixcham mavzular to'plamini ko'rdik. Differensial tenglamalarga qaytganimizda, ushbu mavzularning ko'pchiligi vaqti-vaqti bilan paydo bo'ladi va siz hech bo'lmaganda so'zlarning ma'nosini bilishingiz kerak bo'ladi.
Chiziqli algebradan siz bilishingiz kerak bo'lgan asosiy mavzu, ammo siz differentsial tenglamalar sistemasini echishga qodirmisiz, keyingi bo'limning mavzusi.