tolong yang no. 4 (a & b)
Matematika
Bachtiarbayuaziz
Pertanyaan
tolong yang no. 4 (a & b)
1 Jawaban
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1. Jawaban Rega32
FAVORITIN DONG :D
A.
*kalikan saja dulu yang ada 3 nya, hasilnya:
[tex]\left[\begin{array}{ccc}x&y\\-11&-1\\\end{array}\right] + \left[\begin{array}{ccc}3y&-6\\3&-9\\\end{array}\right] = \left[\begin{array}{ccc}-2&5\\1&-3\\\end{array}\right]\left[\begin{array}{ccc}1&2\\3&4\\\end{array}\right][/tex]
*kemudian yang ruas kanan kalikan, hasilnya:
[tex]\left[\begin{array}{ccc}x&y\\-11&-1\\\end{array}\right] + \left[\begin{array}{ccc}3y&-6\\3&-9\\\end{array}\right] = \left[\begin{array}{ccc}-2(1)+5(3)&-2(2)+5(4)\\1(1)+(-3)(3)&1(2)+(-3)(4)\\\end{array}\right][/tex]
*Cari x atau y, disini yang paling mudah mencari y terlebih dahulu, maka:
[tex]y+(-6)=16 [/tex]
[tex]y=16+6 [/tex]
[tex]y=22 [/tex]
*kemudian cari x:
[tex]x+3y=13 [/tex]
[tex]x+3(22)=13 [/tex]
[tex]x+66=13 [/tex]
[tex]x=13-66 [/tex]
[tex]x=-53[/tex]
Soal B
*kalikan terlebih dahulu Matriks kirinya:
[tex] \left[\begin{array}{ccc}3(x)+7(y)\\-2(x)+1(y)\\\end{array}\right] = \left[\begin{array}{ccc}16\\-5\end{array}\right] [/tex]
hasilnya:
[tex] \left[\begin{array}{ccc}3x+7y\\(-2x)+y\\\end{array}\right] = \left[\begin{array}{ccc}16\\-5\end{array}\right][/tex]
*cari x atau y:
[tex]-2x+y=-5 [/tex]
[tex]y=2x-5[/tex]
*Setelah y didapat, masukan nilai y di bagian atas:
[tex]3x+7(2x-5)=16 [/tex]
[tex]3x+14x-35=16 [/tex]
[tex]17x=16+35 [/tex]
[tex]17x=51 [/tex]
[tex]x= \frac{51}{17} [/tex]
[tex]x=3 [/tex]
*cari lagi nilai y
[tex]-2x+y=-5 [/tex]
[tex]-2(3)+y=-5 [/tex]
[tex]-6+y=-5 [/tex]
[tex]y=-5+6 [/tex]
[tex]y=1[/tex]