Tolong bagaimana cara mengerjakannya?

Penjumlahan Matriks 2 × 2 ⇒ A + B
[tex] \left[\begin{array}{cc}a&b\\c&d\end{array}\right] + \left[\begin{array}{cc}p&q\\r&s\end{array}\right] = \left[\begin{array}{cc}a+p&b+q\\c+r&d+s\end{array}\right] \\ \\ \\ \left[\begin{array}{cc}-2c&4\\2&5\end{array}\right] + \left[\begin{array}{cc}-4&-4\\(-b-5)&a\end{array}\right] = \left[\begin{array}{cc}(-2c-4)&(4-4)\\(2-b-5)&(5+a)\end{array}\right] \\ \\ \\ \left[\begin{array}{c}A\end{array}\right]+ \left[\begin{array}{c}B\end{array}\right]= \left[\begin{array}{cc}(-2c-4)&0\\(-b-3)&(5+a)\end{array}\right][/tex]

Perkalian Matriks 2 × 2 ⇒ C × D
[tex] \left[\begin{array}{cc}a&b\\c&d\end{array}\right] . \left[\begin{array}{cc}p&q\\r&s\end{array}\right] = \left[\begin{array}{cc}a.p+b.r&a.q+b.s\\c.p+d.r&c.q+d.s\end{array}\right] \\ \\ \\ \left[\begin{array}{cc}-1&3\\0&2\end{array}\right]. \left[\begin{array}{cc}4&1\\-2&3\end{array}\right] = \left[\begin{array}{cc}(-1.4)+(3.-2)&(-1.1)+(3.3)\\(0.4)+(2.-2)&(0.1)+(2.3)\end{array}\right] \\ \\ \\ \left[\begin{array}{c}C\end{array}\right] . \left[\begin{array}{c}D\end{array}\right] = \left[\begin{array}{cc}(-4)+(-6)&(-1)+(9)\\(0)+(-4)&(0)+(6)\end{array}\right] \\ \\ \\ \left[\begin{array}{c}C\end{array}\right] . \left[\begin{array}{c}D\end{array}\right] = \left[\begin{array}{cc}-10&8\\-4&6\end{array}\right] [/tex]

A + B = C.D
[tex] \left[\begin{array}{cc}(-2c-4)&0\\(b-3)&(5+a)\end{array}\right] = \left[\begin{array}{cc}-10&8\\-4&6\end{array}\right] \\ \\ \\ \left[\begin{array}{cc}(-2c-4)&0\\(-b-3)&(5+a)\end{array}\right] - \left[\begin{array}{cc}-10&8\\-4&6\end{array}\right] = 0 [/tex]

[tex]-2c-4-(-10)=0 \\ -2c-4+10=0 \\ -2c+6=0 \\ -2c=-6 \\ c= \frac{-6}{-2} \\ c=3 \\ \\ -b-3-(-4)=0 \\ -b-3+4=0 \\ -b+1=0 \\ -b=-1 \\ b=1 \\ \\ 5+a-6=0 \\ a-1=0 \\ a=1 \\ \\ a+b+c=1+1+3=5 [/tex]

Semoga asumsi saya untuk posisi variabel "a" sudah sesuai. Mohon maaf jika ada salah.