September 2017

Minggu, 24 September 2017

Simak UI 2011 Kode 211 Materi Eksponen

zulnining September 24, 2017 0 Comments

Jika solusi dari persamaan $\begin{array}{lcl} 5^{x+5} =7^{x} \end{array}$ dapat dinyatakan dalam bentuk $\begin{array}{lcl} x=\; ^{a}\textrm{log}\; 5^{5} \end{array}$, maka nilai $a$ adalah ....

$\begin{array}{lcl} A.\; \frac{5}{12}\\ B.\; \frac{5}{7}\\ C.\; \frac{7}{5}\\ D.\; \frac{12}{7}\\ E.\; \frac{12}{5}\\ \end{array}$

Pembahasan:
KonseR

$\begin{array}{lcl} \; \; \; \; \; \; \; \: a\: ^{f\left ( x \right )}=b\: ^{g\left ( x \right )}\\ f\left ( x \right )log\: a=g\left ( x \right )log\: b\\\\ log\: a+log\: b=log\: ab\\ log\: a-log\: b=log\: \frac{a}{b}\\ ^{a}\textrm{log}\: b=\frac{log\: b}{log\: a} \end{array}$

$\begin{array}{lcl} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; 5^{x+5}=7^{x}\\ \; \; \; \; \, \left ( x+5 \right )\: log\: 5=x\: log\: 7\\ x\: log\: 5+5\: log\: 5=x\: log\: 7\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; log\: 5=x\: log\: 7-x\: log\: 5\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \: log\: 5^{5}=x\left (log\: 7-\: log\: 5 \right )\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; log\: 5^{5}=x\left ( log\: \frac{7}{5} \right )\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; x=\frac{log\: 5^{5}}{log\: \frac{7}{5}}\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; =\: ^{\frac{7}{5}}\textrm{log}\: log\: 5^{5} \end{array}$
Nilai $\begin{array}{lcl} a=\frac{7}{5} \end{array}$

Jawaban ______________________________________ (C)

SNMPTN 2010 Kode 336 Materi Eksponen

zulnining September 24, 2017 0 Comments

Jika $n$ memenuhi $\begin{array}{lcl} \underbrace{25^{0,25}\; x\; 25^{0,25}\; x\; 25^{0,25}\; x\; ....x\; 25^{0,25}}=125\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; n\; faktor \end{array}$ maka $(n-3)(n+2)$ = ....
A. 24
B. 26
C. 28
D. 32
E. 36

Pembahasan:

$\begin{array}{lcl} \; \; \; \; \: \underbrace{25^{0,25}\; x\; 25^{0,25}\; x\; 25^{0,25}\; x\; ....x\; 25^{0,25}}=125\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; n\; faktor\\ \underbrace{\left (5^{2} \right )^{\frac{1}{4}}\; x\; \left (5^{2} \right )^{\frac{1}{4}}\; x\; \left (5^{2} \right )^{\frac{1}{4}}\; x\; ....x\; \left (5^{2} \right )^{\frac{1}{4}}}=125\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; n\; faktor\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \underbrace{5^{\frac{1}{2}}\; x\; 5^{\frac{1}{2}}\; x\; 5^{\frac{1}{2}}\; x\; ....x\; 5^{\frac{1}{2}}}=125\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; n\; faktor\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; 5_{\; \; \; \; \; \; \; \; \; \; \; \; n\; faktor}^{\underbrace{\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+....+\frac{1}{2}}}=125\\ \end{array}$
$\begin{array}{lcl} 5^{\frac{1}{2}n}= 5^{3}\\ \; \frac{1}{2}n = 3\\ \; \; \; n=6\\\\ \left ( n-3 \right )\left ( n+2 \right )=\left ( 6-3 \right )\left ( 6+2 \right )\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \: =\left ( 3 \right )\left ( 8 \right )\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; =24 \end{array}$

Jawaban ___________________________ (A)

SPMB 2006 Materi Eksponen

zulnining September 24, 2017 0 Comments

Jumlah semua nilai $x$yang memenuhi persamaan $9^{x^{2}-3x+1}+9^{x^{2}-3x}=20-10\left ( 3^{x^{2}-3x} \right )$ ....
A. 0
B. 1
C. 2
D. 3
E. 4

Pembahasan:

$\begin{array}{lcl} \; \; \; \; \; \; \; \; \; \; \; \; \; 9^{x^{2}-3x+1}+9^{x^{2}-3x}=20-10\left ( 3^{x^{2}-3x} \right )\\ \left ( 3^{2} \right )^{x^{2}-3x}\; \cdot \: 9^{1}+\left (3^{2} \right )^{x^{2}-3x}=20-10\left ( 3^{x^{2}-3x} \right )\\ \; \; 9\left ( 3^{x^{2}-3x} \right )^{2}+\left ( 3^{x^{2}-3x} \right )^{2}=20-10\left ( 3^{x^{2}-3x} \right )\\\\ Misal \; 3^{x^{2}-3x}=y\\\\ \; \; \; \; \; \; \; \; \; \; \; \; 9y^{2}+y^{2}=20-10y\\ 10y^{2}+10y-20=0\; \; \; \; \; \; \; \;\; \; \; \; \; \; \; \; \; (dibagi\; 10)\\ \; \; \; \; \; \; \; \; \; y^{2}+y-2=0\\ \; \; \left ( y+2 \right )(y-1)=0 \end{array}$ $\begin{array}{lcl} y=-2\; atau\; y=1\; \; \; \; \; \; \; \; \; (ambil \; y=1,\; y=-2\; TM\; tidak\; memenuhi)\\\\ 3^{x^{2}-3x}=1\\ 3^{x^{2}-3x}=3^{0}\\ x^{2}-3x=0\\ x\left ( x-3 \right )=0\\ x=0\; atau\; x=3\\\\ x_{1}=0\\ x_{2}=3\\\\ x_{1}+x_{2}=0+3\\ \; \; \; \; \; \; \; \; \; \; \; \;=3 \end{array}$

Jawaban __________________________ (D)

Jumat, 22 September 2017

Soal dan Pembahasan Eksponen

zulnining September 22, 2017 0 Comments

1. UM UGM 2005 Kode 821
Jika $\sqrt{0,3+\sqrt{0,08}}=\sqrt{a}+\sqrt{b}$, maka $\frac{1}{a}+\frac{1}{b}$ = ....
    A. 25
    B. 20
    C. 15
    D. 10
    E. 5

    Pembahasan:
    KonseR
$\begin{array}{lcl} \bigstar \; \sqrt{\left ( a+b \right )+\sqrt{a\cdot b}}=\sqrt{a}+\sqrt{b}\\\\ \; \; \; \; \; \; \; \; \; \; \; \;\; \; \; \; \; \; \; \; \; \sqrt{\left ( 0,3 \right )+\sqrt{0,08}}=\sqrt{a}+\sqrt{b}\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \: \sqrt{\left ( 0,3 \right )+\sqrt{4}\cdot \sqrt{0,02}}=\sqrt{a}+\sqrt{b}\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \sqrt{\left ( 0,3 \right )+2\sqrt{0,02}}=\sqrt{a}+\sqrt{b}\\ \sqrt{\left ( 0,2+0,1 \right )+2\sqrt{\left (0,2 \right )\left ( 0,1 \right )}}=\sqrt{a}+\sqrt{b}\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \sqrt{0,2}+\sqrt{0,1}=\sqrt{a}+\sqrt{b} \end{array}$
   $\begin{array}{lcl} \sqrt{a}&=&\sqrt{0,2}\\ a&=&0,2\\ &=&\frac{2}{10}\\ &=&\frac{1}{5}\\\\ \sqrt{b}&=&\sqrt{0,1}\\ b&=&0,1\\ &=&\frac{1}{10}\\\\ \end{array}$
   $\begin{array}{lcl} \frac{1}{a}+\frac{1}{b}&=&\frac{1}{\frac{1}{5}}+\frac{1}{\frac{1}{10}}\\ &=&5+10\\ &=&15 \end{array}$

   Jawaban ____________________________________ (C)

2. SPMB 2006 Kode320
    Diketahui$\begin{array}{lcl} 4^{x}=25\;\; \; dan \; 5^{y}=\frac{1}{8} \end{array}$. Bila $y$
    nyatakan dalam $x$, diperoleh $y$ = ....
    A. $\begin{array}{lcl} -\frac{3}{x} \end{array}$
B. $\begin{array}{lcl} -\frac{2}{x} \end{array}$
C. $\begin{array}{lcl} -\frac{x}{3} \end{array}$
D. $\begin{array}{lcl} -\frac{x}{2} \end{array}$
    E. $\begin{array}{lcl} -\frac{2x}{3} \end{array}$

Pembahasan:
$\begin{array}{lcl} 4^{x}&=&25\\ \left ( 2^{2} \right )^{x}&=&25\\ \left (2^{x} \right )^{2}&=&25\\ 2^{x}&=&5\\\\ 5^{y}&=&\frac{1}{8}\\ \left ( 2^{x} \right )^{y}&=&8^{-1}\\ 2^{xy}&=&\left ( 2^{3} \right )^{-1}\\ 2^{xy}&=&2^{-3}\\ xy&=&-3\\ y&=&-\frac{3}{x} \end{array}$

Jawaban ____________________________________ (A)

Kamis, 21 September 2017

Kumpulan Soal dan Pembahasan Matriks (Bagian 1)

zulnining September 21, 2017 0 Comments

1. SPMB 2004 Regional III
Jika matriks $\begin {array}{lcl} A=\begin{pmatrix} 2 \; &1 \\ -2\; &3 \end{pmatrix},\; B=\begin{pmatrix} a\\ 1 \end{pmatrix} \end {array}$, dan $\begin {array}{lcl} C=\begin{pmatrix} 11\\ 1-4b \end{pmatrix} \end {array}$ memenuhi $AB=C$, maka $|a-b|$
    = ....
    A. 2
B. 3
C. 4
D. 5
E. 6

Pembahasan:
$\begin {array}{lcl} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; AB=C\\ \; \; \; \; \; \; \; \; \begin{pmatrix} 2 \; &1 \\ -2 \; &3 \end{pmatrix}\begin{pmatrix} a\\ 1 \end{pmatrix}=\begin{pmatrix} 11\\ 1-4b \end{pmatrix}\\ \begin{pmatrix} \left ( 2 \right )\left ( a \right )+\left ( 1 \right )\left ( 1 \right )\\ \left ( -2 \right )\left ( a \right ) +\left ( 3 \right )\left ( 1 \right ) \end{pmatrix}=\begin{pmatrix} 11\\ 1-4b \end{pmatrix}\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \begin{pmatrix} 2a+1\\ -2a+3 \end{pmatrix}=\begin{pmatrix} 11\\ 1-4b \end{pmatrix}\\\\ 2a+1=11\\ \; \; \; \; \; \; 2a=10\\ \; \; \; \; \; \; \; \; a=5 \end {array}$
  $\begin {array}{lcl} \; \; \; -2a+3=1-4b\\ -2\left ( 5 \right )+3=1-4b\\ \; \; \; \; \; \; \; \; \; \; \; -7=1-4b\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; b=2\\\\ |a-b|=|5-2|\\ \; \; \; \;\; \; \; \; \; \; \: =3 \end {array}$

Jawaban ___________________________________ (B)

2. SPMB 2004 Regional III
    Transpos dari matriks $P$ adalah $P^{T}$. Jika matriks $\begin {array}{lcl} A=\begin{pmatrix} 3\; &7 \\ 1\: &2 \end{pmatrix},\; B=\begin{pmatrix} 4\; &1 \end{pmatrix},\; dan\; C=\begin{pmatrix} x\\ y \end{pmatrix} \end {array}$
    memenuhi $A^{-1}B^{T}=C$, maka $x+y$ = ....
A. -2
B. -1
C. 0
D. 1
E. 2

Pembahasan:
  $\begin {array}{lcl} A=\begin{pmatrix} 3\; &7 \\ 1\: &2 \end{pmatrix}\\ A^{-1}=\frac{1}{\left ( 3 \right )\left ( 2 \right )-\left ( 7 \right )\left ( 1 \right )}\begin{pmatrix} 2\; &-7 \\ -1 & 3 \end{pmatrix}\\ \; \; \; \; \; \; \; =-\begin{pmatrix} 2\; &-7 \\ -1\; &3 \end{pmatrix} \\ \; \; \; \; \; \; \; =\begin{pmatrix} -2\; &7 \\ 1\; &-3 \end{pmatrix}\\\\ \; \; B=\begin{pmatrix} 4\; &1 \end{pmatrix}\\ B^{T}=\begin{pmatrix} 4\\ 1 \end{pmatrix} \end {array}$
$\begin {array}{lcl} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; A^{-1}B^{T}=C\\ \; \; \; \; \; \begin{pmatrix} -2\; &7 \\ 1\; &-3 \end{pmatrix}\begin{pmatrix} 4\\ 1 \end{pmatrix}=\begin{pmatrix} x\\ y \end{pmatrix}\\ \begin{pmatrix} \left (-2 \right )\left ( 4 \right )+\left ( 7 \right )\left ( 1 \right )\\ \left (1 \right )\left ( 4 \right )+\left ( -3 \right )\left ( 1 \right ) \end{pmatrix}=\begin{pmatrix} x\\ y \end{pmatrix}\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;\begin{pmatrix} -1\\ 1 \end{pmatrix}=\begin{pmatrix} x\\ y \end{pmatrix}\\\\ x=-1\\ y=1\\\\ x+y=\left ( -1 \right )+1\\ \; \; \; \; \; \; \; \; \; =0 \end {array}$

    Jawaban ___________________________________ (C)

3. SPMB 2004 Regional I
Jika matriks $\begin {array}{lcl} A=\begin{pmatrix} 2x+1\; &3 \\ 6x-1\; & 5 \end{pmatrix} \end {array}$ tidak mempunyai invers, maka nilai $x$ adalah ....
    A. -2
    B. -1
    C. 0
    D. 1
    E. 2

    Pembahasan:
    Matriks $A$ tidak mempunyai invers, maka $detA=0$
  $\begin {array}{lcl} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \: \begin{vmatrix} 2x+1\; &3 \\ 6x-1\; & 5 \end{vmatrix}=0\\ \left ( 2x+1 \right )\left ( 5 \right )-\left ( 3 \right )\left ( 6x-1 \right )=0\\ \; \; \; \; \; \; \; \; \; \; \; \; 10x+5-18x+3=0\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; -8x+8=0\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; -8x=-8\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \: x=1 \end {array}$

Jawaban ___________________________________ (B)

4. SPMB 2004 Regional I
Jika matriks $\begin {array}{lcl} A=\begin{pmatrix} a\; &1-a \\ 0\; & 1 \end{pmatrix}\; dan\; A^{-1}=\begin{pmatrix} 2\; &b \\ 0\; & 1 \end{pmatrix} \end {array}$ maka nilai $b$ adalah ....
    A. -1
    B. $-\frac{1}{2}$
C. 0
D. $\frac{1}{2}$
E. 1

Pembahasan:
  $\begin {array}{lcl} \; \; \; \: A=\begin{pmatrix} a\; &1-a \\ 0\; & 1 \end{pmatrix}\\ A^{-1}=\frac{1}{\left ( a \right )\left ( 1 \right )-\left ( 1-a \right )\left ( 0 \right )} \begin{pmatrix} 1\; &-1+a \\ 0\; & a \end{pmatrix}\\ \; \; \; \; \; \; \; =\frac{1}{a} \begin{pmatrix} 1\; &-1+a \\ 0\; & a \end{pmatrix}\\ \; \; \; \; \; \; \; =\begin{pmatrix} \frac{1}{a}\; &\frac{-1+a}{a} \\ 0\; & 1 \end{pmatrix}\\\\ A^{-1}=\begin{pmatrix} 2\; &b \\ 0\; & 1 \end{pmatrix} \end {array}$
  $\begin {array}{lcl} \begin{pmatrix} \frac{1}{a}\; &\frac{-1+a}{a} \\ 0\; & 1 \end{pmatrix}=\begin{pmatrix} 2\; &b \\ 0\; & 1 \end{pmatrix}\\\\ \frac{1}{a}=2\\\\ \; \; \; \: \frac{-1+a}{a}=b\\ -\frac{1}{a}+1=b\\ -2+1=b\\ \; \; \; \; \; \; \; \; \; b=-1 \end {array}$

Jawaban ___________________________________ (A)

5. SPMB 2004 Regional I
Jika matriks $\begin {array}{lcl} A=\begin{pmatrix} a\; & 2\; &3 \\ 1\; & a\; &4\\ a\; & 2\; &5 \end{pmatrix} \end {array}$ tidak mempunyai invers, maka nilai $a$ adalah ....
A. -2 atau 2
B. $-\sqrt{2}$ atau $\sqrt{2}$
C. -1 atau 1
D. 2
E. $2\sqrt{2}$

Pembahasan:
Matriks A tidak mempunyai invers, maka
$\begin {array}{lcl} detA=0\\ \begin{vmatrix} a\; & 2\; &3 \\ 1\; & a\; &4\\ a\; & 2\; &5 \end{vmatrix}=0\\\\ \begin{vmatrix} a\; & 2\; &3 \\ 1\; & a\; &4\\ a\; & 2\; &5 \end{vmatrix}\begin{matrix} a\; &2 \\ 1\; &a \\ a\; &2 \end{matrix}=0\\ \left ( a \right )\left ( a \right )\left ( 5 \right )+\left ( 2 \right )\left ( 4 \right )\left ( a \right )+\left ( 3 \right )\left ( 1 \right )\left ( 2 \right )-\left ( a \right )\left ( a \right )\left ( 3 \right )-\left ( 2 \right )\left ( 4 \right )\left ( a \right )-\left ( 5 \right )\left ( 1 \right )\left ( 2 \right )=0\\ 5a^{2}+8a+6-3a^{2}-8a-10=0\\ 2a^{2}-4=0\\ \; \; a^{2}-2=0 \end {array}$
$\begin {array}{lcl} \; \; a^{2}-2=0\\ \; \; \; \; \; \; \; \; a^{2}=2\\ \; \; \; \; \; \; \; \; \; \: a=\pm \sqrt{2}\\\\ a=-\sqrt{2}\; atau \; a=\sqrt{2} \end {array}$

   Jawaban ___________________________________ (B)

Matriks-3 (Determinan & Invers Matriks berordo 3 x 3)

zulnining September 21, 2017 1 Comments

# Determinan Matriks Persegi Berordo 3 x 3 Menggunakan Aturan Sarrus
Misal $A=\begin{pmatrix} a_{11}\; &a_{12}\; &a_{13} \\ a_{21}\; &a_{22} \; &a_{23} \\ a_{31}\;&a_{32}\; & a_{33} \end{pmatrix}$
   $\begin{array}{lcl} detA=|A|=\begin{vmatrix} a_{11}\; &a_{12}\; &a_{13} \\ a_{21}\; &a_{22} \; &a_{23} \\ a_{31}\;&a_{32}\; & a_{33} \end{vmatrix}\begin{matrix} a_{11}\; &a_{12} \\ a_{21}\; & a_{22}\\ a_{31}\; & a_{32} \end{matrix}\\ \; \; \; \; \; \; \; \; =a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{31}a_{22}a_{13}-a_{32}a_{23}a_{11}-a_{33}a_{21}a_{12} \end{array}$

Contoh:
Hitunglah determinan matriks $\begin{array}{lcl} A=\begin{pmatrix} 1\; & 0\; &1 \\ -1\; & 3\; &0 \\ 1\; &0\; &2 \end{pmatrix} \end{array}$

Pembahasan:
$\begin{array}{lcl} detA=\begin{vmatrix} 1\; & 0\; &1 \\ -1\; & 3\; &0 \\ 1\; &0\; &2 \end{vmatrix}\begin{matrix} 1\; &0 \\ -1\; & 3\\ 1\; &0 \end{matrix}\\ \; \; \; \; \; \; \; \; =\left ( 1 \right )\left ( 3 \right )\left ( 2 \right )+\left ( 0 \right )\left ( 0 \right )\left ( 1 \right )+\left ( 1 \right )\left ( -1 \right )\left ( 0 \right )-\left ( 1 \right )\left ( 3 \right )\left ( 1 \right )-\left ( 0 \right )\left ( 0 \right )\left ( 1 \right )-\left ( 2 \right )\left ( -1 \right )\left ( 0 \right )\\ \; \; \; \; \; \; \; \; =6+0+0-3-0-0\\ \; \; \; \; \; \; \; \; =3 \end{array}$

# Determinan Matriks Berordo 3 x 3 Menggunakan Kofaktor
   Menentukan minor $(M)$ dan kofaktor $(C)$ dari matriks berordo 3
   Misal matriks $\begin{array}{lcl} A=\begin{pmatrix} a_{11}\; &a_{12}\; &a_{13} \\ a_{21}\; &a_{22}\; & a_{23}\\ a_{31}\; &a_{32}\; & a_{33} \end{pmatrix} \end{array}$
   Minor dan kofaktor dari matriks $A$
   1. $\begin{array}{lcl} M_{11}=\begin{vmatrix} a_{22 }\; & a_{23}\\ a_{32}\; & a_{33} \end{vmatrix}\\ C_{11}=\left ( -1 \right )^{1+1}M_{11}=\begin{vmatrix} a_{22 }\; & a_{23}\\ a_{32}\; & a_{33} \end{vmatrix}\\ \end{array}$

   2. $\begin{array}{lcl} M_{12}=\begin{vmatrix} a_{21 }\; & a_{23}\\ a_{31}\; & a_{33} \end{vmatrix}\\ C_{12}=\left ( -1 \right )^{1+2}M_{12}=-\begin{vmatrix} a_{21 }\; & a_{23}\\ a_{31}\; & a_{33} \end{vmatrix}\\ \end{array}$

   3. $\begin{array}{lcl} M_{13}=\begin{vmatrix} a_{21 }\; & a_{22}\\ a_{31}\; & a_{32} \end{vmatrix}\\ C_{13}=\left ( -1 \right )^{1+3}M_{13}=\begin{vmatrix} a_{21 }\; & a_{22}\\ a_{31}\; & a_{32} \end{vmatrix}\\ \end{array}$

   4. $\begin{array}{lcl} M_{21}=\begin{vmatrix} a_{12 }\; & a_{13}\\ a_{32}\; & a_{33} \end{vmatrix}\\ C_{21}=\left ( -1 \right )^{2+1}M_{21}=-\begin{vmatrix} a_{12 }\; & a_{13}\\ a_{32}\; & a_{33} \end{vmatrix}\\ \end{array}$

   5. $\begin{array}{lcl} M_{22}=\begin{vmatrix} a_{11 }\; & a_{13}\\ a_{31}\; & a_{33} \end{vmatrix}\\ C_{22}=\left ( -1 \right )^{2+2}M_{21}=\begin{vmatrix} a_{11 }\; & a_{13}\\ a_{31}\; & a_{33} \end{vmatrix}\\ \end{array}$

   6. $\begin{array}{lcl} M_{23}=\begin{vmatrix} a_{11 }\; & a_{12}\\ a_{31}\; & a_{32} \end{vmatrix}\\ C_{23}=\left ( -1 \right )^{2+3}M_{23}=-\begin{vmatrix} a_{11 }\; & a_{12}\\ a_{31}\; & a_{32} \end{vmatrix}\\ \end{array}$

   7. $\begin{array}{lcl} M_{31}=\begin{vmatrix} a_{12 }\; & a_{13}\\ a_{22}\; & a_{23} \end{vmatrix}\\ C_{31}=\left ( -1 \right )^{3+1}M_{31}=\begin{vmatrix} a_{12 }\; & a_{13}\\ a_{22}\; & a_{23} \end{vmatrix}\\ \end{array}$

   8. $\begin{array}{lcl} M_{32}=\begin{vmatrix} a_{11 }\; & a_{13}\\ a_{21}\; & a_{23} \end{vmatrix}\\ C_{32}=\left ( -1 \right )^{3+2}M_{32}=-\begin{vmatrix} a_{11 }\; & a_{13}\\ a_{21}\; & a_{23} \end{vmatrix}\\ \end{array}$

Determinan dari matriks $A$ (pilih salah satu saja)
   $\begin{array}{lcl} 1.\; detA=|A|=a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}\\ 2.\; detA=|A|=a_{21}C_{21}+a_{22}C_{22}+a_{23}C_{23}\\ 3.\; detA=|A|=a_{31}C_{31}+a_{32}C_{32}+a_{33}C_{33}\\ 4.\; detA=|A|=a_{11}C_{11}+a_{21}C_{21}+a_{31}C_{31}\\ 5.\; detA=|A|=a_{12}C_{12}+a_{22}C_{22}+a_{32}C_{32}\\ 6.\; detA=|A|=a_{13}C_{13}+a_{23}C_{23}+a_{33}C_{33}\\ \ \end{array}$

Contoh:
Hitunglah determinan matriks menggunakan kofaktor $\begin{array}{lcl} A=\begin{pmatrix} 1 \; & 0\; &1 \\ -1\; &3\; &0 \\ 1\; & 0\; &2 \end{pmatrix} \ \end{array}$
Pembahasan:
Untuk menghitung determinan matriks $A$ dengan menggunakan kofaktor,
kita menggunakan rumus 1 (bisa dipilih)
$\begin{array}{lcl} a_{11}=1\\ C_{11}=\begin{vmatrix} 3\; &0 \\ 0\; & 2 \end{vmatrix}\\ C_{11}=6\\\\ a_{12}=0\\ C_{12}=-\begin{vmatrix} -1\; &0 \\ 1\; & 2 \end{vmatrix}\\ C_{12}=2\\ \ \end{array}$
$\begin{array}{lcl} a_{13}=1\\ C_{13}=\begin{vmatrix} -1\; &3 \\ 1\; & 0 \end{vmatrix}\\ C_{13}=0-3\\ C_{13}=-3\\\\ detA=|A|=a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}\\ \; \; \; \; \; \; \; \; =\left ( 1 \right )\left ( 6 \right )+\left ( 0 \right )\left ( 2 \right )+\left ( 1 \right )\left ( -3 \right )\\ \; \; \; \; \; \; \; \; =6+0-3\\ \; \; \; \; \; \; \; \; =3 \ \end{array}$

# Invers Matriks Berordo 3 x 3
   Misal $A$ adalah matriks berordo 3 x 3, invers matriks $A$ adalah:
   $\begin{array} {lcl} A=\begin{pmatrix} a_{11}\; & a_{12}\; &a_{13} \\ a_{21}\; &a_{22} \; &a_{23} \\ a_{31}\; &a_{32} \; &a_{33} \end{pmatrix}\\\\ Matriks\; kofaktor\;A=\begin{pmatrix} C_{11}\; & C_{12}\; &C_{13} \\ C_{21}\; &C_{22} \; &C_{23} \\ C_{31}\; &C_{32} \; &C_{33} \end{pmatrix} \\\\ Adj\: A=\begin{pmatrix} C_{11}\; & C_{21}\; &C_{31} \\ C_{12}\; &C_{22} \; &C_{32} \\ C_{13}\; &C_{23} \; &C_{33} \end{pmatrix},\; adj\: A\; merupakan\; transpose \; dari\; matriks\; kofaktor \\\\ A^{-1}=\frac{1}{det\: A}\: adj\: A \end{array}$

Contoh:
Carilah invers dari matriks $\begin{array} {lcl} A=\begin{pmatrix} 1\; & 0\; &1 \\ -1\; &3 \; &0 \\ 1\; &0 \; &2 \end{pmatrix} \end{array}$

Pembahasan:
$\begin{array} {lcl} A=\begin{pmatrix} 1\; & 0\; &1 \\ -1\; &3 \; &0 \\ 1\; &0 \; &2 \end{pmatrix}\\ detA=\begin{vmatrix} 1\; & 0\; &1 \\ -1\; &3 \; &0 \\ 1\; &0 \; &2 \end{vmatrix}\begin{matrix} 1\; & 0\\ -1\; &3 \\ 1\; &0 \end{matrix}\\ \; \; \; \; \; \; \; \; =\left ( 1 \right )\left ( 3 \right )\left ( 2 \right )+\left ( 0 \right )\left ( 0 \right )\left ( 1 \right )+\left ( 1 \right )\left ( -1 \right )\left ( 0 \right )-\left ( 1 \right )\left ( 3 \right )\left ( 1 \right )-\left ( 0 \right )\left ( 0 \right )\left ( 1 \right )-\left ( 2 \right )\left ( -1 \right )\left ( 0 \right )\\ \; \; \; \; \; \; \; \; =6+0+0-3-0-0\\ \; \; \; \; \; \; \; \; =3 \end{array}$
$\begin{array} {lcl} C_{11}=\begin{vmatrix} 3\; & 0\\ 0\; &2 \end{vmatrix}=6\\ C_{12}=-\begin{vmatrix} -1\; & 0\\ 1\; &2 \end{vmatrix}=2\\ C_{13}=\begin{vmatrix} -1\; & 3\\ 1\; &0 \end{vmatrix}=-3\\ C_{21}=-\begin{vmatrix} 0\; & 0\\ 0\; &2 \end{vmatrix}=0\\ C_{22}=\begin{vmatrix} 1\; & 1\\ 1\; &2 \end{vmatrix}=1\\ C_{23}=-\begin{vmatrix} 1\; & 0\\ 1\; &0 \end{vmatrix}=0\\ \end{array}$
$\begin{array} {lcl} C_{31}=\begin{vmatrix} 0\; & 1\\ 3\; &0 \end{vmatrix}=-3\\ C_{32}=-\begin{vmatrix} 1\; & 1\\ -1\; &0 \end{vmatrix}=-1\\ C_{33}=\begin{vmatrix} 1\; & 0\\ -1\; &3 \end{vmatrix}=3\\ \end{array}$
  $\begin{array} {lcl} Adj\: A=\begin{pmatrix} C_{11}\; & C_{21}\; &C_{31} \\ C_{12}\; &C_{22} \; &C_{32} \\ C_{13}\; &C_{23} \; &C_{33} \end{pmatrix}\\\\ \; \; \; \; \; \; \; \; \; \; =\begin{pmatrix} 6\; & 0\; &-3 \\ 2\; &1 \; &-1 \\ -3\; &0 \; &3 \end{pmatrix}\\\\ A^{-1}=\frac{1}{detA}\, adjA\\ \; \; \; \; \; \; \; =\frac{1}{3}\begin{pmatrix} 6\; & 0\; &-3 \\ 2\; &1 \; &-1 \\ -3\; &0 \; &3 \end{pmatrix}\\ \end{array}$
$\begin{array} {lcl} \; \; \; \; \; \; \; =\begin{pmatrix} 2\; & 0\; &-1 \\ \frac{2}{3}\; &\frac{1}{3} \; &-\frac{1}{3} \\ -1\; &0 \; &1 \end{pmatrix}\\ \end{array}$

Rabu, 20 September 2017

Soal dan Pembahasan Matriks-2

zulnining September 20, 2017 0 Comments

5. SPMB 2003 Regional III
Diketahui matriks $\begin{array} {lcl} P=\begin{pmatrix} a\; & b\\ c\; & d\\ e\; & f \end{pmatrix},\; Q=\begin{pmatrix} u\; &v \\ w\; & z \end{pmatrix} \end {array}$, $P^{T}$ transpos dari $P$. Operasi yang dapat
    dilakukan pada $P$ dan $Q$ = ....
$\begin{array} {lcl} A.\; P+Q\; dan\; PQ\\ B.\; P^{T}Q\; dan\; QP\\ C.\; PQ\; dan\; Q^{-1}P\\ D.\; PQ\; dan \; QP\\ E.\; PQ\; dan\; QP^{T} \end {array}$

Pembahasan:
* Operasi penjumlahan dan pengurangan pada matriks syaratnya ordo kedua
   matriks itu sama
* Operasi perkalian pada matriks banyak kolom dari matriks pertama harus
   sama dengan banyaknya baris pada matriks kedua
$\begin{array} {lcl} A_{m\, x\, n}\cdot B_{n\, x\, p}=AB_{m\, x\, p} \end {array}$

$\begin{array} {lcl} P=\begin{pmatrix} a\; & b\\ c\; &d \\ e\; & f \end{pmatrix}\\ P_{3\, x\, 2}\\ P^{T}= \begin{pmatrix} a\; & c\; &e \\ b\; & d\; & f \end{pmatrix}\\ P^{T}_{2\, x\, 3}\\ Q=\begin{pmatrix} u\; & v\\ w\; & z \end{pmatrix}\\ Q_{2\, x\, 2} \end{array}$

$\begin{array} {lcl} P_{3\, x\, 2}\cdot Q_{2\, x\, 2} \end{array}$, (benar karena banyak kolom pada matriks pertama sama dengan banyaknya baris pada matriks kedua)
$\begin{array} {lcl} Q_{3\, x\, 2}\cdot P^{T}_{2\, x\, 3} \end{array}$, (benar karena banyaknya kolom pada baris pertama sama dengan banyaknya baris pada matriks kedua)

Jawaban _______________________________ (E)

6. SPMB 2004 Regional II
   Transpos dari matriks $P$ adalah $P^{T}$. Jika $\begin{array} {lcl} P=\begin{pmatrix} 2\; & 3\\ 5\; & 7 \end{pmatrix} \end{array}$ maka matriks $\begin{array} {lcl} \left ( P^{T} \right )^{-1} \end{array}$ adalah ....
   $\begin{array} {lcl} A.\; \begin{pmatrix} -7\; & 3\\ 5\; & -2 \end{pmatrix}\\ B.\; \begin{pmatrix} 5\; & -2\\ -7\; & 3 \end{pmatrix}\\ C.\; \begin{pmatrix} -7\; & 5\\ 3\; & -2 \end{pmatrix}\\ D.\; \begin{pmatrix} -5\; & 7\\ 2\; & -3 \end{pmatrix}\\ E.\; \begin{pmatrix} 2\; & -5\\ -3\; & 7 \end{pmatrix}\\ \end{array}$

Pembahasan:
$\begin{array} {lcl} P= \begin{pmatrix} 2\; & 3\\ 5\; & 7 \end{pmatrix}\\ P^{T}= \begin{pmatrix} 2\; & 5\\ 3\; & 7 \end{pmatrix}\\ \left ( P^{T} \right )^{-1}=\frac{1}{\left ( 2 \right )\left ( 7 \right )-\left ( 5 \right )\left ( 3 \right )}\begin{pmatrix} 7\; &-5 \\ -3& 2 \end{pmatrix}\\ \; \; \; \; \; \; \; \; \; \; \; \; =-1\begin{pmatrix} 7\; & -5\\ -3\; & 2 \end{pmatrix}\\ \; \; \; \; \; \; \; \; \; \; \; \; =\begin{pmatrix} -7\; & 5\\ 3\; &-2 \end{pmatrix} \end{array}$

Jawaban _______________________________ (C)

7. SPMB 2004 Regional III
Jika $\begin{array} {lcl} A= \begin{pmatrix} 1\; & 2\\ 3\; & 5 \end{pmatrix}\\ \end{array}$ dan $\begin{array} {lcl} A^{-1}B= \begin{pmatrix} -2\; & 1\\ 2\; & 0 \end{pmatrix}\\ \end{array}$, maka matriks B adalah ....
   $\begin{array} {lcl} A.\; \begin{pmatrix} -2\; & 1\\ 0\; & 3 \end{pmatrix}\\ B.\; \begin{pmatrix} 1\; & -4\\ 3\; & 0 \end{pmatrix}\\ C.\; \begin{pmatrix} 2\; & 1\\ 4\; & 3 \end{pmatrix}\\ D.\; \begin{pmatrix} 2\; & 0\\ 3\; & -4 \end{pmatrix}\\ E.\; \begin{pmatrix} 1\; & 2\\ 0\; & 3 \end{pmatrix}\\ \end{array}$

Pembahasan:

$\begin{array} {lcl} A^{-1}B= \begin{pmatrix} -2\; & 1\\ 2\; & 0 \end{pmatrix}\\ \; \; \; \; \; \; B=\left ( A^{-1} \right )^{-1}\begin{pmatrix} -2\; & 1\\ 2\; & 0 \end{pmatrix}\\ \; \; \; \; \; \; \; \; \; \; =A\begin{pmatrix} -2\; & 1\\ 2\; & 0 \end{pmatrix}\\ \; \; \; \; \; \; \; \; \; \; =\begin{pmatrix} 1\; & 2\\ 3\; & 5 \end{pmatrix}\begin{pmatrix} -2\; & 1\\ 2\; & 0 \end{pmatrix}\\ \; \; \; \; \; \; \; \; \; \; =\begin{pmatrix} \left (1 \right )\left ( -2 \right )+\left ( 2 \right )\left ( 2 \right )\; & \left (1 \right )\left ( 1 \right )+\left ( 2 \right )\left ( 0 \right )\\ \left (3 \right )\left ( -2 \right )+\left ( 5 \right )\left ( 2 \right )\; & \left (3 \right )\left ( 1 \right )+\left ( 5 \right )\left ( 0 \right ) \end{pmatrix}\\ \; \; \; \; \; \; \; \; \; \; =\begin{pmatrix} 2\; & 1\\ 4\; & 3 \end{pmatrix} \end{array}$

Jawaban _______________________________ (C)

Soal dan Pembahasan Matematika STIS 2017 No. 35-41

zulnining September 20, 2017 0 Comments

Berbagi M@th - Silahkan didownload soal dan pembahasan matematika STIS 2017 No. 35-41. Materi yang diujikan integral, limit, dan logika. Didalam soal dan pembahasan matematika STIS 2017 sudah ada KonseR (Konsep dasaR), semoga bermanfaat bagi yang membutuhkan dan dapat menambah atau memperkaya soal-soal yang sering diujikan untuk masuk perguruan tinggi negeri atau kedinasan.

Klik ----->Soal dan Pembahasan Matematika STIS 2017 No. 35-41

Selasa, 19 September 2017

Soal dan Pembahasan Matriks-2

zulnining September 19, 2017 0 Comments

3. UM UGM 2015 Kode 622
Diberikan matriks $\begin{array}{lcl} P=\begin{pmatrix} 2\; &-1 \\ 4\; & 3 \end{pmatrix}\; dan \; Q=\begin{pmatrix} 2r\; & 1\\ r\; & p+1 \end{pmatrix} \end{array}$ dengan $\begin{array}{lcl} r\neq 0 \end{array}$ dan $\begin{array}{lcl} p\neq 0 \end{array}$. Matriks
    $PQ$ tidak mempunyai invers apabila nilai $p$ = ....
   $\begin{array}{lcl} A.\; -\frac{3}{2}\\ B.\; -\frac{1}{2}\\ C.\; -\frac{1}{4}\\ D.\; \frac{1}{2}\\ E.\; \frac{8}{7}\\ \end{array}$

Pembahasan:
Matriks $PQ$ tidak mempunyai invers maka $detPQ=0$
$\begin{array} {lcl} P=\begin{pmatrix} 2\; &-1 \\ 4\; & 3 \end{pmatrix}\\ detP=\left ( 2 \right )\left ( 3 \right )-\left ( -1 \right )\left ( 4 \right )\\ \; \; \; \; \; \; \; \; \, =6+4\\ \; \; \; \; \; \; \; \; \, =10\\\\ Q=\begin{pmatrix} 2r\; &1 \\ r\; &p+1 \end{pmatrix}\\ detQ=\left ( 2r \right )\left ( p+1 \right )-\left ( 1 \right )\left ( r \right )\\ \; \; \; \; \; \; \; \; \, =2rp+2r-r\\ \; \; \; \; \; \; \; \; \, =2rp+r\\ \; \;\; \; \; \; \; \; \, =r\left ( 2p+1 \right ) \end{array}$

$\begin{array} {lcl} \; \; \; \; \; \; \; \; \; \; \; \; \; \: detPQ=0\\ \; \; \; \left ( detP \right )\left ( detQ \right )=0\\ \left ( 10 \right )\left ( r\left ( 2p+1 \right ) \right )=0\\ \; \; \; \; \; \; \; \; \; \: r\left ( 2p+1 \right )=0\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \: 2p+1=0\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; 2p=-1\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \: p=-\frac{1}{2} \end{array}$

Jawaban _______________________________ (B)

4. SBMPTN 2015 Kode 618
Jika $\begin{array} {lcl} A=\begin{pmatrix} 1\; &2a \\ a\; & 9 \end{pmatrix} \end{array}$ merupakan matriks yang mempunyai invers, maka hasil kali semua
nilai $a$ yang mungkin sehingga $\begin{array} {lcl} det\left ( A \right )=det\left ( A^{-1} \right ) \end{array}$ adalah ....
    A. 10
    B. 20
    C. 30
    D. 40
    E. 50

Pembahasan:

$\begin{array} {lcl} A=\begin{pmatrix} 1\; &2a \\ a\; & 9 \end{pmatrix}\\ detA=\left ( 1 \right )\left ( 9 \right )-\left ( 2a \right )\left ( a \right )\\ \; \; \; \; \; \; \; \; =9-2a^{2}\\\\ \; \; \; \; \; \; \; \; \; \; \; det\left ( A \right )=det\left ( A^{-1} \right )\\ \; \; \; \; \; \; \; \; \; \; \; det\left ( A \right )=\frac{1}{det\left ( A \right )}\\ det\left ( A \right )det\left ( A \right )=1\\ \; \; \; \; \; \; \; \; \; \; \left ( detA \right )^{2}=1 \end{array}$

$\begin{array} {lcl} \; \; \; \; \; \; \left ( 9-2a^{2} \right )^{2}=1\\ \; \; \; \; \; \;\; \; \; \; 9-2a^{2}= \pm 1\\\\ \; \; \; \; \; \;\; \; \; \; 9-2a^{2}= 1\\ \; \; \; \; \; \;\; \; \; \;\; \; \; \; \; \; 2a^{2}= 8\\ \; \; \; \; \; \;\; \; \; \;\; \; \; \; \; \;\; \; a^{2}= 4\\ \; \; \; \; \; \;\; \; \; \;\; \; \; \; \; \;\; \; \; \, a= \pm 2\\\\ \; \; \; \; \; \;\; \; \; \; 9-2a^{2}=- 1\\ \; \; \; \; \; \;\; \; \; \;\; \; \; \; \; \; 2a^{2}= 10\\ \; \; \; \; \; \;\; \; \; \;\; \; \; \; \; \;\; \; a^{2}= 5\\ \; \; \; \; \; \;\; \; \; \;\; \; \; \; \; \;\; \; \; \, a= \pm \sqrt{5} \end{array}$

Hasil kali semua nilai $a$ yang mungkin

$\begin{array} {lcl} =\left ( 2 \right )\left ( -2 \right )\left ( \sqrt{5} \right )\left (- \sqrt{5} \right )\\ =20 \end{array}$

Jawaban _______________________________ (B)

Soal dan Pembahasan Matriks-2

zulnining September 19, 2017 0 Comments

1. Simak UI 2009 Kode 941
Diketahui $\begin{array}{lcl} A=\begin{pmatrix} x+2\; &3 \\ 3\; & 3 \end{pmatrix},\; B=\begin{pmatrix} 3\; &0 \\ 5\; & x+2 \end{pmatrix} \end{array}$, maka perkalian $x$ yang memenuhi
   $det(AB)=36$ adalah ....
   A. -8
   B. -7
   C. -6
   D. 2
   E. 6

Pembahasan:
$\begin{array}{lcl} A=\begin{pmatrix} x+2\; &3 \\ 3\; & 3 \end{pmatrix}\\ detA=\left ( x+2 \right )\left ( 3 \right )-\left ( 3 \right )\left ( 3 \right )\\ \; \; \; \; \; \; \; \; =3x+6-9\\ \; \; \; \; \; \; \; \; =3x-3\\\\ B=\begin{pmatrix} 3\; &0 \\ 5\; & x+2 \end{pmatrix}\\ detB=\left ( 3 \right )\left ( x+2 \right )-\left ( 0 \right )\left ( 5 \right )\\ \; \; \; \; \; \; \; \; =3x+6 \end{array}$
$\begin{array}{lcl} det\left ( AB \right )=36\\ \left ( detA \right )\left ( detB \right )=36\\ \left ( 3x-3 \right )\left ( 3x+6 \right )=36\\ 9x^{2}+18x-9x-18-36=0\\ 9x^{2}+9x-54=0 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; (sama-sama\; dibagi\; 9)\\ x^{2}+x-6=0\\ \left ( x+3 \right )\left ( x-2 \right )=0\\ x=-3\; atau\;x=2 \end{array}$
Perkalian nilai
$\begin{array}{lcl} x&=&\left ( -3 \right ) \left ( 2 \right )\\ &=&-6 \end{array}$

Jawaban _______________________________ (C)

2. Simak UI 2010 Kode 208
Diketahui $\begin{array}{lcl} A=\begin{pmatrix} -1\; &50 \\ -2\; & 105 \end{pmatrix} \end{array}$, maka $detA^{3}$ = ....
    A. -125
    B. -25
C. 5
D. 25
E. 125

Pembahasan:
$\begin{array}{lcl} A=\begin{pmatrix} -1\; &50 \\ -2\; & 105 \end{pmatrix}\\\\ detA=\left ( -1 \right )\left ( 105 \right )-\left ( 50 \right )\left ( -2 \right )\\ \; \; \; \; \; \; \; \; =-105+100\\ \; \; \; \; \; \; \; \; =-5\\\\ detA^{3}=\left ( detA \right )^{3}\\ \; \; \; \; \; \; \; \; \; \; =\left ( -5 \right )^{3}\\ \; \; \; \; \; \; \; \; \; \; =-125 \end{array}$

Jawaban _______________________________ (A)

Matriks-2 (Determinan dan Invers Matriks Berordo 2)

zulnining September 19, 2017 0 Comments

# Determinan Matriks Persegi Berordo 2

Determinan hanya dimiliki oleh matriks persegi, seperti matriks berordo 2x2, 3x3, dan sebagainya
Determinan matriks A dapat ditulis $det A$ atau $|A|$
Misalkan matriks

       $\begin{array}{lcl} A=\begin{pmatrix} a\; & b\\ c\; & d \end{pmatrix}\\ maka\; detA=|A|=\begin{vmatrix} a\; &b \\ c\; & d\\ \end{vmatrix}\\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; =ad-bc \end{array}$
Contoh:
    Tentukan nilai determinan matriks-matriks
  $\begin{array}{lcl} A=\begin{pmatrix} 5\; & -2\\ 4\; & 8 \end{pmatrix}\\ \end{array}$ dan $\begin{array}{lcl} B=\begin{pmatrix} 7\; & 4\\ 6\; & -1 \end{pmatrix}\\ \end{array}$
Pembahasan:
  $\begin{array}{lcl} A=\begin{pmatrix} 5\; & -2\\ 4\; & 8 \end{pmatrix}\\ detA=\begin{vmatrix} 5\; & -2\\ 4\; & 8 \end{vmatrix}\\ \; \; \; \; \; \; \; \; =\left ( 5 \right )\left ( 8 \right )-\left ( -2 \right )\left ( 4 \right )\\ \; \; \; \; \; \; \; \; =40+8\\ \; \; \; \; \; \; \; \; =48 \end{array}$
$\begin{array}{lcl} B=\begin{pmatrix} 7\; & 4\\ 6\; & -1 \end{pmatrix}\\ detA=\begin{vmatrix} 7\; & 4\\ 6\; & -1 \end{vmatrix}\\ \; \; \; \; \; \; \; \; =\left ( 7 \right )\left ( -1 \right )-\left (4 \right )\left ( 6 \right )\\ \; \; \; \; \; \; \; \; =-7-24\\ \; \; \; \; \; \; \; \; =-31 \end{array}$

# Invers Matriks Persegi Berordo 2
   Jika matriks $\begin{array}{lcl} A=\begin{pmatrix} a\; & b\\ c\; & d \end{pmatrix}\\ \end{array}$, maka invers matriks A
   $\begin{array}{lcl} A^{-1}=\frac{1}{ad-bc}\begin{pmatrix} d \; &-b \\ -c\; & a \end{pmatrix}\\\\ \; \; \; \; \; \; \; =\frac{1}{|A|}\begin{pmatrix} d\; & -b\\ -c\; &a \end{pmatrix} \end{array}$
   Contoh:
   Tentukan invers matriks $\begin{array}{lcl} A=\begin{pmatrix} 5 \; &3 \\ -7\; & -4 \end{pmatrix} \end{array}$
   Pembahasan:
   $\begin{array}{lcl} A=\begin{pmatrix} 5 \; &3 \\ -7\; & -4 \end{pmatrix}\\\\ |A|=\left ( 5 \right )\left ( -4 \right )-\left ( 3 \right )\left ( -7 \right )\\ \: \: \: \: \: \: \: =-20+21\\ \: \: \: \: \: \: \: =1\\ A^{-1}=\frac{1}{|A|}\begin{pmatrix} -4\; & -3\\ 7\; & 5 \end{pmatrix}\\ \; \; \; \; \; \; \; =\frac{1}{1}\begin{pmatrix} -4\; &-3 \\ 7\; & 5 \end{pmatrix}\\ \; \; \; \; \; \; \; =\begin{pmatrix} -4\; &-3 \\ 7\; & 5 \end{pmatrix} \end{array}$

Matriks yang tidak memiliki invers dinamakan matriks singular, $det=0$
Jika $\begin{array}{lcl} AB=I;\; I=\begin{pmatrix} 1\; &0 \\ 0\; & 1 \end{pmatrix} \end{array}$, maka $A$ dan $B$ dikatakan saling invers
$\begin{array}{lcl} AX=B\Rightarrow X=A^{-1}B\\ XA=B\Rightarrow X=AB^{-1} \end{array}$
$\begin{array}{lcl} det\left ( AB \right )=\left ( detA \right )\left ( detB \right )\\ detA^{-1}=\frac{1}{detA}\\ detA^{T}=detA\\ detA^{n}=\left ( detA \right )^{n}\\ det\left ( kA^{n} \right )=k^{ordo\; matriks}\left ( detA \right )^{n} \end{array}$
Untuk soal dan pembahasan Klik ----> 1 2 3

Senin, 18 September 2017

Soal Tambahan Matriks-1

zulnining September 18, 2017 0 Comments

7. UM UGM 2017 Kode 723
Jika $a$ memenuhi $\begin{pmatrix} a^{2}\; &3 \\ 0\; &6a \end{pmatrix}=\begin{pmatrix} a\; &5 \\ 1\; & 0 \end{pmatrix}+\begin{pmatrix} 20\; &-1 \\ -2\; & a^{2}+5 \end{pmatrix}^{T}$ dengan $A^{T}$ menyatakan transpose
   matriks $A$, maka $a^{2}+a$ = ....
   A. 2
   B. 12
   C. 20
   D. 30
   E. 42

Pembahasan:

$\begin{array}{lcl} \begin{pmatrix} a^{2}\; &3 \\ 0\; &6a \end{pmatrix}&=&\begin{pmatrix} a\; &5 \\ 1\; & 0 \end{pmatrix}+\begin{pmatrix} 20\; &-1 \\ -2\; & a^{2}+5 \end{pmatrix}^{T}\\\\ \begin{pmatrix} a^{2}\; &3 \\ 0\; &6a \end{pmatrix}&=&\begin{pmatrix} a\; &5 \\ 1\; & 0 \end{pmatrix}+\begin{pmatrix} 20\; &-2 \\ -1\; & a^{2}+5 \end{pmatrix}\\\\ \begin{pmatrix} a^{2}\; &3 \\ 0\; &6a \end{pmatrix}&=&\begin{pmatrix} a+20\; &3 \\ 0\; & a^{2}+5 \end{pmatrix} \end{array}$

$\begin{array}{lcl} a^{2}&=&a+20\\ 6a&=&a^{2}+5\\ 6a&=&a+20+5\\ 5a&=&25\\ a&=&5 \end{array}$
Maka
$\begin{array}{lcl} a^{2}+a&=&5^{2}+5\\ &=&25+5\\ &=&30 \end{array}$

Jawaban ________________________ (D)

8. SBMPTN 2014 Kode 663
Jika matriks $\begin{array}{lcl} A=\begin{pmatrix} 2x\; &-2 \\ x\; & 3y+2 \end{pmatrix},\; B=\begin{pmatrix} 9\; & 3x\\ 8 \; & -4 \end{pmatrix} \end{array}$ dan $\begin{array}{lcl} C=\begin{pmatrix} 5\; & 6\\ -8 \; & 7 \end{pmatrix} \end{array}$ memenuhi $A+B=C^{t}$ dengan $C^{t}$    transpose matriks $C$, maka $2x+3y$ = ....
   A. 3
   B. 4
   C. 5
   D. 6
   E. 7

Pembahasan:
$\begin{array}{lcl} C=\begin{pmatrix} 5\; & 6\\ -8 \; & 7 \end{pmatrix}\\\\ C^{t}=\begin{pmatrix} 5\; & -8\\ 6 \; & 7 \end{pmatrix}\\\\ \end{array}$
$\begin{array}{lcl} A+B=C^{t}\\\\ \begin{pmatrix} 2x\; & -2\\ x \; & 3y+2 \end{pmatrix}+\begin{pmatrix} 9\; & 3x\\ 8 \; & -4 \end{pmatrix}&=&\begin{pmatrix} 5\; & -8\\ 6 \; & 7 \end{pmatrix}\\\\ \begin{pmatrix} 2x+9\; & -2+3x\\ x+8 \; & 3y-2 \end{pmatrix}&=&\begin{pmatrix} 5\; & -8\\ 6 \; & 7 \end{pmatrix}\\\\ \end{array}$
$\begin{array}{lcl} 2x+9&=&5\\ 2x&=&-4\\\\ 3y-2&=&7\\ 3y&=&9 \end{array}$

Maka
$\begin{array}{lcl} 2x+3y&=&-4+9\\ &=&5 \end{array}$

Jawaban ________________________ (C)

9. SIMAK UI 2012 Kode 224
Jika matriks $\begin{array}{lcl} A=\begin{pmatrix} 2\; & 1\\ 3\; & 5 \end{pmatrix} \end{array}$, maka matriks $B$ yang memenuhi $\begin{array}{lcl} A+B^{T}=\left ( A-B \right )^{T} \end{array}$ adalah ....
    A. $\begin{array}{lcl} \begin{pmatrix} 2\; &3 \\ 1\; & 5 \end{pmatrix} \end{array}$
B. $\begin{array}{lcl} \begin{pmatrix} 0\; &2 \\ -2\; & 0 \end{pmatrix} \end{array}$
C. $\begin{array}{lcl} \begin{pmatrix} 0\; &-2 \\ 2\; & 0 \end{pmatrix} \end{array}$
D. $\begin{array}{lcl} \begin{pmatrix} 0\; &1 \\ -1\; & 0 \end{pmatrix} \end{array}$
E. $\begin{array}{lcl} \begin{pmatrix} 0\; &-1 \\ 1\; & 0 \end{pmatrix} \end{array}$

Pembahasan:
Misal matriks
$\begin{array}{lcl} B&=&\begin{pmatrix} a\; &b \\ c\; & d \end{pmatrix}\\\\ B^{T}&=&\begin{pmatrix} a\; &c \\ b\; & d \end{pmatrix} \end{array}$
$\begin{array}{lcl} A+B^{T}&=&\left ( A-B \right )^{T}\\ \begin{pmatrix} 2\; &1 \\ 3\; & 5 \end{pmatrix}+\begin{pmatrix} a\; & c\\ b\; & d \end{pmatrix}&=&\left ( \begin{pmatrix} 2\; &1 \\ 3\; & 5 \end{pmatrix}-\begin{pmatrix} a\; &b \\ c\; & d \end{pmatrix} \right )^{T}\\\\ \begin{pmatrix} 2+a\; &1+c \\ 3+b\; &5+d \end{pmatrix}&=&\begin{pmatrix} 2-a\; &1-b \\ 3-c\; & 5-d \end{pmatrix}^{T}\\\\ \begin{pmatrix} 2+a\; &1+c \\ 3+b\; &5+d \end{pmatrix}&=&\begin{pmatrix} 2-a\; &3-c \\ 1-b\; & 5-d \end{pmatrix} \end{array}$

$\begin{array}{lcl} 2+a&=&2-a\\ {\color{Red} a}&=&{\color{Red} 0}\\ 3+b&=&1-b\\ 2b&=&-2\\ {\color{Red} b}&=&{\color{Red} -1}\\ 1+c&=&3-c\\ 2c&=&2\\ {\color{Red} c}&=&{\color{Red} 1}\\ 5+d&=&5-d\\ {\color{Red} d}&=&{\color{Red} 0} \end{array}$

Maka matriks $\begin{array}{lcl} B=\begin{pmatrix} a\; &b \\ c\; & d \end{pmatrix}=\begin{pmatrix} 0\; &-1 \\ 1\; & 0 \end{pmatrix} \end{array}$

M@th

Minggu, 24 September 2017

Simak UI 2011 Kode 211 Materi Eksponen

SNMPTN 2010 Kode 336 Materi Eksponen

SPMB 2006 Materi Eksponen

Jumat, 22 September 2017

Soal dan Pembahasan Eksponen

Kamis, 21 September 2017

Kumpulan Soal dan Pembahasan Matriks (Bagian 1)

Matriks-3 (Determinan & Invers Matriks berordo 3 x 3)

Rabu, 20 September 2017

Soal dan Pembahasan Matriks-2

Soal dan Pembahasan Matematika STIS 2017 No. 35-41

Selasa, 19 September 2017

Soal dan Pembahasan Matriks-2

Soal dan Pembahasan Matriks-2

Matriks-2 (Determinan dan Invers Matriks Berordo 2)

Senin, 18 September 2017

Soal Tambahan Matriks-1

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