Contoh Soal 6 Matriks

$\begin{array}{l}\\ 26.& (\mathbf{\text{UM UGM 2004}})\\ & \text{Nilai-nilai}x\text{agar matriks}\\ & \left(\begin{array}{cc}5x& 5\\ 4& x\end{array}\right)\\ & \text{tidak memiliki invers adalah}....\\ & \begin{array}{l}\\ \text{a}.& 4\text{atau}5\\ \text{b}.& -2\text{atau}2\\ \text{c}.& -4\text{atau}5\\ \text{d}.& -6\text{atau}4\\ \text{e}.& 0\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{supaya matriks}\left(\begin{array}{cc}5x& 5\\ 4& x\end{array}\right)\\ & \text{tidak memiliki invers},\text{maka}\\ & \text{determinan matriks}\left(\begin{array}{cc}5x& 5\\ 4& x\end{array}\right)=0\\ & \text{Sehingga}\\ & \left|\begin{array}{cc}5x& 5\\ 4& x\end{array}\right|=0\\ & \iff 5x^2-20=0\\ & \iff x^2=4\\ & \iff x=\pm 2\end{array}\end{array}$

$\begin{array}{l}\\ 27.& (\mathbf{\text{UM UGM 2005}})\\ & \text{Matriks}\left(\begin{array}{cc}x& 1\\ -2& 1-x\end{array}\right)\\ & \text{tidak memiliki invers untuk}\\ & \text{nilai}x=....\\ & \begin{array}{l}\\ \text{a}.& -1\text{atau}-2\\ \text{b}.& -1\text{atau}0\\ \text{c}.& -1\text{atau}1\\ \text{d}.& -1\text{atau}2\\ \text{e}.& 1\text{atau}2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \text{Mirip dengan pembahasan no. 26}\\ & \begin{array}{rl}& \text{Nilai}\left|\begin{array}{cc}x& 1\\ -2& 1-x\end{array}\right|=0\\ & \iff x-x^2-(-2)=0\\ & \iff 2+x-x^2=0\\ & \iff x^2-x-2=0\\ & \iff (x-2)(x+1)=0\\ & \iff x=2\text{atau}x=-1\end{array}\end{array}$

$\begin{array}{l}\\ 28.& (\mathbf{\text{Mat Das SIMAK UI 2014}})\\ & \text{Jika matriks}\text{A}\text{adalah invers}\\ & \text{dari matriks}{\displaystyle \frac{1}{3}.\left(\begin{array}{cc}-1& -3\\ 4& 5\end{array}\right)\text{dan}}\\ & \text{A}\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}1\\ 3\end{array}\right)\text{maka nilai}2x+y\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -{\displaystyle \frac{10}{3}}\\ \text{b}.& -{\displaystyle \frac{1}{3}}\\ \text{c}.& 1\\ \text{d}.& {\displaystyle \frac{9}{7}}\\ \text{e}.& {\displaystyle \frac{20}{3}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Misalkan diketahui matriks}\\ & \text{B}={\displaystyle \frac{1}{3}.\left(\begin{array}{cc}-1& -3\\ 4& 5\end{array}\right),}\\ & \text{maka}\text{A}={\left({\displaystyle \frac{1}{3}.\left(\begin{array}{cc}-1& -3\\ 4& 5\end{array}\right)}\right)}^{-1}\\ & \text{selanjutnya}\text{A}\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}1\\ 3\end{array}\right)\\ & \left(\begin{array}{c}x\\ y\end{array}\right)=A^{-1}\left(\begin{array}{c}1\\ 3\end{array}\right),\\ & \text{ingat bahwa}{\left({\mathbf{\text{A}}}^{-1}\right)}^{-1}=\mathbf{\text{A}}\\ & \left(\begin{array}{c}x\\ y\end{array}\right)={\left({\left({\displaystyle \frac{1}{3}.\left(\begin{array}{cc}-1& -3\\ 4& 5\end{array}\right)}\right)}^{-1}\right)}^{-1}\left(\begin{array}{c}1\\ 3\end{array}\right)\\ & \left(\begin{array}{c}x\\ y\end{array}\right)={\displaystyle \frac{1}{3}.\left(\begin{array}{cc}-1& -3\\ 4& 5\end{array}\right)\left(\begin{array}{c}1\\ 3\end{array}\right)}\\ & \left(\begin{array}{c}x\\ y\end{array}\right)={\displaystyle \frac{1}{3}\left(\begin{array}{c}-1-9\\ 4+15\end{array}\right)=\left(\begin{array}{c}{\displaystyle -\frac{10}{3}}\\ {\displaystyle \frac{19}{3}}\end{array}\right)}\\ & 2x+y=2(-{\displaystyle \frac{10}{3}})+\frac{19}{3}\\ & ={\displaystyle \frac{-20+19}{3}=-{\displaystyle \frac{1}{3}}}\end{array}\end{array}$