9.7: Kutatua Mifumo na Inverses
- Page ID
- 181026
- Pata inverse ya tumbo.
- Tatua mfumo wa equations linear kwa kutumia tumbo inverse
Nancy mipango ya kuwekeza\($10,500\) katika vifungo viwili tofauti ili kueneza hatari yake. dhamana ya kwanza ina kurudi kila mwaka ya\(10%\), na dhamana ya pili ina kurudi kila mwaka ya\(6%\). Kupokea\(8.5%\) kurudi kutoka vifungo viwili, ni kiasi gani Nancy kuwekeza katika kila dhamana? Njia bora ya kutatua tatizo hili ni nini? Kuna njia kadhaa tunaweza kutatua tatizo hili. Kama tulivyoona katika sehemu zilizopita, mifumo ya equations na matrices ni muhimu katika kutatua matatizo halisi ya ulimwengu yanayohusisha fedha. Baada ya kujifunza sehemu hii, tutakuwa na zana za kutatua tatizo la dhamana kwa kutumia inverse ya tumbo.
Kutafuta Inverse ya Matrix
Tunajua kwamba inverse multiplicative ya idadi halisi\(a\) ni\(a^{−1}\), hivyo
\[aa^{−1}=a^{−1}a=\left(\dfrac{1}{a}\right)a=1 \label{eq0}\]
Kwa mfano, fikiria hali ya kuzidisha scalar
\[2^{−1}=\dfrac{1}{2} \nonumber\]
kwa hiyo kutoka Equation\ ref {eq0}
\[\left(\dfrac{1}{2}\right)2=1. \nonumber\]
Inverse ya kuzidisha ya tumbo ni sawa na dhana, isipokuwa kwamba bidhaa ya tumbo\(A\) na inverse yake\(A^{−1}\) ni sawa na tumbo la utambulisho. Matrix ya utambulisho ni tumbo la mraba iliyo na chini ya diagonal kuu na zero kila mahali pengine. Tunatambua matrices ya utambulisho na\(I_n\) wapi\(n\) inawakilisha mwelekeo wa tumbo. Ulinganifu\ ref {eq1} na\ ref {eq2} ni matrices ya utambulisho kwa\(2×2\) tumbo na\(3×3\) tumbo, mtawalia:
\[I_2=\begin{bmatrix}1&0 \\ 0&1 \end{bmatrix} \label{eq1}\]
\[I_3=\begin{bmatrix}1&0&0 \\ 0&1&0 \\ 0&0&1\end{bmatrix} \label{eq2}\]
Matrix ya utambulisho hufanya kama\(1\) algebra ya tumbo. Kwa mfano,
\[AI=IA=A\nonumber\]
Matrix ambayo ina inverse ya kuzidisha ina mali
\[AA^{−1}=I\]
\[A^{−1}A=I\]
Matrix ambayo ina inverse ya kuzidisha inaitwa tumbo la invertible. Tu tumbo mraba inaweza kuwa na inverse multiplicative, kama reversibility,
\[AA^{−1}=A^{−1}A=I\]
ni mahitaji. Sio matrices yote ya mraba\(A\) yana inverse, lakini ikiwa haiwezi kuingizwa, basi\(A^{−1}\) ni ya pekee. Tutaangalia njia mbili za kutafuta inverse ya\(2 × 2\) tumbo na njia ya tatu ambayo inaweza kutumika kwa wote\(2 × 2\) na\(3 × 3\) matrices.
Matrix ya utambulisho\(I_n\),, ni tumbo la mraba lililo na ndio chini ya diagonal kuu na zero kila mahali pengine.
\[I_2=\begin{bmatrix}1&0 \nonumber \\ 0&1\end{bmatrix}\]
kama kwa tumbo la\(2 × 2\) utambulisho
\[I_3=\begin{bmatrix}1&0&0 \nonumber \\ 0&1&0 \nonumber \\ 0&0&1\end{bmatrix}\]
kama kwa tumbo la\(3 × 3\) utambulisho
Ikiwa\(A\) ni\(n × n\) tumbo na\(B\) ni\(n × n\) tumbo kama hiyo\(AB=BA=I_n\), basi\(B=A−1\), inverse ya kuzidisha ya tumbo\(A\).
Kutokana na tumbo\(A\), onyesha hilo\(AI=IA=A\).
\[A=\begin{bmatrix}3&4 \nonumber \\ −2&5\end{bmatrix}\]
Suluhisho
Tumia kuzidisha kwa tumbo ili kuonyesha kwamba bidhaa\(A\) na matriki ya utambulisho ni sawa na bidhaa ya tumbo la utambulisho na\(A\).
\[\begin{align*} AI&=\begin{bmatrix}3&4 \nonumber \\ −2&5\end{bmatrix}\begin{bmatrix}1&0 \nonumber \\ 0&1\end{bmatrix} \nonumber \\[4pt] &=\begin{bmatrix}3⋅1+4⋅0&3⋅0+4⋅1 \nonumber \\ −2⋅1+5⋅0&−2⋅0+5⋅1\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}3&4 \nonumber \\ −2&5\end{bmatrix} \end{align*}\]
\[\begin{align*} AI&=\begin{bmatrix}1&0 \nonumber \\ 0&1\end{bmatrix}\begin{bmatrix}3&4 \nonumber \\ −2&5\end{bmatrix} \nonumber \\[4pt] &=\begin{bmatrix}1⋅3+0⋅(−2)&1⋅4+0⋅5 \nonumber \\ 0⋅3+1⋅(−2)&0⋅4+1⋅5\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}3&4 \nonumber \\ −2&5\end{bmatrix} \end{align*}\]
- Kutokana na tumbo\(A\) la utaratibu\(n × n\) na tumbo\(B\) la utaratibu\(n × n\) kuzidisha\(AB\).
- Ikiwa\(AB=I\), basi pata bidhaa\(BA\). Ikiwa\(BA=I\), basi\(B=A^{−1}\) na\(A=B^{−1}\).
Onyesha kwamba matrices zilizopewa ni inverses nyingi za kila mmoja.
\[A=\begin{bmatrix}1&5 \nonumber \\ −2&−9\end{bmatrix}\]
na
\[B=\begin{bmatrix}−9&−5 \nonumber \\ 2&1\end{bmatrix}\]
Suluhisho
Kuzidisha\(AB\) na\(BA\). Ikiwa bidhaa zote mbili zinafanana na utambulisho, basi matrices mbili ni inverses ya kila mmoja.
\[\begin{align*} AB &= \begin{bmatrix}1&5 \nonumber \\ −2&−9\end{bmatrix}·\begin{bmatrix}−9&−5 \nonumber \\ 2&1\end{bmatrix} \nonumber \\[4pt] &=\begin{bmatrix}1(−9)+5(2)&1(−5)+5(1) \nonumber \\ −2(−9)−9(2)&−2(−5)−9(1)\end{bmatrix} \nonumber \\[4pt] &=\begin{bmatrix}1&0 \nonumber \\ 0&1\end{bmatrix} \end{align*}\]
na
\[\begin{align*} BA &= \begin{bmatrix}−9&−5 \nonumber \\ 2&1\end{bmatrix}·\begin{bmatrix}1&5 \nonumber \\ −2&−9\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}−9(1)−5(−2)&−9(5)−5(−9) \nonumber \\ 2(1)+1(−2)&2(−5)+1(−9)\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}1&0 \nonumber \\0&1\end{bmatrix} \end{align*}\]
\(A\)na\(B\) ni inverses ya kila mmoja.
Onyesha kwamba matrices mbili zifuatazo ni inverses ya kila mmoja.
\[A=\begin{bmatrix}1&4 \nonumber \\[4pt] −1&−3\end{bmatrix}\]
na
\[B=\begin{bmatrix}−3&−4 \nonumber \\[4pt] 1&1\end{bmatrix}\]
- Jibu
-
\(\begin{align*} AB&=\begin{bmatrix}1&4 \nonumber \\[4pt] −1&−3\end{bmatrix}\begin{bmatrix}−3&−4 \nonumber \\[4pt] 1&1\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}1(−3)+4(1)&1(−4)+4(1) \nonumber \\[4pt] −1(−3)+−3(1)&−1(−4)+−3(1)\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}1&0 \nonumber \\[4pt] 0&1\end{bmatrix} \end{align*}\)
\(\begin{align*} BA&=\begin{bmatrix}−3&−4 \nonumber \\[4pt] 1&1\end{bmatrix}\begin{bmatrix}1&4 \nonumber \\[4pt] −1&−3\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}−3(1)+−4(−1)&−3(4)+−4(−3) \nonumber \\[4pt] 1(1)+1(−1)&1(4)+1(−3)\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}1&0 \nonumber \\[4pt] 0&1\end{bmatrix} \end{align*}\)
Kutafuta Inverse ya Kuzidisha Kutumia Kuzidisha Matrix
Sasa tunaweza kuamua kama matrices mbili ni inverses, lakini tutawezaje kupata inverse ya tumbo iliyotolewa? Kwa kuwa tunajua kwamba bidhaa ya tumbo na inverse yake ni tumbo la utambulisho, tunaweza kupata inverse ya tumbo kwa kuanzisha equation kwa kutumia kuzidisha matrix.
Tumia kuzidisha matrix ili kupata inverse ya tumbo iliyotolewa.
\[A=\begin{bmatrix}1&−2 \nonumber \\[4pt] 2&−3\end{bmatrix}\]
Suluhisho
Kwa njia hii, tunazidisha\(A\) na tumbo iliyo na vipindi visivyojulikana na kuiweka sawa na utambulisho.
\(\begin{bmatrix}1&−2 \nonumber \\[4pt] 2&−3\end{bmatrix}\begin{bmatrix}a&b \nonumber \\[4pt] c&d\end{bmatrix}=\begin{bmatrix}1&0 \nonumber \\[4pt] 0&1\end{bmatrix}\)
Pata bidhaa ya matrices mbili upande wa kushoto wa ishara sawa.
\[\begin{bmatrix}1&−2 \nonumber \\[4pt] 2&−3\end{bmatrix}\begin{bmatrix}a&b \nonumber \\[4pt] c&d\end{bmatrix}=\begin{bmatrix}1a−2c&1b−2d \nonumber \\[4pt] 2a−3c&2b−3d\end{bmatrix}\]
Kisha, fungua mfumo wa equations na kuingia katika mstari wa 1, safu ya 1 ya tumbo mpya sawa na kuingia kwanza ya utambulisho,\(1\). Weka kuingia katika mstari wa 2, safu ya 1 ya tumbo mpya sawa na kuingia sambamba ya utambulisho, yaani\(0\).
\(1a−2c=1\space R_1\)
\(2a−3c=0\space R_2\)
Kutumia shughuli za mstari, kuzidisha na kuongeza kama ifuatavyo:\((−2)R_1+R_2\rightarrow R_2\). Kuongeza milinganyo, na kutatua kwa\(c\).
\[ \begin{align*} 1a−2c &=1 \nonumber \\[4pt] 0+1c &=−2 \nonumber \\[4pt] c=−2 \nonumber \end{align*} \nonumber\]
Back-mbadala ya kutatua kwa\(a\).
\[ \begin{align*} a−2(−2)&=1 \nonumber \\[4pt] a+4&=1 \nonumber \\[4pt] a&=−3 \nonumber\end{align*} \nonumber\]
Andika mfumo mwingine wa equations kuweka kuingia katika mstari wa 1, safu ya 2 ya tumbo mpya sawa na kuingia sambamba ya utambulisho,\(0\). Weka kuingia katika mstari wa 2, safu ya 2 sawa na kuingia sambamba ya utambulisho.
\(1b−2d=0\space R_1\)
\(2b−3d=1\space R_2\)
Kutumia shughuli za mstari, kuzidisha na kuongeza kama ifuatavyo:\((−2)R_1+R_2=R_2\). Kuongeza milinganyo mbili na kutatua kwa\(d\).
\[ \begin{align*} 1b−2d&=0 \nonumber \\[4pt] 0+1d&=1 \nonumber \\[4pt] d&=1 \nonumber \end{align*} \nonumber\]
Mara nyingine zaidi, nyuma-mbadala na kutatua kwa\(b\).
\[ \begin{align*} b−2(1)&=0 \nonumber \\[4pt] b&−2=0 \nonumber \\[4pt] b &=2 \nonumber \end{align*} \nonumber\]
\[A^{−1}=\begin{bmatrix}−3&2 \nonumber \\[4pt] −2&1\end{bmatrix}\]
Kutafuta Inverse ya kuzidisha kwa Kuongeza na Identity
Njia nyingine ya kupata inverse ya kuzidisha ni kwa kuongeza na utambulisho. Wakati tumbo\(A\) limebadilishwa\(I\), tumbo la kuongezeka\(I\) linabadilika\(A^{−1}\).
Kwa mfano, kutokana na
\(A=\begin{bmatrix}2&1 \nonumber \\[4pt] 5&3\end{bmatrix}\)
kuongeza\(A\) na utambulisho
\(\left[ \begin{array}{cc|cc} 2&1&1&0 \\ 5&3&0&1\end{array} \right]\)
Fanya shughuli za mstari kwa lengo la kugeuza A katika utambulisho.
- Badilisha mstari wa 1 na mstari wa 2.
\(\left[ \begin{array}{cc|cc} 5&3&0&1 \nonumber \\[4pt] 2&1&1&0\end{array} \right]\)
- Panua mstari wa 2 na -1 na uongeze mstari wa 1.
\(\left[ \begin{array}{cc|cc} 1&1&-2&1 \nonumber \\[4pt] 2&1&1&0\end{array} \right]\)
- Panua mstari 1 na -1 na uongeze mstari wa 2.
\(\left[ \begin{array}{cc|cc} 1&1&-2&1 \nonumber \\[4pt] 0&-1&5&-2\end{array} \right]\)
- Ongeza mstari 2 kwa mstari 1.
\(\left[ \begin{array}{cc|cc} 1&0&3&-1 \nonumber \\[4pt] 0&-1&5&-2\end{array} \right]\)
- Panua mstari wa 2 na-1. -1.
\(\left[ \begin{array}{cc|cc} 1&0&3&-1 \nonumber \\[4pt] 0&1&-5&2\end{array} \right]\)
Matrix tumepata ni\(A^{−1}\).
\(A^{−1}=\begin{bmatrix}3&−1 \nonumber \\[4pt] −5&2\end{bmatrix}\)
Kutafuta Inverse ya Kuzidisha ya\(2×2\) Matrices Kutumia Mfumo
Tunapohitaji kupata inverse ya kuzidisha ya\(2 × 2\) tumbo, tunaweza kutumia formula maalum badala ya kutumia kuzidisha matrix au kuongeza na utambulisho.
Ikiwa\(A\) ni\(2×2\) tumbo, kama vile
\(A=\begin{bmatrix}a&b \nonumber \\[4pt] c&d\end{bmatrix}\)
inverse ya multiplicative\(A\) inatolewa na formula
\(A^{−1}=\dfrac{1}{ad−bc}\begin{bmatrix}d&−b \nonumber \\[4pt] −c&a\end{bmatrix}\)
wapi\(ad−bc≠0\). Ikiwa\(ad−bc=0\), basi\(A\) haina inverse.
Tumia formula ili kupata inverse ya kuzidisha
\[A=\begin{bmatrix}1&−2 \nonumber \\[4pt] 2&−3\end{bmatrix}\]
Suluhisho
Tunaweza kuangalia kwamba formula yetu inafanya kazi kwa kutumia moja ya njia nyingine kuhesabu inverse. Hebu tuongeze\(A\) na utambulisho.
\(\left[ \begin{array}{cc|cc} 1&-2&1&0 \nonumber \\[4pt] 2&-3&0&1\end{array}\right]\)
Fanya shughuli za mstari kwa lengo la\(A\) kugeuka katika utambulisho.
- Panua mstari wa 1\(−2\) na uongeze mstari wa 2.
\(\left[ \begin{array}{cc|cc} 1&-2&1&0 \nonumber \\[4pt] 0&1&-2&1\end{array} \right]\)
- Panua mstari wa 1\(2\) na uongeze mstari wa 1.
\(\left[ \begin{array}{cc|cc} 1&0&-3&2 \nonumber \\[4pt] 0&1&-2&1\end{array} \right]\)
Kwa hiyo, tumehakikishia ufumbuzi wetu wa awali.
\(A^{−1}=\begin{bmatrix}−3&2 \nonumber \\[4pt] −2&1\end{bmatrix}\)
Tumia formula ili kupata inverse ya tumbo\(A\). Thibitisha jibu lako kwa kuongeza na tumbo la utambulisho.
\(A=\begin{bmatrix}1&−1 \nonumber \\[4pt] 2&3\end{bmatrix}\)
- Jibu
-
\(A^{−1}=\begin{bmatrix}\dfrac{3}{5}&\dfrac{1}{5} \nonumber \\[4pt] −\dfrac{2}{5}&\dfrac{1}{5}\end{bmatrix}\)
Pata inverse, ikiwa iko, ya tumbo iliyotolewa.
\(A=\begin{bmatrix}3&6 \nonumber \\[4pt] 1&2\end{bmatrix}\)
Suluhisho
Tutatumia njia ya kuongeza na utambulisho.
\(\left[ \begin{array}{cc|cc} 3&6&1&0 \nonumber \\[4pt] 1&3&0&1\end{array} \right]\)
- Badilisha mstari wa 1 na mstari wa 2.
\(\left[ \begin{array}{cc|cc} 1&3&0&1 \nonumber \\[4pt] 3&6&1&0\end{array} \right]\)
- Panua mstari 1 na -3 na uongeze kwenye mstari wa 2.
\(\left[ \begin{array}{cc|cc} 1&2&1&0 \nonumber \\[4pt] 0&0&-3&1\end{array} \right]\)
- Hakuna kitu zaidi tunaweza kufanya. Zero katika mstari wa 2 zinaonyesha kwamba tumbo hili halina inverse.
Kutafuta Inverse ya Multiplicative ya\(3×3\) Matrices
Kwa bahati mbaya, hatuna formula sawa na ile ya\(2×2\) tumbo ili kupata inverse ya\(3×3\) tumbo. Badala yake, sisi kuongeza tumbo awali na utambulisho Matrix na matumizi ya shughuli mstari kupata inverse.
Kutokana na\(3 × 3\) tumbo
\[A=\begin{bmatrix}2&3&1 \nonumber \\[4pt] 3&3&1 \nonumber \\[4pt] 2&4&1\end{bmatrix}\]
kuongeza\(A\) na tumbo la utambulisho
\[\begin{array}{c|c}A&I\end{array}=\left[ \begin{array}{ccc|ccc}2&3&1&1&0&0 \nonumber \\[4pt] 3&3&1&0&1&0 \nonumber \\[4pt] 2&4&1&0&0&1\end{array} \right]\]
Kuanza, tunaandika tumbo la kuongezeka na utambulisho upande wa kulia na\(A\) wa kushoto. Kufanya shughuli za mstari wa msingi ili tumbo la utambulisho linaonekana upande wa kushoto, tutapata tumbo la kinyume upande wa kulia. Tutapata inverse ya tumbo hili katika mfano unaofuata.
- Andika tumbo la awali lililoongezwa na tumbo la utambulisho upande wa kulia.
- Tumia shughuli za mstari wa msingi ili utambulisho uonekane upande wa kushoto.
- Nini kinachopatikana kwa haki ni inverse ya tumbo la awali.
- Tumia kuzidisha kwa tumbo ili kuonyesha kwamba\(AA^{−1}=I\) na\(A^{−1}A=I\).
Kutokana na\(3 × 3\) tumbo\(A\), tafuta inverse.
\(A=\begin{bmatrix}2&3&1 \nonumber \\[4pt] 3&3&1 \nonumber \\[4pt] 2&4&1\end{bmatrix}\)
Suluhisho
Kuongeza\(A\) na tumbo utambulisho, na kisha kuanza shughuli mstari mpaka utambulisho Matrix nafasi\(A\). Matrix upande wa kulia itakuwa inverse ya\(A\).
\(\left[ \begin{array}{ccc|ccc}2&3&1&1&0&0 \nonumber \\[4pt] 3&3&1&0&1&0 \nonumber \\[4pt] 2&4&1&0&0&1 \end{array} \right] \xrightarrow{Interchange\space R_2\space and\space R_1} \left[ \begin{array}{ccc|ccc}3&3&1&0&1&0 \nonumber \\[4pt] 2&3&1&1&0&0 \nonumber \\[4pt] 2&4&1&0&0&1 \end{array} \right]\)
\(−R_2+R_1=R_1\rightarrow \left[ \begin{array}{ccc|ccc}1&0&0&-1&1&0 \nonumber \\[4pt] 2&3&1&1&0&0 \nonumber \\[4pt] 2&4&1&0&0&1\end{array} \right]\)
\(−R_2+R_3=R_3\rightarrow \left[ \begin{array}{ccc|ccc}1&0&0&-1&1&0 \nonumber \\[4pt] 2&3&1&1&0&0 \nonumber \\[4pt] 0&1&0&-1&0&1\end{array} \right]\)
\(R_2\leftrightarrow R_3\rightarrow \left[ \begin{array}{ccc|ccc}1&0&0&-1&1&0 \nonumber \\[4pt] 0&1&0&-1&0&1 \nonumber \\[4pt] 2&3&1&1&0&0\end{array} \right]\)
\(−2R_1+R_3=R_3\rightarrow \left[ \begin{array}{ccc|ccc}1&0&0&-1&1&0 \nonumber \\[4pt] 0&1&0&-1&0&1 \nonumber \\[4pt] 0&3&1&3&-2&0\end{array} \right]\)
\(−3R_2+R_3=R_3\rightarrow \left[ \begin{array}{ccc|ccc}1&0&0&-1&1&0 \nonumber \\[4pt] 0&1&0&-1&0&1 \nonumber \\[4pt] 0&0&1&6&-2&-3\end{array} \right]\)
Hivyo,
\(A^{−1}=B=\begin{bmatrix}−1&1&0 \nonumber \\[4pt] −1&0&1 \nonumber \\[4pt] 6&−2&−3\end{bmatrix}\)
Uchambuzi
Ili kuthibitisha kwamba\(B=A^{−1}\), hebu tuzidishe matrices mbili pamoja ili kuona kama bidhaa ni sawa na utambulisho, ikiwa\(AA^{−1}=I\) na\(A^{−1}A=I\).
\[\begin{align*} AA^{−1} & =\begin{bmatrix}2&3&1 \nonumber \\[4pt] 3&3&1 \nonumber \\[4pt] 2&4&1\end{bmatrix}\begin{bmatrix}−1&1&0 \nonumber \\[4pt] −1&0&1 \nonumber \\[4pt] 6&−2&−3\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}2(−1)+3(−1)+1(6)&2(1)+3(0)+1(−2)&2(0)+3(1)+1(−3) \nonumber \\[4pt] 3(−1)+3(−1)+1(6)& 3(1)+3(0)+1(−2)& 3(0)+3(1)+1(−3) \nonumber \\[4pt] 2(−1)+4(−1)+1(6)& 2(1)+4(0)+1(−2)& 2(0)+4(1)+1(−3)\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}1&0&0&0&1&0 \nonumber \\[4pt] 0&0&1\end{bmatrix} \nonumber \\[4pt] A^{−1}A &= \begin{bmatrix}−1&1&0 \nonumber \\[4pt] −1&0&1 \nonumber \\[4pt] 6&−2&−3\end{bmatrix}\begin{bmatrix}&2&31 \nonumber \\[4pt] 3&3&1 \nonumber \\[4pt] 2&4&1\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}−1(2)+1(3)+0(2)& −1(3)+1(3)+0(4)& −1(1)+1(1)+0(1) \nonumber \\[4pt] −1(2)+0(3)+1(2)& −1(3)+0(3)+1(4)& −1(1)+0(1)+1(1) \nonumber \\[4pt] 6(2)+−2(3)+−3(2)& 6(3)+−2(3)+−3(4)& 6(1)+−2(1)+−3(1)\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}1&0&0 \nonumber \\[4pt] 0&1&0 \nonumber \\[4pt] 0&0&1\end{bmatrix} \end{align*}\]
Pata inverse ya\(3×3\) tumbo.
\(A=\begin{bmatrix}2&−17&11 \nonumber \\[4pt] −1&11&−7 \nonumber \\[4pt] 0&3&−2\end{bmatrix}\)
- Jibu
-
\(A^{−1}=\begin{bmatrix}1&1&2 \nonumber \\[4pt] 2&4&−3 \nonumber \\[4pt] 3&6&−5\end{bmatrix}\)
Kutatua Mfumo wa Ulinganisho wa Mstari Kutumia Inverse ya Matrix
Kutatua mfumo wa equations linear kwa kutumia inverse ya tumbo inahitaji ufafanuzi wa matrices mbili mpya:\(X\) ni tumbo anayewakilisha vigezo vya mfumo, na\(B\) ni tumbo anayewakilisha constants. Kutumia kuzidisha matrix, tunaweza kufafanua mfumo wa equations na idadi sawa ya equations kama vigezo kama
\(AX=B\)
Ili kutatua mfumo wa equations linear kwa kutumia tumbo inverse, hebu\(A\) kuwa tumbo la mgawo, basi\(X\) iwe tumbo la kutofautiana, na\(B\) iwe tumbo la mara kwa mara. Hivyo, tunataka kutatua mfumo\(AX=B\). Kwa mfano, angalia mfumo wafuatayo wa equations.
\(a_1x+b_1y=c_1\)
\(a_2x+b_2y=c_2\)
Kutoka kwa mfumo huu, tumbo la mgawo ni
\(A=\begin{bmatrix}a_1&b_1 \nonumber \\[4pt] a_2&b_2\end{bmatrix}\)
Matrix ya kutofautiana ni
\(X=\begin{bmatrix}x \nonumber \\[4pt] y\end{bmatrix}\)
Na tumbo la mara kwa mara ni
\(B=\begin{bmatrix}c_1 \nonumber \\[4pt] c_2\end{bmatrix}\)
Kisha\(AX=B\) inaonekana kama
\(\begin{bmatrix}a_1&b_1 \nonumber \\[4pt] a_2&b_2\end{bmatrix}\begin{bmatrix}x \nonumber \\[4pt] y\end{bmatrix}=\begin{bmatrix}c_1 \nonumber \\[4pt] c_2\end{bmatrix}\)
Kumbuka majadiliano mapema katika sehemu hii kuhusu kuzidisha idadi halisi kwa inverse yake,\((2^{−1}) 2=\left(\dfrac{1}{2}\right) 2=1\). Ili kutatua single linear equation\(ax=b\) kwa\(x\), tunataka tu kuzidisha pande zote mbili za equation na inverse multiplicative (kubadilishana) ya\(a\). Hivyo,
\[\begin{align*} ax&= b\\ \left(\dfrac{1}{a}\right)ax&= \left(\dfrac{1}{a}\right)b\\ \left(a^{-1}\right)ax&= \left(a^{-1}\right)b\\ \left[\left(a^{-1}\right)a\right]x&= \left(a^{-1}\right)b\\ 1x&= \left(a^{-1}\right)b\\ x&= \left(a^{-1}\right)b \end{align*}\]
Tofauti tu kati ya kutatua equation linear na mfumo wa milinganyo imeandikwa katika mfumo Matrix ni kwamba kutafuta inverse ya tumbo ni ngumu zaidi, na Matrix kuzidisha ni mchakato tena. Hata hivyo, lengo ni sawa-kujitenga variable.
Sisi kuchunguza wazo hili kwa undani, lakini ni muhimu kwa kuanza na\(2 × 2\) mfumo na kisha kuendelea na\(3 × 3\) mfumo.
Kutokana na mfumo wa equations, andika tumbo la mgawo\(A\), tumbo la kutofautiana\(X\), na tumbo la mara kwa mara\(B\). Kisha
\(AX=B\)
Kuzidisha pande zote mbili kwa inverse ya\(A\) kupata ufumbuzi.
\[\begin{align*} \left(A^{-1}\right)AX&= \left(A^{-1}\right)B\\ \left[\left(A^{-1}\right)A \right]X&= \left(A^{-1}\right)B\\ IX&= \left(A^{-1}\right)B\\ X&= \left(A^{-1}\right)B \end{align*}\]
Hapana, ikiwa tumbo la mgawo hauwezi kuingizwa, mfumo hauwezi kuwa sawa na hauna suluhisho, au kuwa tegemezi na kuwa na ufumbuzi mkubwa sana.
Tatua mfumo uliotolewa wa equations kwa kutumia inverse ya tumbo.
\[\begin{align*} 3x+8y&= 5\\ 4x+11y&= 7 \end{align*}\]
Suluhisho
Andika mfumo kwa suala la tumbo la mgawo, tumbo la kutofautiana, na tumbo la mara kwa mara.
\(A=\begin{bmatrix}3&8 \nonumber \\[4pt] 4&11\end{bmatrix}\),\(X=\begin{bmatrix}x \nonumber \\[4pt] y\end{bmatrix}\),\(B=\begin{bmatrix}5 \nonumber \\[4pt] 7\end{bmatrix}\)
Kisha
\(\begin{bmatrix}3&8 \nonumber \\[4pt] 4&11\end{bmatrix}\begin{bmatrix}x \nonumber \\[4pt] y\end{bmatrix}=\begin{bmatrix}5 \nonumber \\[4pt] 7\end{bmatrix}\)
Kwanza, tunahitaji kuhesabu\(A^{−1}\). Kutumia formula ili kuhesabu inverse ya\(2\)\(2\) tumbo, tuna:
\[\begin{align*} A^{−1} &= \dfrac{1}{ad−bc}\begin{bmatrix}d&−b \nonumber \\[4pt] −c&a\end{bmatrix} \\ &= \dfrac{1}{3(11)−8(4)}\begin{bmatrix}11&−8 \nonumber \\[4pt] −4&3\end{bmatrix} \\ &=\dfrac{1}{1}\begin{bmatrix}11&−8 \nonumber \\[4pt] −4&3\end{bmatrix} \end{align*}\]
Hivyo,
\(A^{−1}=\begin{bmatrix}11&−8 \nonumber \\[4pt] −4 &3\end{bmatrix}\)
Sasa tuko tayari kutatua. Kuzidisha pande zote mbili za equation na\(A^{−1}\).
\[\begin{align*} \left(A^{−1}\right)AX&=\left(A^{−1}\right)B \\[4pt] \begin{bmatrix}11&−8 \nonumber \\[4pt] −4&3\end{bmatrix}\begin{bmatrix}3&8 \nonumber \\[4pt] 4&11\end{bmatrix}\begin{bmatrix}x \nonumber \\[4pt] y\end{bmatrix}&=\begin{bmatrix}11&−8 \nonumber \\[4pt] −4&3\end{bmatrix}\begin{bmatrix}5 \nonumber \\[4pt] 7\end{bmatrix} \\[4pt] \begin{bmatrix}1&0 \nonumber \\[4pt] 0&1\end{bmatrix}\begin{bmatrix}x \nonumber \\[4pt] y\end{bmatrix}&=\begin{bmatrix}11(5)+(−8)7 \nonumber \\[4pt] −4(5)+3(7)\end{bmatrix} \\[4pt] \begin{bmatrix}x \nonumber \\[4pt] y\end{bmatrix}&=\begin{bmatrix}−1 \nonumber \\[4pt] 1\end{bmatrix} \end{align*}\]
Suluhisho ni\((−1,1)\).
Hapana, kumbuka kuwa kuzidisha kwa tumbo sio kubadilisha, hivyo\(A^{−1}B≠BA^{−1}\). Fikiria hatua zetu za kutatua equation ya tumbo.
\[\begin{align*} \left(A^{-1}\right)AX&= \left(A^{-1}\right)B\\ \left[ \left(A^{-1}\right)A \right]X&= \left(A^{-1}\right)B\\ IX&= \left(A^{-1}\right)B\\ X&= \left(A^{-1}\right)B \end{align*}\]
Taarifa katika hatua ya kwanza sisi tele pande zote mbili za equation na\(A^{−1}\), lakini\(A^{−1}\) ilikuwa\(A\) upande wa kushoto wa upande wa kushoto na upande wa kushoto wa\(B\) upande wa kulia. Kwa sababu Matrix kuzidisha si commutative, ili mambo.
Tatua mfumo wafuatayo kwa kutumia inverse ya tumbo.
\[\begin{align*} 5x+15y+56z&= 35\\ -4x-11y-41z&= -26\\ -x-3y-11z&= -7 \end{align*}\]
Suluhisho
Andika equation\(AX=B\).
\(\begin{bmatrix}5&15&56 \nonumber \\[4pt] −4&−11&−41 \nonumber \\[4pt] −1&−3&−11\end{bmatrix}\begin{bmatrix}x \nonumber \\[4pt] y \nonumber \\[4pt] z\end{bmatrix}=\begin{bmatrix}35 \nonumber \\[4pt] −26 \nonumber \\[4pt] −7\end{bmatrix}\)
Kwanza, tutapata inverse ya\(A\) kwa kuongeza na utambulisho.
\(\left[ \begin{array}{ccc|ccc}5&15&56&1&0&0 \nonumber \\[4pt] −4&−11&−41&0&1&0 \nonumber \\[4pt] −1&−3&−11&0&0&1\end{array} \right]\)
Kuzidisha mstari 1 na\(\dfrac{1}{5}\).
\(\left[ \begin{array}{ccc|ccc}1&3&\dfrac{56}{5}&\dfrac{1}{5}&0&0 \nonumber \\[4pt] −4&−11&−41&0&1&0 \nonumber \\[4pt] −1&−3&−11&0&0&1\end{array} \right]\)
Panua mstari wa 1\(4\) na uongeze mstari wa 2.
\(\left[ \begin{array}{ccc|ccc}1&3&\dfrac{56}{5}&\dfrac{1}{5}&0&0 \nonumber \\[4pt] 0&1&\dfrac{19}{5}&\dfrac{4}{5}&1&0 \nonumber \\[4pt] −1&−3&−11&0&0&1\end{array} \right]\)
Ongeza mstari 1 kwa mstari 3.
\(\left[ \begin{array}{ccc|ccc}1&3&\dfrac{56}{5}&\dfrac{1}{5}&0&0 \nonumber \\[4pt] 0&1&\dfrac{19}{5}&\dfrac{4}{5}&1&0 \nonumber \\[4pt] 0&0&\dfrac{1}{5}&\dfrac{1}{5}&0&1\end{array} \right]\)
Panua mstari wa 2\(−3\) na uongeze mstari wa 1.
\(\left[ \begin{array}{ccc|ccc}1&0&-\dfrac{1}{5}&-\dfrac{11}{5}&-3&0 \nonumber \\[4pt] 0&1&\dfrac{19}{5}&\dfrac{4}{5}&1&0 \nonumber \\[4pt] 0&0&\dfrac{1}{5}&\dfrac{1}{5}&0&1\end{array} \right]\)
Kuzidisha mstari 3 na\(5\).
\(\left[ \begin{array}{ccc|ccc}1&0&-\dfrac{1}{5}&-\dfrac{11}{5}&-3&0 \nonumber \\[4pt] 0&1&\dfrac{19}{5}&\dfrac{4}{5}&1&0 \nonumber \\[4pt] 0&0&1&1&0&5\end{array} \right]\)
Panua mstari wa 3\(\dfrac{1}{5}\) na uongeze mstari wa 1.
\(\left[ \begin{array}{ccc|ccc}1&0&0&-2&-3&1 \nonumber \\[4pt] 0&1&\dfrac{19}{5}&\dfrac{4}{5}&1&0 \nonumber \\[4pt] 0&0&1&1&0&5\end{array} \right]\)
Panua mstari wa 3\(−\dfrac{19}{5}\) na uongeze mstari wa 2.
\(\left[ \begin{array}{ccc|ccc}1&0&0&-2&-3&1 \nonumber \\[4pt] 0&1&0&-3&1&-19 \nonumber \\[4pt] 0&0&1&1&0&5\end{array} \right]\)
Hivyo,
\(A^{−1}=\begin{bmatrix}−2&−3&1 \nonumber \\[4pt] −3&1&−19 \nonumber \\[4pt] 1&0&5\end{bmatrix}\)
Kuzidisha pande zote mbili za equation na\(A^{−1}\). Tunataka\(A^{−1}AX=A^{−1}B\):
\(\begin{bmatrix}−2&−3&1 \nonumber \\[4pt] −3&1&−19 \nonumber \\[4pt] 1&0&5\end{bmatrix}\begin{bmatrix}5&15&56 \nonumber \\[4pt] −4&−11&−41 \nonumber \\[4pt] −1&−3&−11\end{bmatrix}\begin{bmatrix}x \nonumber \\[4pt] y \nonumber \\[4pt] z\end{bmatrix}=\begin{bmatrix}−2&−3&1 \nonumber \\[4pt] −3&1&−19 \nonumber \\[4pt] 1&0&5\end{bmatrix}\begin{bmatrix}35 \nonumber \\[4pt] −26 \nonumber \\[4pt] −7\end{bmatrix}\)
Hivyo,
\(A^{−1}B=\begin{bmatrix}−70+78−7 \nonumber \\[4pt] −105−26+133 \nonumber \\[4pt] 35+0−35\end{bmatrix}=\begin{bmatrix}1 \nonumber \\[4pt] 2 \nonumber \\[4pt] 0\end{bmatrix}\)
Suluhisho ni\((1,2,0)\).
Tatua mfumo kwa kutumia inverse ya tumbo la mgawo.
\[\begin{align*} 2x-17y+11z&= 0\\ -x+11y-7z&= 8\\ 3y-2z&= -2 \end{align*}\]
- Jibu
-
\(X=\begin{bmatrix}4 \nonumber \\[4pt] 38 \nonumber \\[4pt] 58\end{bmatrix}\)
- Hifadhi tumbo la mgawo na tumbo la mara kwa mara kama vigezo vya tumbo\([ A ]\) na\([ B ]\).
- Ingiza kuzidisha ndani ya calculator, wito up kila variable Matrix kama inahitajika.
- Ikiwa tumbo la mgawo hauwezi kuingizwa, calculator itawasilisha tumbo la suluhisho; ikiwa tumbo la mgawo hauwezi kuingizwa, calculator itawasilisha ujumbe wa kosa.
Tatua mfumo wa equations na inverses ya tumbo kwa kutumia calculator
\[\begin{align*} 2x+3y+z&= 32\\ 3x+3y+z&= -27\\ 2x+4y+z&= -2 \end{align*}\]
Suluhisho
Kwenye ukurasa wa tumbo wa calculator, ingiza tumbo la mgawo kama variable ya tumbo\([ A ]\), na uingie tumbo la mara kwa mara kama kutofautiana kwa tumbo\([ B ]\).
\([A]=\begin{bmatrix}2&3&1 \nonumber \\[4pt] 3&3&1 \nonumber \\[4pt] 2&4&1\end{bmatrix}\),\([B]=\begin{bmatrix}32 \nonumber \\[4pt] −27 \nonumber \\[4pt] −2\end{bmatrix}\)
Kwenye skrini ya nyumbani ya calculator, funga katika kuzidisha ili kutatua\(X\), wito juu ya kila variable ya tumbo kama inahitajika.
\([A]^{−1}×[B]\)
Tathmini maneno.
\(\begin{bmatrix}−59 \nonumber \\[4pt] −34 \nonumber \\[4pt] 252\end{bmatrix}\)
Fikia rasilimali hizi za mtandaoni kwa maelekezo ya ziada na mazoezi na mifumo ya kutatua na inverses.
Mlinganyo muhimu
| Matrix ya utambulisho kwa\(2 × 2\) tumbo | \(I_2=\begin{bmatrix}1&0 \nonumber \\[4pt] 0&1\end{bmatrix}\) |
| Matrix ya utambulisho kwa\(3 × 3\) tumbo | \(I_3=\begin{bmatrix}1&0&0 \nonumber \\[4pt] 0&1&0 \nonumber \\[4pt] 0&0&1\end{bmatrix}\) |
| Inverse ya kuzidisha ya\(2 × 2\) tumbo | \(A^{−1}=\dfrac{1}{ad−bc}\begin{bmatrix}d&−b \nonumber \\[4pt] −c&a\end{bmatrix}\), wapi\(ad−bc≠0\) |
Dhana muhimu
- Matrix ya utambulisho ina mali\(AI=IA=A\). Angalia Mfano\(\PageIndex{1}\).
- Matrix invertible ina mali\(AA^{−1}=A^{−1}A=I\). Angalia Mfano\(\PageIndex{2}\).
- Tumia kuzidisha tumbo na utambulisho ili kupata inverse ya\(2×2\) tumbo. Angalia Mfano\(\PageIndex{3}\).
- Inverse ya kuzidisha inaweza kupatikana kwa kutumia formula. Angalia Mfano\(\PageIndex{4}\).
- Njia nyingine ya kutafuta inverse ni kwa kuongeza na utambulisho. Angalia Mfano\(\PageIndex{5}\).
- Tunaweza kuongeza\(3×3\) tumbo na utambulisho juu ya haki na kutumia shughuli mfululizo kurejea tumbo awali katika utambulisho, na tumbo upande wa kulia inakuwa inverse. Angalia Mfano\(\PageIndex{6}\).
- Andika mfumo wa equations kama\(AX=B\), na kuzidisha pande zote mbili kwa inverse ya\(A\):\(A^{−1}AX=A^{−1}B\). Angalia Mfano\(\PageIndex{7}\) na Mfano\(\PageIndex{8}\).
- Tunaweza pia kutumia calculator kutatua mfumo wa equations na inverses ya tumbo. Angalia Mfano\(\PageIndex{9}\).