Contoh Soal 3 Matriks

$\begin{array}{l}\\ 11.& \text{Diketahui matriks}\\ & \text{A}=\left(\begin{array}{cc}{}^a\log 6& 2\\ 1& a+3b\end{array}\right),\\ & \text{B}=\left(\begin{array}{cc}0& -5\\ -6& 3a-5b\end{array}\right),\text{dan}\\ & \text{C}=\left(\begin{array}{cc}{}^a\log 2& -{\displaystyle \frac{1}{2}}\\ -2(b+c)& 3\end{array}\right),\\ & \text{serta}\text{I}\text{adalah matriks identitas}.\\ & \text{Jika}2\text{A}+\text{B}-2\text{C}=2\text{I},\\ & \text{maka nilai}4a+b+c\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 1\\ \text{b}.& 5\\ \text{c}.& 7\\ \text{d}.& 11\\ \text{e}.& 13\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \begin{array}{rl}& 2\text{A}+\text{B}-2\text{C}=2\text{I}\\ & 2\left(\begin{array}{cc}{}^a\log 6& 2\\ 1& a+3b\end{array}\right)+\left(\begin{array}{cc}0& -5\\ -6& 3a-5b\end{array}\right)\\ & -2\left(\begin{array}{cc}{}^a\log 2& -{\displaystyle \frac{1}{2}}\\ -2(b+c)& 3\end{array}\right)=2\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\\ & \left(\begin{array}{cc}2.{}^a\log 6-2.{}^2\log 2& 2.2-5-2(-{\displaystyle \frac{1}{2}})\\ 2.1-6-2(-2(b+c))& 2(a+3b)+3a-5b-2.3\end{array}\right)=\left(\begin{array}{cc}2& 0\\ 0& 2\end{array}\right)\\ & \{\begin{array}{ll}2& =2.{}^a\log 6-2.{}^2\log 2\\ 0& =2.1-6-2(-2(b+c))\\ 2& =2(a+3b)+3a-5b-2.3\end{array}\\ & \begin{array}{|cc|}\hline \text{dari persamaan}(1)& \text{dari persamaan}(2)\\ \begin{array}{rl}2.{}^a\log 6-2.{}^2\log 2& =2\\ {}^a\log 6^2-{}^2\log 2^2& =2\\ {}^a\log {\displaystyle \frac{6^2}{2^2}}& =2\\ {}^a\log 9& =2\\ 9& =a^2\\ 3& =a\\ 12& =4a\end{array}& \begin{array}{rl}2.1-6-2(-2(b+c))& =0\\ 2-6+4(b+c)& =0\\ 4(b+c)& =4\\ b+c& =1\\ & \\ \text{sehingga diperoleh}& ,\\ 4a+b+c=12+1& \\ =13& \end{array}\\ \hline\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 12.& \text{Jika}\left(\begin{array}{cc}-4x& 2y\\ y& x\end{array}\right)\left(\begin{array}{c}2\\ -3\end{array}\right)=\left(\begin{array}{c}2\\ -12\end{array}\right),\\ & \text{maka nilai}xy=....\\ & \begin{array}{l}\\ \text{a}.& -6\\ \text{b}.& -3\\ \text{c}.& 2\\ \text{d}.& 3\\ \text{e}.& 6\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{|cc|}\hline \begin{array}{rl}\left(\begin{array}{cc}-4x& 2y\\ y& x\end{array}\right)\left(\begin{array}{c}2\\ -3\end{array}\right)& =\left(\begin{array}{c}2\\ -12\end{array}\right)\\ \left(\begin{array}{c}-4x.2+2y.-3\\ y.2+x.-3\end{array}\right)& =\left(\begin{array}{c}2\\ -12\end{array}\right)\\ \left(\begin{array}{c}-8x-6y\\ -3x+2y\end{array}\right)& =\left(\begin{array}{c}2\\ -12\end{array}\right)\\ \\ \mathbf{\text{SPLDV}}& \end{array}& \begin{array}{rl}-8x-6y& =2(\times 1)\\ -3x+2y& =-12(\times 3)\\ \text{menjadi}& \\ -8x-6y& =2\\ -9x+6y& =-36{}_{+}\\ ----& ---\\ -17x& =-34\\ x& =2\end{array}\\ \begin{array}{rl}-8x-6y& =2\\ -8(2)-6y& =2\\ -16-6y& =2\\ -6y& =2+16\\ -6y& =18\\ y& =-3\\ \text{sehingga}& \\ xy& =2.(-3)=-6\end{array}& \\ \hline\end{array}\end{array}$

$\begin{array}{l}\\ 13.& \text{Diketahui}\text{N}=\left(\begin{array}{cc}-2& 3\\ -1& 4\end{array}\right)\\ & \text{dan}\text{M}=\left(\begin{array}{cc}-1& 3\\ -1& 5\end{array}\right).\\ & \text{Jika}{\text{N}}^2=p\text{N}-q\text{M},\\ & :\text{maka nilai}p-q=....\\ & \begin{array}{l}\\ \text{a}.& 2\\ \text{b}.& 3\\ \text{c}.& 4\\ \text{d}.& 5\\ \text{e}.& 6\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}& {\text{N}}^2=p\text{N}-q\text{M}\\ & \left(\begin{array}{cc}-2& 3\\ -1& 4\end{array}\right)\times \left(\begin{array}{cc}-2& 3\\ -1& 4\end{array}\right)=p\left(\begin{array}{cc}-2& 3\\ -1& 4\end{array}\right)-q\left(\begin{array}{cc}-1& 3\\ -1& 5\end{array}\right)\\ & \left(\begin{array}{cc}-2.-2+3.-1& -2.3+3.4\\ -1.-2+4.-1& -1.3+4.4\end{array}\right)=\left(\begin{array}{cc}-2p+q& 3p-3q\\ -p+q& 4p-5q\end{array}\right)\\ & \left(\begin{array}{cc}4-3& -6+12\\ 2-4& -3+16\end{array}\right)=\left(\begin{array}{cc}-2p+q& 3p-3q\\ -p+q& 4p-5q\end{array}\right)\\ & \left(\begin{array}{cc}1& 6\\ -2& 13\end{array}\right)=\left(\begin{array}{cc}-2p+q& 3p-3q\\ -p+q& 4p-5q\end{array}\right)\\ \\ & \begin{array}{|cc|}\hline \begin{array}{rl}-2p+q& =1\\ -p+q& =-2{}_{-}\\ ----& ---\\ -p& =3\\ p& =-3\\ & \\ & \end{array}& \begin{array}{rl}-p+q& =-2\\ -(-3)+q& =-2\\ q& =-2-3\\ q& =-5\\ \text{sehingga}& \text{didapatkan}\\ p-q& =-3-(-5)\\ & =2\end{array}\\ \hline\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 14.& \text{Diketahui matriks}\text{Z}=\left(\begin{array}{cc}-2& 6\\ -3& 5\end{array}\right)\\ & \text{dan}f(x)=x^2-x.\\ & \text{Jika}f(\text{Z})=\left(\begin{array}{cc}-3p-8q& 12\\ -6& -2(p+q)\end{array}\right),\\ & \text{maka nilai}{p}^2-q^2=....\\ & \begin{array}{l}\\ \text{a}.& 5\\ \text{b}.& 7\\ \text{c}.& 9\\ \text{d}.& 12\\ \text{e}.& 15\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}f(\text{Z})& =\left(\begin{array}{cc}-3p-8q& 12\\ -6& -2(p+q)\end{array}\right)\\ {\text{Z}}^2-\text{Z}& =\left(\begin{array}{cc}-3p-8q& 12\\ -6& -2(p+q)\end{array}\right)\\ \left(\begin{array}{cc}-2& 6\\ -3& 5\end{array}\right)\times \left(\begin{array}{cc}-2& 6\\ -3& 5\end{array}\right)-\left(\begin{array}{cc}-2& 6\\ -3& 5\end{array}\right)& =\left(\begin{array}{cc}-3p-8q& 12\\ -6& -2(p+q)\end{array}\right)\\ \left(\begin{array}{cc}4-18& -12+30\\ 6-15& -18+25\end{array}\right)-\left(\begin{array}{cc}-2& 6\\ -3& 5\end{array}\right)& =\left(\begin{array}{cc}-3p-8q& 12\\ -6& -2(p+q)\end{array}\right)\\ \left(\begin{array}{cc}-12& 12\\ -6& 2\end{array}\right)& =\left(\begin{array}{cc}-3p-8q& 12\\ -6& -2(p+q)\end{array}\right)\end{array}\\ & \begin{array}{rl}\text{Sehingga}& \\ -12& =-3p-8q.................(1)\\ -1& =p+q......................(2)\\ \text{persamaan}& (2)\text{ke persamaan}(1)\\ -12& =-3p-3q-5q=-3(p+q)-5q\\ -12& =-3(-1)-5q\\ -12& =3-5q\\ 5q& =3+12\\ q& =3........................(3)\\ \text{persamaan}& (3)\text{ke persamaan}(2)\\ p+q& =-1\\ p& =-1-q=-1-3=-4\\ p^2-q^2& =(-4{)}^2-3^2=16-9\\ & =7\end{array}\end{array}$

$\begin{array}{l}\\ 15.& (\mathbf{\text{SBMPTN 2013}})\\ & \text{Jika}A=\left(\begin{array}{ccc}2& -1& 1\\ a& b& c\end{array}\right),\\ & B=\left(\begin{array}{cc}-2& 1\\ 1& -1\\ 0& 2\end{array}\right)\text{dan}\\ & AB=\left(\begin{array}{cc}-5& 5\\ 3& -3\end{array}\right)\\ & \text{maka nilai}2c-a=....\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& 2\\ \text{c}.& 4\\ \text{d}.& 5\\ \text{e}.& 6\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}& AB=\left(\begin{array}{cc}-5& 5\\ 3& -3\end{array}\right)\\ & \left(\begin{array}{ccc}2& -1& 1\\ a& b& c\end{array}\right)\left(\begin{array}{cc}-2& 1\\ 1& -1\\ 0& 2\end{array}\right)=\left(\begin{array}{cc}-5& 5\\ 3& -3\end{array}\right)\\ & \left(\begin{array}{cc}-5& 5\\ -2a+b& a-b+2c\end{array}\right)=\left(\begin{array}{cc}-5& 5\\ 3& -3\end{array}\right)\\ & \begin{array}{l}\\ -2a+b& =3& \\ a-b+2c& =-3& +\\ 2c-a& =0\end{array}\end{array}\end{array}$

DAFTAR PUSTAKA

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