Tolong Bantu Bray
Matriks A = 1 -2
2k+1 4
B= 5 3
-3 -2
C= 4 -1
3k² 3
Tentukan nilai K yang memenuhi persamaan 2A + B^{-1} = C
Matriks A = 1 -2
2k+1 4
B= 5 3
-3 -2
C= 4 -1
3k² 3
Tentukan nilai K yang memenuhi persamaan 2A + B^{-1} = C
Takamori37
![\displaystyle A=\left[\begin{array}{cc}1&-2\\2k+1&4\end{array}\right] \\ \\ B=\left[\begin{array}{cc}5&3\\-3&-2\end{array}\right] \\ \\ C=\left[\begin{array}{cc}4&-1\\3k^2&3\end{array}\right] \\ \\ 2A+B^{-1}=C \\ 2\left[\begin{array}{cc}1&-2\\2k+1&4\end{array}\right]+\frac{1}{-10+9}\left[\begin{array}{cc}-2&-3\\3&5\end{array}\right]=\left[\begin{array}{cc}4&-1\\3k^2&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cdisplaystyle+A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26-2%5C%5C2k%2B1%264%5Cend%7Barray%7D%5Cright%5D+%5C%5C+%5C%5C+B%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D5%263%5C%5C-3%26-2%5Cend%7Barray%7D%5Cright%5D+%5C%5C+%5C%5C+C%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26-1%5C%5C3k%5E2%263%5Cend%7Barray%7D%5Cright%5D+%5C%5C+%5C%5C+2A%2BB%5E%7B-1%7D%3DC+%5C%5C+2%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26-2%5C%5C2k%2B1%264%5Cend%7Barray%7D%5Cright%5D%2B%5Cfrac%7B1%7D%7B-10%2B9%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%26-3%5C%5C3%265%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26-1%5C%5C3k%5E2%263%5Cend%7Barray%7D%5Cright%5D "\displaystyle A=\left[\begin{array}{cc}1&-2\\2k+1&4\end{array}\right] \\ \\ B=\left[\begin{array}{cc}5&3\\-3&-2\end{array}\right] \\ \\ C=\left[\begin{array}{cc}4&-1\\3k^2&3\end{array}\right] \\ \\ 2A+B^{-1}=C \\ 2\left[\begin{array}{cc}1&-2\\2k+1&4\end{array}\right]+\frac{1}{-10+9}\left[\begin{array}{cc}-2&-3\\3&5\end{array}\right]=\left[\begin{array}{cc}4&-1\\3k^2&3\end{array}\right]")
![\left[\begin{array}{cc}2&-4\\4k+2&8\end{array}\right]-1\left[\begin{array}{cc}-2&-3\\3&5\end{array}\right]=\left[\begin{array}{cc}4&-1\\3k^2&3\end{array}\right] \\ \\ \left[\begin{array}{cc}4&-1\\4k-1&3\end{array}\right]=\left[\begin{array}{cc}4&-1\\3k^2&3\end{array}\right] \\ \\ $Didapat:$ \\ 4k-1=3k^2 \\ 3k^2-4k+1=0 \\ (3k-1)(k-1)=0 \\ k=\frac{1}{3}$ dan $k=1](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%26-4%5C%5C4k%2B2%268%5Cend%7Barray%7D%5Cright%5D-1%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%26-3%5C%5C3%265%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26-1%5C%5C3k%5E2%263%5Cend%7Barray%7D%5Cright%5D+%5C%5C+%5C%5C+%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26-1%5C%5C4k-1%263%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26-1%5C%5C3k%5E2%263%5Cend%7Barray%7D%5Cright%5D+%5C%5C+%5C%5C+%24Didapat%3A%24+%5C%5C+4k-1%3D3k%5E2+%5C%5C+3k%5E2-4k%2B1%3D0+%5C%5C+%283k-1%29%28k-1%29%3D0+%5C%5C+k%3D%5Cfrac%7B1%7D%7B3%7D%24+dan+%24k%3D1 "\left[\begin{array}{cc}2&-4\\4k+2&8\end{array}\right]-1\left[\begin{array}{cc}-2&-3\\3&5\end{array}\right]=\left[\begin{array}{cc}4&-1\\3k^2&3\end{array}\right] \\ \\ \left[\begin{array}{cc}4&-1\\4k-1&3\end{array}\right]=\left[\begin{array}{cc}4&-1\\3k^2&3\end{array}\right] \\ \\ $Didapat:$ \\ 4k-1=3k^2 \\ 3k^2-4k+1=0 \\ (3k-1)(k-1)=0 \\ k=\frac{1}{3}$ dan $k=1")
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