Whakaoti i te {l}{A+B+1=0}{A-2B=3}

$$\left. \begin{array} { l } { A + B + 1 = 0 } \\ { A - 2 B = 3 } \end{array} \right.$$

Whakaoti mō A, B

$$A=\frac{1}{3}\approx 0.333333333$$

$$B = -\frac{4}{3} = -1\frac{1}{3} \approx -1.333333333$$

Ngā Upane e Whakamahi ana i te Whakakapinga

Hei whakaoti i ētahi whārite takirua mā te whakakapinga, me whakaoti tētahi whārite i te tuatahi mō tētahi o ngā taurangi. Ka whakakapi i te otinga mō taua taurangi ki tērā o ngā whārite.

$$A+B+1=0,A-2B=3$$

Kōwhiria tētahi o ngā whārite ka whakaotia mō te A mā te wehe i te A i te taha mauī o te tohu ōrite.

$$A+B+1=0$$

Me tango 1 mai i ngā taha e rua o te whārite.

$$A+B=-1$$

Me tango B mai i ngā taha e rua o te whārite.

$$A=-B-1$$

Whakakapia te -B-1 mō te A ki tērā atu whārite, A-2B=3.

$$-B-1-2B=3$$

Tāpiri -B ki te -2B.

$$-3B-1=3$$

Me tāpiri 1 ki ngā taha e rua o te whārite.

$$-3B=4$$

Whakawehea ngā taha e rua ki te -3.

$$B=-\frac{4}{3}$$

Whakaurua te -\frac{4}{3} mō B ki A=-B-1. I te mea kotahi anake te taurangi kei te whārite i puta, ka taea e koe te whakaoti mō A hāngai tonu.

$$A=-\left(-\frac{4}{3}\right)-1$$

Whakareatia -1 ki te -\frac{4}{3}.

$$A=\frac{4}{3}-1$$

Tāpiri -1 ki te \frac{4}{3}.

$$A=\frac{1}{3}$$

Kua oti te pūnaha te whakatau.

$$A=\frac{1}{3},B=-\frac{4}{3}$$

Ngā Upane e Whakamahi ana i Ngā Poukapa

Tuhia ngā whārite ki te tānga ngahuru ka whakamahi i ngā poukapa hei whakaoti i te pūnaha o ngā whārite.

$$A+B+1=0,A-2B=3$$

Tuhia ngā whārite ki te tikanga tātai poukapa.

$$\left(\begin{matrix}1&1\\1&-2\end{matrix}\right)\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}-1\\3\end{matrix}\right)$$

Whakarea mauī i te whārite ki te poukapa kōaro o \left(\begin{matrix}1&1\\1&-2\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\1&-2\end{matrix}\right))\left(\begin{matrix}1&1\\1&-2\end{matrix}\right)\left(\begin{matrix}A\\B\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\1&-2\end{matrix}\right))\left(\begin{matrix}-1\\3\end{matrix}\right)$$

Ko te hua o tētahi poukapa me te kōaro ko te poukapa tuakiri.

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}A\\B\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\1&-2\end{matrix}\right))\left(\begin{matrix}-1\\3\end{matrix}\right)$$

Whakareatia ngā poukapa kei te taha mauī o te tohu ōrite.

$$\left(\begin{matrix}A\\B\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\1&-2\end{matrix}\right))\left(\begin{matrix}-1\\3\end{matrix}\right)$$

Mō te poukapa 2\times 2 \left(\begin{matrix}a&b\\c&d\end{matrix}\right), ko te \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right) te poukapa kōaro, nō reira ka taea te tuhi anō te whārite poukapa hei rapanga whakarea poukapa.

$$\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{-2-1}&-\frac{1}{-2-1}\\-\frac{1}{-2-1}&\frac{1}{-2-1}\end{matrix}\right)\left(\begin{matrix}-1\\3\end{matrix}\right)$$

Mahia ngā tātaitanga.

$$\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}\frac{2}{3}&\frac{1}{3}\\\frac{1}{3}&-\frac{1}{3}\end{matrix}\right)\left(\begin{matrix}-1\\3\end{matrix}\right)$$

Whakareatia ngā poukapa.

$$\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}\frac{2}{3}\left(-1\right)+\frac{1}{3}\times 3\\\frac{1}{3}\left(-1\right)-\frac{1}{3}\times 3\end{matrix}\right)$$

Mahia ngā tātaitanga.

$$\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}\frac{1}{3}\\-\frac{4}{3}\end{matrix}\right)$$

Tangohia ngā huānga poukapa A me B.

$$A=\frac{1}{3},B=-\frac{4}{3}$$

Ngā Upane e Whakamahi i te Tangohanga

Hei whakaoti mā te tangohanga, ko ngā tau whakarea o tētahi o ngā taurangi me mātua ōrite i ngā whārite e rua kia whakakorehia ai te taurangi ina tangohia tētahi whārite mai i tētahi atu.

$$A+B+1=0,A-2B=3$$

Me tango A-2B=3 mai i A+B+1=0 mā te tango i ngā kīanga tau ōrite i ia taha o te tohu ōrite.

$$A-A+B+2B+1=-3$$

Tāpiri A ki te -A. Ka whakakore atu ngā kupu A me -A, ka toe he whārite me tētahi taurangi kotahi ka taea te whakaoti.

$$B+2B+1=-3$$

Tāpiri B ki te 2B.

$$3B+1=-3$$

Me tango 1 mai i ngā taha e rua o te whārite.

$$3B=-4$$

Whakawehea ngā taha e rua ki te 3.

$$B=-\frac{4}{3}$$

Whakaurua te -\frac{4}{3} mō B ki A-2B=3. I te mea kotahi anake te taurangi kei te whārite i puta, ka taea e koe te whakaoti mō A hāngai tonu.

$$A-2\left(-\frac{4}{3}\right)=3$$

Whakareatia -2 ki te -\frac{4}{3}.

$$A+\frac{8}{3}=3$$

Me tango \frac{8}{3} mai i ngā taha e rua o te whārite.

$$A=\frac{1}{3}$$

Kua oti te pūnaha te whakatau.

$$A=\frac{1}{3},B=-\frac{4}{3}$$