Bdw bagaimna caranya in soal. Ad yg tau ngak . Tolong bantu aku dong Plissssss ,, d.kunpul ini hari loh

1.) (a)
[tex] \left[\begin{array}{ccc}a+2&4\\-3&1\end{array}\right] = \left[\begin{array}{ccc}6&4\\-3&2b-1\end{array}\right] \\ a+2 = 6 \\ a = 4 \\ \\ 2b-1=1\\ 2b=2 \\ b=1[/tex]

(b)
[tex] \left[\begin{array}{ccc}a-b\\a+b\end{array}\right] = \left[\begin{array}{ccc}6\\4\end{array}\right] \\ \\ a-b=6 \\ a=6+b \\ \\ a+b = 4 \\ (6+b)+b=4 \\ 6+2b = 4 \\ 2b = -2 \\ b= -1 \\ \\ a = 6+(-1) \\ a = 5[/tex]

2.)
[tex]A = \left[\begin{array}{ccc}2&3\\4&6\end{array}\right] \\ B = \left[\begin{array}{ccc}3&7\\1&5\end{array}\right] \\ \\[/tex]

(a)
[tex]A+B = \left[\begin{array}{ccc}2&3\\4&6\end{array}\right] +\left[\begin{array}{ccc}3&7\\1&5\end{array}\right] \\ A+B = \left[\begin{array}{ccc}5&10\\5&11\end{array}\right] [/tex]

(b)
[tex]B-A = \left[\begin{array}{ccc}3&7\\1&5\end{array}\right] - \left[\begin{array}{ccc}2&3\\4&6\end{array}\right] \\ B - A = \left[\begin{array}{ccc}1&4\\-3&-1\end{array}\right] [/tex]

(c)
[tex]A \times B = \left[\begin{array}{ccc}2&3\\4&6\end{array}\right] \times \left[\begin{array}{ccc}3&7\\1&5\end{array}\right] \\ A \times B = \left[\begin{array}{ccc}(2\times 3 + 3 \times 1)&(2\times7+3\times5)\\(4\times3+6\times1)&(4\times7+6\times5)\end{array}\right] \\ A \times B = \left[\begin{array}{ccc}9&29\\18&58\end{array}\right] [/tex]

3.)
[tex]A = \left[\begin{array}{ccc}3&4\\2&5\end{array}\right] [/tex]
Tentukan invers matriks A

[tex]A^{-1} = \frac{1}{|A|} \times adj(A) \\ A^{-1} = \frac{1}{3\times5-4\times2}\times \left[\begin{array}{ccc}5&-4\\-2&3\end{array}\right] \\ A^{-1} = \frac{1}{-8} \left[\begin{array}{ccc}5&-4\\-2&3\end{array}\right][/tex]

4.)
Tentukan penyelesaian [tex] \left \{ {{3x+4y=11} \atop {2x-5y=15}} \right. [/tex] dengan matriks.

[tex]D = \left[\begin{array}{ccc}3&4\\2&-5\end{array}\right] = 3\times(-5)-4\times2 = -23 \\ D_x = \left[\begin{array}{ccc}11&4\\15&-5\end{array}\right] = 11\times(-5)-15\times4 = -115\\ D_y = \left[\begin{array}{ccc}3&11\\2&15\end{array}\right] = 3\times15 - 2\times11 = 23 \\ \\ x = \frac{D_{x}}{D} \\ x = \frac{-115}{-23} \\ x = 5 \\ \\ y = \frac{D_y}{D} \\ y = \frac{23}{-23} \\ y = -1[/tex]

5.)
Tentukan determinan dari matriks [tex] \left[\begin{array}{ccc}3&7&4\\1&5&6\\2&8&9\end{array}\right] [/tex]

[tex]det(A) = | \left[\begin{array}{ccc}3&7&4\\1&5&6\\2&8&9\end{array}\right] | \\ = (3)(5)(9) + (7)(6)(2) + (4)(1)(8) - ((2)(5)(4) + (8)(6)(3) + (9)(1)(7)) \\ = (135) + (84) + (32) - ((40)+(144) + (63)) \\ = (251) - (247) \\ = 4[/tex]

Mohon diteliti kembali.
Karena keterbatasan kemampuan, untuk bagian yang menuliskan determinan, garisnya bukan kurung siku, melainkan garis lurus di antara elemen matriksnya.