Pembahasan Soal UM UGM Matematika Dasar tahun 2013 nomor 16 sampai 20

Nomor 16

Jika

$A = \left( \begin{matrix} 2 & 1 \\ -1 & 0 \end{matrix} \right) , \, B = \left( \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right) \,$ dan

$I \,$ matriks identitas, maka

$AB^{-1} + BA^{-1} = ....$

$\spadesuit \,$ Konsep Matriks
invers :

$A = \left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \rightarrow A^{-1} = \frac{1}{a.d-b.c} \left( \begin{matrix} d & -b \\ -c & a \end{matrix} \right)$

$\spadesuit \,$ Menentukan invers dan hasil

$AB^{-1} + BA^{-1}$

$\begin{align} A & = \left( \begin{matrix} 2 & 1 \\ -1 & 0 \end{matrix} \right) \rightarrow A^{-1} = \frac{1}{0.2 - (-1).1} \left( \begin{matrix} 0 & -1 \\ 1 & 2 \end{matrix} \right) \\ A & = \left( \begin{matrix} 0 & -1 \\ 1 & 2 \end{matrix} \right) \\ B & = \left( \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right) \rightarrow B^{-1} = \frac{1}{1.1 - 0.(-1)} \left( \begin{matrix} 1 & -1 \\ 0 & 1 \end{matrix} \right) \\ B & = \left( \begin{matrix} 1 & -1 \\ 0 & 1 \end{matrix} \right) \\ AB^{-1} + BA^{-1} & = \left( \begin{matrix} 2 & 1 \\ -1 & 0 \end{matrix} \right).\left( \begin{matrix} 1 & -1 \\ 0 & 1 \end{matrix} \right) + \left( \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right). \left( \begin{matrix} 0 & -1 \\ 1 & 2 \end{matrix} \right) \\ AB^{-1} + BA^{-1} & = \left( \begin{matrix} 2 & -1 \\ -1 & 1 \end{matrix} \right) + \left( \begin{matrix} 1 & 1 \\ 1 & 2 \end{matrix} \right) \\ AB^{-1} + BA^{-1} & = \left( \begin{matrix} 3 & 0 \\ 0 & 3 \end{matrix} \right) = 3 \left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right) = 3 I \end{align}$
Jadi, nilai

$AB^{-1} + BA^{-1} = 3I . \heartsuit$

Nomor 17

Jika

$x \,$ dan

$y \,$ memenuhi sistem persamaan

$\begin{align} \frac{2}{x-1} - \frac{1}{y+2} & = 10 \\ \frac{3}{y+2} + \frac{1}{x-1} & = -9 \end{align}$
maka

$x + y = ....$

$\clubsuit \,$ Misalkan

$a = \frac{1}{x-1} \, \,$ dan

$\, b = \frac{1}{y+2}$
Sehingga persamaannya menjadi :

$\frac{2}{x-1} - \frac{1}{y+2} = 10 \rightarrow 2a - b = 10 \,$ ....pers(i)

$\frac{3}{y+2} + \frac{1}{x-1} = -9 \rightarrow 3b + a = -9 \,$ ....pers(ii)

$\clubsuit \,$ Menentukan nilai

$a \,$ dan

$b$

$\begin{array}{c|c|cc} 2a - b = 10 & \text{kali 1} & 2a - b = 10 & \\ 3b + a = -9 & \text{kali 1} & 6b + 2a = -18 & - \\ \hline & & -7b = 28 & \\ & & b = -4 & \end{array}$
Pers(i) :

$2a - b = 10 \rightarrow 2a - (-4) = 10 \rightarrow a = 3$

$\clubsuit \,$ Menentukan nilai

$x \,$ dan

$y$

$\begin{align} a=3 \rightarrow a & = \frac{1}{x-1} \\ 3 & = \frac{1}{x-1} \\ x-1 & = \frac{1}{3} \\ x & = \frac{1}{3} + 1 = \frac{4}{3} \\ b=-4 \rightarrow b & = \frac{1}{y+2} \\ -4 & = \frac{1}{y+2} \\ y+2 & = -\frac{1}{4} \\ y & = - \frac{1}{4} - 2 = -\frac{9}{4} \end{align}$
Sehingga nilai :

$x + y = \frac{4}{3} + (-\frac{9}{4}) = \frac{16 - 27}{12} = -\frac{11}{12}$
Jadi, nilai

$x + y = -\frac{11}{12} . \heartsuit$

Nomor 18

Jika (

$b+c, \, b, \, c$ ) memenuhi sistem persamaan

$\begin{align} 3x-y+2z & = -1 \\ -2x+y+3z & = -3 \end{align}$
maka

$b+ c = ....$

$\spadesuit \,$ Bentuk (

$b+c, \, b, \, c$ ) , artinya

$x = b+c, \, y = b, \, z = c$

$\spadesuit \,$ Substitusi ke sistem persamaan
persamaan pertama :

$\begin{align} 3x-y+2z = -1 \rightarrow 3(b+c)-b+2c & = -1 \\ 2b + 5c & = -1 \, \, \, \text{...peris(i)} \end{align}$
persamaan kedua :

$\begin{align} -2x+y+3z = -3 \rightarrow -2(b+c)+b+3c & = -3 \\ -b + c & = -3 \, \, \, \text{...peris(ii)} \end{align}$

$\spadesuit \,$ Eliminasi pers(i) dan pers(ii)

$\begin{array}{c|c|cc} 2b + 5c = -1 & \text{(kali 1)} & 2b + 5c = -1 & \\ -b + c = -3 & \text{(kali 2)} & -2b + 2c & = -36 & + \\ \hline & & 7c = -7 & \\ & & c = -1 & \end{array}$
Pers(ii) :

$-b + c = -3 \rightarrow -b + c(-1) = -3 \rightarrow b = 2$
Sehingga nilai

$b + c = 2 + (-1) = 1$
Jadi, nilai

$b + c = 1 . \heartsuit$

Nomor 19

Jika

$m > 0 , \,$ maka himpunan semua penyelesaian pertidaksamaan

$\sqrt{m^2 - x^2} \leq x \,$ adalah ....

$\clubsuit \,$ Syarat bentuk akar :
*). Dalam akar positif :

$\sqrt{m^2 - x^2}$

$m^2 - x^2 \geq 0 \rightarrow (m-x)(m+x) \geq 0 \rightarrow x = m \vee x = -m$

HP1 =

$\{ -m \leq x \leq m \}$
*). Karena

$x \geq \sqrt{m^2 - x^2} \,$ yang mana

$\sqrt{m^2 - x^2} \geq 0 \,$ , maka haruslah nilai

$x \geq 0 \,$ (nilai

$x \,$ juga positif). Sehingga HP2 =

$\{ x \geq 0 \}$

$\clubsuit \,$ Kuadratkan pertidaksamaannya

$\begin{align} \sqrt{m^2 - x^2} & \leq x \\ (\sqrt{m^2 - x^2})^2 & \leq x^2 \\ m^2 - x^2 & \leq x^2 \\ m^2 - 2x^2 & \leq 0 \\ m^2 - 2x^2 & = 0 \rightarrow 2x^2 = m^2 \rightarrow x = \pm \sqrt{\frac{m^2}{2}} = \pm \frac{m}{\sqrt{2}} \end{align}$

HP3 =

$\{ x \leq -\frac{m}{\sqrt{2}} \vee x \geq \frac{m}{\sqrt{2}} \}$
Sehingga solusi totalnya :
HP =

$HP1 \cap HP2 \cap HP3 = \{ \frac{m}{\sqrt{2}} \leq x \leq m \}$
Jadi, penyelesaiannya adalah

$HP = \{ \frac{m}{\sqrt{2}} \leq x \leq m \} . \heartsuit$

Nomor 20

Semua nilai

$x \,$ yang memenuhi

$\frac{3\sqrt{2-x}}{x-1} < 2 \,$ adalah ....

$\spadesuit \,$ Syarat bentuk akar dan pecahan :

$\frac{3\sqrt{2-x}}{x-1}$
*).

$2-x \geq 0 \rightarrow -x \geq -2 \rightarrow x \leq 2 \,$
*). Penyebut :

$x - 1 \neq 0 \rightarrow x \neq 1$
dari kedua syarat, diperoleh : HP1 =

$\{ x \leq 2 , \, \, x \neq 1 \}$

$\spadesuit \,$ Menyelesaiakan pertidaksamaan :

$\begin{align} \frac{3\sqrt{2-x}}{x-1} & < 2 \\ \frac{3\sqrt{2-x}}{x-1} -2 & < 0 \\ \frac{3\sqrt{2-x} - 2(x-1) }{x-1} & < 0 \\ \text{akar-akarnya : } & \\ x-1 = 0 \vee & \, \, \, 3\sqrt{2-x} - 2(x-1) = 0 \\ x-1=0 \rightarrow x & = 1 \\ 3\sqrt{2-x} - 2(x-1) & = 0 \\ 3\sqrt{2-x} & = 2(x-1) \, \, \, \, \text{(kuadratkan)} \\ 9(2-x) & = 4(x^2-2x+1) \\ 4x^2 + x - 14 & = 0 \\ (4x-7)(x+2) & = 0 \\ x = \frac{7}{4} \vee x = -2 \end{align}$
untuk

$x = -2 \,$ tidak memenuhi karena