tolong di bantu yaa. beserta caranya juga . terima kasih sebelumnya

No. 18

Diketahui:

[tex]A=\left[\begin{array}{ccc} \frac{3}{7} & \frac{2}{7} & a\\ b & \frac{3}{7} & \frac{2}{7}\\ \frac{2}{7} & \frac{6}{7} & c \end{array}\right][/tex]

Karena A orthogonal maka berlaku [tex]A^t=A^{-1}[/tex] sedemikian hingga [tex]AA^t=I[/tex], yaitu:

[tex]\left[ \begin {array}{ccc} 3/7&2/7&a\\ \noalign{\medskip}b&3/7&2/7 \\ \noalign{\medskip}2/7&6/7&c\end {array} \right] \left[ \begin {array}{ccc} 3/7&b&2/7\\ \noalign{\medskip}2/7&3/7&6/7 \\ \noalign{\medskip}a&2/7&c\end {array} \right]=\left[ \begin {array}{ccc} 1&0&0\\ \noalign{\medskip}0&1&0 \\ \noalign{\medskip}0&0&1\end {array} \right] [/tex]

[tex]AA^t=\left[ \begin {array}{ccc} {\frac{13}{49}}+{a}^{2}&3/7\,b+{\frac{6}{ 49}}+2/7\,a&{\frac{18}{49}}+ac\\ \noalign{\medskip}3/7\,b+{\frac{6}{49 }}+2/7\,a&{b}^{2}+{\frac{13}{49}}&2/7\,b+{\frac{18}{49}}+2/7\,c \\ \noalign{\medskip}{\frac{18}{49}}+ac&2/7\,b+{\frac{18}{49}}+2/7\,c& {\frac{40}{49}}+{c}^{2}\end {array} \right][/tex]

Slanjutnya diperoleh:

[tex] \frac{13}{49}+a^2=1 \\ b^2+ \frac{13}{49}=1 \\ \frac{40}{49}+c^2=1 [/tex]

dari persamaan-persamaan diatas diperoleh: [tex]a= \frac{6}{7},\;b= \frac{6}{7} ,\;c= \frac{3}{7} [/tex], sehingga

[tex]a^2+b^2+c^2=\dfrac{81}{49}[/tex]

Soal No 19.

diketahui A adalah matriks yang orthogonal, dengan

[tex]A=\left[ \begin {array}{cc} a&-b\\ \noalign{\medskip}b&a\end {array}\right],\quad \text{dan}\quad A^t= \left[ \begin {array}{cc} a&b\\ \noalign{\medskip}-b&a\end {array}\right] [/tex]
Karena A orthogonal maka berlaku [tex]AA^t=I[/tex] yaitu

[tex]\left[ \begin {array}{cc} {a}^{2}+{b}^{2}&0\\ \noalign{\medskip}0&{a}^{2}+{b}^{2}\end {array} \right]=\left[ \begin {array}{cc} 1&0\\ \noalign{\medskip}0&1\end {array} \right] [/tex]

Maka A akan orthogonal jika [tex]a^2+b^2=1[/tex]

Soal No. 23

[tex]A=\left[ \begin {array}{ccc} -1&-1&0\\ \noalign{\medskip}-1&1&2 \end {array} \right]\quad\text{dan}\quad B=\left[ \begin {array}{cc} -1&x\\ \noalign{\medskip}1&y \\ \noalign{\medskip}0&z\end {array} \right] [/tex]

[tex]AB=\left[ \begin {array}{cc} 0&-x-y\\ \noalign{\medskip}2&-x+y+2\,z \end {array} \right] [/tex]

[tex](AB)^{-1}=\left[ \begin {array}{cc} -{\frac {x-y-2\,z}{2(x+y)}}& \frac{1}{2} \\ \noalign{\medskip}- \frac{1}{x+y}&0\end {array} \right] [/tex]

Karena [tex](AB)^{-1}=\left[ \begin {array}{cc} -1&1/2\\ \noalign{\medskip}1/2&0 \end {array} \right][/tex] maka diperoleh:

[tex]-{\frac {x-y-2\,z}{2(x+y)}}=-1 \qquad \text{dan}\quad - \frac{1}{x+y} = \frac{1}{2}[/tex]

dengan menyelesaikan sistem diatas diperoleh [tex]z-x=3[/tex]