Diketahui matriks A=(U1U4−U2U3) dan Un adalah suku ke-n barisan geometri. Jika U1+U3=p1 dan U2+U4=q1 dengan p,q=0 maka determinan A sama dengan...

Pertanyaan

Diketahui matriks $A=\begin{pmatrix}U_1&-U_2\\U_4&U_3\end{pmatrix}$ dan ${\mathrm U}_{\mathrm n}$ adalah suku ke-n barisan geometri. Jika $U_1+U_3=\frac1{\mathrm p}$ dan $U_2+U_4=\frac1{\mathrm q}$ dengan $\mathrm p,\mathrm q\neq0$ maka determinan $A$ sama dengan $...$

$fraction numerator 1 over denominator straight p squared plus straight q squared end fraction$
$fraction numerator straight p squared over denominator straight p squared plus straight q squared end fraction$
$fraction numerator pq squared over denominator straight p squared plus straight q squared end fraction$
$fraction numerator 1 over denominator straight p plus straight q end fraction$
$fraction numerator straight q over denominator straight p plus straight q end fraction$

O. Rahmawati

Master Teacher

Mahasiswa/Alumni UIN Sunan Gunung Djati Bandung

Jawaban terverifikasi

Pembahasan

Diketahui:

$A=\begin{pmatrix}U_1&-U_2\\U_4&U_3\end{pmatrix}$
$U_1+U_3=\frac1{\mathrm p}$
$U_2+U_4=\frac1{\mathrm q}$

Ditanya: $\det(A)=...$

Jawab:

Menentukan rasio:

$\begin{array}{rcl}U_1+U_3&=&\frac1{\mathrm p}\\a+ar^2&=&\frac1{\mathrm p}\\a\left(1+r^2\right)&=&\frac1{\mathrm p},\;\mathrm p\neq0\end{array}$

dan

$U_2+U_4=\frac1{\mathrm q}\\ar+ar^3=\frac1{\mathrm q}\\ar\left(1+r^2\right)=\frac1{\mathrm q},\;\mathrm q\neq0$

Jadi, rasionya adalah

$fraction numerator a r open parentheses 1 plus r squared close parentheses over denominator a open parentheses 1 plus r squared close parentheses end fraction equals fraction numerator begin display style 1 over straight q end style over denominator begin display style 1 over straight p end style end fraction equals straight p over straight q$

Menentukan nilai suku pertama atau $a$ :

$table attributes columnalign right center left columnspacing 0px end attributes row cell a open parentheses 1 plus r squared close parentheses end cell equals cell 1 over straight p end cell row a equals cell fraction numerator 1 over denominator straight p open parentheses 1 plus r squared close parentheses end fraction end cell end table$

Menentukan determinan A:

$table attributes columnalign right center left columnspacing 0px end attributes row cell d e t left parenthesis A right parenthesis end cell equals cell U subscript 1 times U subscript 3 minus open parentheses negative U subscript 2 close parentheses times U subscript 4 end cell row blank equals cell U subscript 1 times U subscript 3 plus U subscript 2 times U subscript 4 end cell row blank equals cell a times a r squared plus a r times a r cubed end cell row blank equals cell a squared r squared open parentheses 1 plus r squared close parentheses end cell row blank equals cell open parentheses fraction numerator 1 over denominator straight p open parentheses 1 plus r squared close parentheses end fraction close parentheses squared r squared open parentheses 1 plus r squared close parentheses end cell row blank equals cell fraction numerator r squared over denominator straight p squared open parentheses 1 plus r squared close parentheses end fraction end cell row blank equals cell fraction numerator open parentheses begin display style straight p over straight q end style close parentheses squared over denominator straight p squared open parentheses 1 plus open parentheses begin display style straight p over straight q end style close parentheses squared close parentheses end fraction end cell row blank equals cell fraction numerator 1 over denominator straight q squared open parentheses 1 plus begin display style straight p squared over straight q squared end style close parentheses end fraction end cell row blank equals cell fraction numerator 1 over denominator straight p squared plus straight q squared end fraction end cell row blank blank blank end table$

Oleh karena itu jawbaan yang benar adalah A.