MATEMATIKA 006 jawablah dengan baik dan benar jawb nomor 1 -4.... Question from @Radar511

December 2018 1 136 Report

MATEMATIKA 006
jawablah dengan baik dan benar
jawb nomor 1 -4

nurhoisam 1. $\left[\begin{array}{ccc}-1&3\\2&1\end{array}\right]$ $\left[\begin{array}{ccc}a\\4\end{array}\right]$ = $\left[\begin{array}{ccc}9\\10\end{array}\right]$
$\left[\begin{array}{ccc}(-1.a)+(3.4)\\(2.a)+(1.4)\end{array}\right]$ = $\left[\begin{array}{ccc}9\\10\end{array}\right]$
$\left[\begin{array}{ccc}-a+12\\2a+4\end{array}\right]$ = $\left[\begin{array}{ccc}9\\10\end{array}\right]$
//mencari nilai a :
----- pers 1
-a + 12 = 9
-a = 9 - 12
-a = -3 maka
a = 3
----- pers 2
2a + 4 = 10
2a = 10 - 4
2a = 6
a = 3
// nilai a-1 = 3 - 1 = 2 (C)

2. $\left[\begin{array}{ccc}5&3&-1\\-2&0&4\\2&1&3\end{array}\right]$ $\left[\begin{array}{ccc}1\\a\\b\end{array}\right]$ = $\left[\begin{array}{ccc}a\\6b\\-4\end{array}\right]$
$\left[\begin{array}{ccc}(5.1)+(3.a)+(-1.b)\\(-2.1)+(0.a)+(4.a)\\(2.1)+(1.a)+(3.b)\end{array}\right]$ = $\left[\begin{array}{ccc}a\\6b\\-4\end{array}\right]$
$\left[\begin{array}{ccc}5+3a-b\\-2+4a\\2+a+3b\end{array}\right] = \left[\begin{array}{ccc}a\\6b\\-4\end{array}\right]$
// menjadi 3 persamaan
1) 5+3a-b=a ⇒pindah ruas
3a-a - b = -5
2a - b = -5 ---------- pers.1
2) -2+4a=6b ⇒ pindah ruas
4a - 6b = 2 --------- pers.2
3) 2+a+3b = -4 ⇒pindah ruas
a + 3b = -4-2
a + 3b = -6 --------- pers 3
// selesaikan dengan dua persamaan. Misalnya ambil persamaan 1 dan 3..
2a - b = -5 | ×1 |
a + 3b = -6 | ×2 | -------- agar nilai a hilang dan mendapat nilai b
maka, menjadi :
2a - b = -5
2a + 6b = -12
____________ (-)
-7b = 7
b = -1
// maka nilai a dapat dicari. Misalnya ambil persamaan 1
2a - b = -5
2a - (-1) = -5
2a + 1 = -5
2a = -5 - 1
2a = -6
a = -3
∴ nilai a + b = -3 + (-1) = -4 (B)

3. $\left[\begin{array}{ccc}3&5\\1&2\end{array}\right] \left[\begin{array}{ccc}a&0\\a+b&c+2\end{array}\right]$ = $\left[\begin{array}{ccc}1&-5\\0&-2\end{array}\right]$
$\left[\begin{array}{ccc}(3.0) + (5.(a+b))&(3.c) + (5.(c+2))\\(1.0) + (2.(a+b))&(1.c)+(2.(c+2))\end{array}\right]$ = $\left[\begin{array}{ccc}1&-5\\0&-2\end{array}\right]$
$\left[\begin{array}{ccc}3a+5a+5b&5c+10\\a+2a+2b&2c+4\end{array}\right] = \left[\begin{array}{ccc}1&-5\\0&-2\end{array}\right]$
$\left[\begin{array}{ccc}8a+5b&8c+10\\3a+3b&3c+4\end{array}\right] = \left[\begin{array}{ccc}1&-5\\0&-2\end{array}\right]$
// menjadi 4 persamaan
1) 8a + 5b = 1 ------------ pers.1
2) 5c+10 = -5 ------------- pers.2
3) 3a+2b = 0 -------------- pers.3
4) 2c + 4 = -2 ------------- pers.4
// mencari nilai a dan b dengan persamaan 1 dan 3
8a + 5b = 1 | ×2 |
3a + 2b = 0 | ×5 |
menjadi :
16a + 10b = 2
15a + 10b = 0
_____________ (-)
a = 2
// mencari nilai b
3a + 2b = 0
3(2) + 2b = 0
6 + 2b = 0
2b = -6
b = -3

// mencari nilai c dengan persamaan 2 atau 4. Misal, pilih persamaan 4
2c + 4 = -2
2c = -2 - 4
2c = -6
c = -3

// nilai 2a + b - c = 2(2) + (-3) - (-3)
= 4 - 3 + 3
= 4 (E)

4. S = 2P - Q + R
S = $\left[\begin{array}{ccc}2&2\\4&6\end{array}\right] - \left[\begin{array}{ccc}-3&7\\2&1\end{array}\right] + \left[\begin{array}{ccc}0&1\\2&-1\end{array}\right]$
S = $\left[\begin{array}{ccc}2+3+0&2-7+1\\4-2+2&6-1-1\end{array}\right]$
S = $\left[\begin{array}{ccc}5&-4\\4&4\end{array}\right]$
det.S = ad - bc
det.S = 5.4 - 4.(-4)
= 20 - (-16)
= 20 + 16
= 36 (D)

Semoga membantu :)

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