Bagaimana cara no 15 & 16

[tex]15.\\ \\ \left[\begin{array}{cc}2+a&2a+b\\3&b+2c\\\end{array}\right] = \left[\begin{array}{cc}4&6\\3&8\\\end{array}\right]\\ \\ 2+a=4 \rightarrow a=2 \\ \\2a+b=6 \rightarrow 2(2)+b=6 \rightarrow b=2 \\ \\ b+2c=8 \rightarrow 2+2c=8 \rightarrow c=3 \\ \\ a=2\\ b=2\\ c=3[/tex]

[tex]16. \\ \\ misalkan \ C = \left[\begin{array}{cc}a&b\\c&d\\\end{array}\right] \\ \\ \left[\begin{array}{cc}-1&2\\1&-3\\\end{array}\right] \times \left[\begin{array}{cc}a&b\\c&d\\\end{array}\right] = \left[\begin{array}{cc}-4&-1\\1&3\\\end{array}\right] \\ \\ \left[\begin{array}{cc}-a+2c&-b+2d\\a-3c&b-3d\\\end{array}\right] = \left[\begin{array}{cc}-4&-1\\1&3\\\end{array}\right] \\ \\ [/tex]

[tex] \\ (1)+(3) \rightarrow (-a+2c)+(a-3c)=-4+1 \rightarrow c=3 \\ substitusi \ c=3 \ ke \ (1) \ atau \ (3) \\ -a+2(3)=-4 \rightarrow a=10 \\ \\ (2)+(4) \rightarrow (-b+2d)+(b-3d)=-1+3 \rightarrow d=-2 \\ substitusi \ d=-2 \ ke \ (2) \ atau \ (4) \\ -b+2(-2)=-1 \rightarrow b=-3 \\ jadi \ C = \left[\begin{array}{cc}10&-3\\3&-2\\\end{array}\right] [/tex]