PAS MATEMATIKA BLOG

Contoh Soal 8 Fungsi Logaritma (Persamaan Logaritma)

$\begin{array}{ll} 36. & Persamaan^{x} \log 2 +^{x} \log (3 x - 4) = 2 \\ mempunyai akar x_{1} dan x_{2}, maka \\ nilai x_{1} + x_{2} adalah . . . . \\ \begin{array}{llll} a . & 2 \\ b . & 3 \\ c . & 4 \\ d . & 6 \\ e . & 8 \end{array} \\ Jawab : d \\ \begin{aligned} Alternatif 1 \\ ^{x} \log 2 +^{x} \log (3 x - 4) = 2 \\ \Leftrightarrow^{x} \log 2 (3 x - 4) = 2 \\ \Leftrightarrow^{x} \log 6 x - 8 = 2 \\ \Leftrightarrow 6 x - 8 = x^{2} \\ \Leftrightarrow x^{2} - 6 x + 8 = 0, dengan {\begin{cases} a & = 1 \\ b & = - 6 \\ c & = 8 \end{cases} \\ \Leftrightarrow x_{1} + x_{2} = - \frac{b}{a} \\ \Leftrightarrow x_{1} + x_{2} = - \frac{- 6}{1} = 6 \\ Alternatif 2 \\ \Leftrightarrow x^{2} - 6 x + 8 = 0 \\ \Leftrightarrow (x - 2) (x - 4) \\ \Leftrightarrow x_{1} = 2 atau x_{2} = 4 \\ \Leftrightarrow x_{1} + x_{2} = 2 + 4 = 6 \end{aligned} \end{array}$

$\begin{array}{ll} 37. & Jika x_{1} dan x_{2} memenuhi \\ (\log x) (2 \log x - 3) = \log 100 \\ maka x_{1} \times x_{2} adalah . . . . \\ \begin{array}{llll} a . & 100 \\ b . & 10 \sqrt{10} \\ c . & \sqrt{10} \\ d . & - \sqrt{10} \\ e . & - 10 \sqrt{10} \end{array} \\ Jawab : b \\ \begin{aligned} (\log x) (2 \log x - 3) = \log 100 \\ \Leftrightarrow (\log x) (2 \log x - 3) = 2 \\ \Leftrightarrow 2 \log^{2} x - 3 \log x - 2 = 0 {\begin{cases} a & = 2 \\ b & = - 3 \\ c & = - 2 \end{cases} \\ \Leftrightarrow \log x_{1} + \log x_{2} = - \frac{- 3}{2} = \frac{3}{2} \\ \Leftrightarrow \log (x_{1} \times x_{2}) = 1 \frac{1}{2} \\ \Leftrightarrow (x_{1} \times x_{2}) = 10^{1 \frac{1}{2}} = 10 \sqrt{10} \end{aligned} \end{array}$

$\begin{array}{ll} 38. & Persamaan \\ 10^{^{^{2}} \log x^{2}} - 7 (10^{^{^{2}} \log x}) + 10 = 0 \\ mempunyai dua akar yaitu x_{1} dan x_{2} \\ Nilai x_{1} \times x_{2} = . . . . \\ \begin{array}{llll} a . & - 2 \\ b . & - 5 \\ c . & 2 \\ d . & 5 \\ e . & 10 \end{array} \\ Jawab : c \\ \begin{aligned} 10^{^{^{2}} \log x^{2}} - 7 (10^{^{^{2}} \log x}) + 10 = 0 \\ 10^{2^{^{2}} \log x} - 7 (10^{^{^{2}} \log x}) + 10 = 0 \\ adalah persamaan kuadrat dalam 10^{^{^{2}} \log x} \\ Misalkan p = 10^{^{^{2}} \log x}, maka persamaan \\ menjadi p^{2} - 7 p + 10 = 0 {\begin{cases} a & = 1 \\ b & = - 7 \\ c & = 10 \end{cases} \\ Karena nilai p_{1} \times p_{2} = \frac{c}{a} maka \\ 10^{^{^{2}} \log x_{1}} \times 10^{^{^{2}} \log x_{2}} = \frac{10}{1} = 10 \\ 10^{^{^{2}} \log x_{1} +^{2} \log x_{2}} = 10 \\ 10^{^{^{2}} \log x_{1} +^{2} \log x_{2}} = 10^{1} \\ \Leftrightarrow^{2} \log x_{1} +^{2} \log x_{2} = 1 \\ \Leftrightarrow^{2} \log x_{1} \times x_{2} = 1 \\ \Leftrightarrow x_{1} \times x_{2} = 2^{1} = 2 \end{aligned} \end{array}$

$\begin{array}{ll} 39. & Nilai x yang memenuhi \\ ^{x} \log (x + 12) - 3.^{x} \log 4 + 1 = 0 \\ adalah . . . . \\ \begin{array}{llll} a . & \frac{1}{2} \\ b . & 2 \\ c . & 4 \\ d . & 8 \\ e . & 16 \end{array} \\ Jawab : c \\ \begin{aligned} ^{x} \log (x + 12) - 3.^{x} \log 4 + 1 = 0 \\ ^{x} \log (x + 12) -^{x} \log 4^{3} = - 1 \\ ^{x} \log \frac{x + 12}{64} = - 1 \\ \frac{x + 12}{64} = x^{- 1} = \frac{1}{x} \\ x + 12 = \frac{64}{x} \\ x^{2} + 12 x - 64 = 0 \\ (x + 16) (x - 4) = 0 \\ x = - 16 atau x = 4 \end{aligned} \end{array}$

$\begin{array}{ll} 40. & Nilai x yang memenuhi \\ \frac{^{2} \log (2 x - 3)}{^{2} \log x} -^{x} \log (x + 6) + \frac{1}{^{(x + 2)} \log x} = 1 \\ adalah . . . . \\ \begin{array}{llll} a . & - 1 dan 6 \\ b . & - 2 dan 6 \\ c . & - 1 \\ d . & - 2 \\ e . & 6 \end{array} \\ Jawab : e \\ \begin{aligned} \frac{^{2} \log (2 x - 3)}{^{2} \log x} -^{x} \log (x + 6) + \frac{1}{^{(x + 2)} \log x} = 1 \\ ^{x} \log (2 x - 3) -^{x} \log (x + 6) +^{x} \log (x + 2) = 1 \\ ^{x} \log (2 x - 3) (x + 2) = 1 + \log (x + 6) \\ ^{x} \log \frac{(2 x^{2} + x - 6)}{(x + 6)} = 1 \\ \frac{(2 x^{2} + x - 6)}{(x + 6)} = x^{1} \\ (2 x^{2} + x - 6) = x^{2} + 6 x \\ x^{2} - 5 x - 6 = 0 \\ (x + 1) (x - 6) = 0 \\ x = - 1 atau x = 6 \end{aligned} \end{array}$

Contoh Soal 7 Fungsi Logaritma (Pertidaksamaan Logaritma)

$\begin{array}{ll} 32. & Agar \log (x^{2} - 1) < 0 maka . . . . \\ \begin{array}{llll} a . & - 1 < x < 1 \\ b . & - \sqrt{2} < x < \sqrt{2} \\ c . & x < - 1 atau x > 1 \\ d . & x < - \sqrt{2} atau x > \sqrt{2} \\ e . & - \sqrt{2} < x < - 1 atau 1 < x < \sqrt{2} \end{array} \\ Jawab : e \\ \begin{aligned} \log (x^{2} - 1) < 0 \\ Diketahui \log f (x) < 0, maka \\ Syarat (1), f (x) > 0 \\ \Leftrightarrow x^{2} - 1 > 0 \\ \Leftrightarrow x < - 1 atau x > 1 \\ Syarat (2), \log (x^{2} - 1) < 0 \\ \log (x^{2} - 1) < \log 1 \\ \Leftrightarrow x^{2} - 1 < 1 \\ \Leftrightarrow x^{2} - 2 < 0 \\ \Leftrightarrow x^{2} - {(\sqrt{2})}^{2} < 0 \\ \Leftrightarrow - \sqrt{2} < x < \sqrt{2} \\ Jadi, - \sqrt{2} < x < - 1 atau 1 < x < \sqrt{2} \end{aligned} \end{array}$

$\begin{array}{ll} 33. & Himpunan penyelesaian dari \\ ^{.^{\frac{1}{2}}} \log (x^{2} - 3) > 0 adalah . . . . \\ \begin{array}{llll} a . & {x | - 2 < x < - \sqrt{3} atau \sqrt{3} < x < 2} \\ b . & {x | - \sqrt{3} < x < - 1 atau \sqrt{3} < x < 2} \\ c . & {x | - 2 < x < - \sqrt{3}} \\ d . & {x | - 2 < x < - \sqrt{3}} \\ e . & {x | \sqrt{3} < x < 2} \end{array} \\ Jawab : a \\ \begin{aligned} ^{.^{\frac{1}{2}}} \log (x^{2} - 3) > 0 \\ Diketahui^{.^{\frac{1}{2}}} \log f (x) > 0, maka \\ Syarat (1), f (x) > 0 \\ \Leftrightarrow x^{2} - 3 > 0 \\ \Leftrightarrow x < - \sqrt{3} atau x > \sqrt{3} \\ Syarat (2),^{.^{\frac{1}{2}}} \log (x^{2} - 3) > 0 \\ ^{.^{\frac{1}{2}}} \log (x^{2} - 3) >^{.^{\frac{1}{2}}} \log 1 \\ \Leftrightarrow x^{2} - 3 < 1 (karena basisnya \frac{1}{2} < 1) \\ \Leftrightarrow x^{2} - 4 < 0 \\ \Leftrightarrow x^{2} - 2^{2} > 0 \\ \Leftrightarrow - 2 < x < 2 \\ Jadi, - 2 < x < - \sqrt{3} atau \sqrt{3} < x < 2 \end{aligned} \end{array}$

$\begin{array}{ll} 34. & Nilai x yang memenuhi \\ ^{2} \log (x^{2} - x) \leq 1 adalah . . . . \\ \begin{array}{llll} a . & x < 0 atau x > 1 \\ b . & - 1 \leq x \leq 2, x \neq 1 atau x \neq 0 \\ c . & - 1 \leq x < 0 atau 1 < x \leq 2 \\ d . & - 1 < x \leq 0 atau 1 \leq x < 2 \\ e . & - 1 \leq x \leq 0 atau 1 \leq x \leq 2 \end{array} \\ Jawab : c \\ \begin{aligned} ^{2} \log (x^{2} - x) \leq 1 \\ Diketahui^{2} \log f (x) > 0, maka \\ Syarat (1), f (x) > 0 \\ \Leftrightarrow x^{2} - x > 0 \Leftrightarrow x (x - 1) > 0 \\ \Leftrightarrow x < 0 atau x > 1 \\ Syarat (2),^{2} \log (x^{2} - x) \leq 1 \\ ^{2} \log (x^{2} - x) \leq^{2} \log 2 \\ \Leftrightarrow x^{2} - x \leq 2 \\ \Leftrightarrow x^{2} - x - 2 \leq 0 \\ \Leftrightarrow (x + 1) (x - 2) \leq 0 \\ \Leftrightarrow - 1 \leq x \leq 2 \\ Jadi, - 1 \leq x < 0 atau 1 < x \leq 2 \end{aligned} \end{array}$

$\begin{array}{ll} 35. & Nilai x yang memenuhi \\ | \log (x + 1) | > 1 adalah . . . . \\ \begin{array}{llll} a . & x < - 0, 9 atau x > 9 \\ b . & x < - 9 atau x > 9 \\ c . & - 1 < x < - 0, 9 atau x > 9 \\ d . & - 9 < x < 0, 9 \\ e . & - 0, 9 < x < 9 \end{array} \\ Jawab : c \\ \begin{aligned} Ingat bahwa \\ | x | > A \Leftrightarrow x < - A atau x > A, A > 0 \\ \Leftrightarrow \log (x + 1) < - 1 atau \log (x + 1) > 1 \\ Syarat (1) buat keduanya, f (x) > 0 \\ (x + 1) > 0 \Leftrightarrow x > - 1 \\ Syarat (2), \log (x + 1) < - 1 \\ \log (x + 1) < \log 10^{- 1} \\ x + 1 < \frac{1}{10} \Leftrightarrow x < - \frac{9}{10} \\ Syarat (3), \log (x + 1) > 1 \\ \log (x + 1) > \log 10^{1} \\ (x + 1) > 10 \Leftrightarrow x > 9 \\ Jadi, - 1 < x < - 0, 9 atau x > 9 \end{aligned} \end{array}$

Lanjutan Materi Fungsi Logaritma (Kelas X Matematika Peminatan)

MENGINGAT KEMBALI

$E. Sifat-Sifat Eksponen dan Logaritma$

$E.1 Persamaan Eksponen dan Logaritma$

$\begin{array}{ll} Eksponens & Logaritma \\ a^{n} = \underset{n f a k t o r}{\underset{⏟}{a \times a \times a \times \dots \times a}} & ^{a} \log b = c \Rightarrow a^{c} = b \\ ∙ a^{p} \times a^{q} = a^{p + q} & ∙^{a} \log x +^{a} \log y =^{a} \log x y \\ ∙ a^{p} : a^{q} = a^{p - q} & ∙^{a} \log x -^{a} \log y =^{a} \log \frac{x}{y} \\ ∙ {(a^{p})}^{q} = a^{p . q} & ∙^{a} \log x = \frac{^{m} \log x}{^{m} \log a} \\ ∙ \sqrt[q]{a^{p}} = a^{(\frac{p}{q})} & ∙^{a} \log b \times^{b} \log c =^{a} \log c \\ ∙ {(a \times b)}^{p} = a^{p} \times b^{p} & ∙^{a^{m}} \log b^{n} = \frac{n}{m} \times^{a} \log b \\ ∙ {(\frac{a}{b})}^{p} = \frac{a^{p}}{b^{p}} & ∙ a^{^{a} \log b} = b \\ ∙ a^{- p} = \frac{1}{a^{p}} & ∙^{a} \log b = \frac{1}{^{b} \log a} \\ ∙ a^{0} = 1, a \neq 0 & ∙^{a} \log 1 = 0 \\ ∙ a^{1} = 1 & ∙^{a} \log a = 1 \\ {\begin{cases} a, b \in R \\ p, q \in Q \end{cases} & {\begin{cases} a \neq 0 \\ a > 0 & (bilangan pokok) \\ x, y > 0 & (numerus) \end{cases} \end{array}$

Selanjutnya

$\begin{array}{lll} No & Bentuk & Syarat \\ 1. & a^{f (x)} = 1 & a \neq 0, maka f (x) = 0 \\ 2. & a^{f (x)} = a^{p} & a > 0, a \neq 1, maka f (x) = p \\ 3. & a^{f (x)} = a^{g (x)} & a > 0, a \neq 1, maka f (x) = g (x) \\ 4. & a^{f (x)} = b^{f (x)} & a \neq 0, b \neq 0, maka f (x) = 0 \\ 5. & f (x)^{g (x)} = 1 & {\begin{cases} f (x) = 1 \\ g (x) = 0, & jika f (x) \neq 0 \\ f (x) = - 1, & jika g (x) = genap \end{cases} \\ 6. & f (x)^{g (x)} = f (x)^{h (x)} & {\begin{cases} (i) . g (x) = h (x) \\ (i i) . f (x) = 1 \\ (i i i) . f (x) = 0, & g (x) > 0, h (x) > 0 \\ (i v) . f (x) = - 1, & g (x) dan h (x) \\ keduanya ganjil \\ atau genap \end{cases} \\ 7. & g (x)^{f (x)} = h (x)^{f (x)} & {\begin{cases} (i) . g (x) = h (x) \\ (i i) . f (x) = 0, & g (x) \neq 0, h (x) \neq 0 \end{cases} \\ 8. & \begin{aligned} A {(a^{f (x)})}^{2} \\ + B (a^{f (x)}) \\ + C = 0 \end{aligned} & a > 0, a \neq 1 \end{array}$

$E.2 Pertidaksamaan Eksponen dan Logaritma$

Berikut sifat pertidaksamaan Eksponen

$\begin{array}{ll} a > 1 & 0 < a < 1 \\ a^{f (x)} \leq a^{g (x)} \Rightarrow f (x) \leq g (x) & a^{f (x)} \leq a^{g (x)} \Rightarrow f (x) \geq g (x) \\ a^{f (x)} < a^{g (x)} \Rightarrow f (x) < g (x) & a^{f (x)} < a^{g (x)} \Rightarrow f (x) > g (x) \\ a^{f (x)} \geq a^{g (x)} \Rightarrow f (x) \geq g (x) & a^{f (x)} \geq a^{g (x)} \Rightarrow f (x) \leq g (x) \\ a^{f (x)} > a^{g (x)} \Rightarrow f (x) > g (x) & a^{f (x)} > a^{g (x)} \Rightarrow f (x) < g (x) \end{array}$

Untuk pertidaksamaan logaritma (dengan syarat $(f (x) > 0 dan g (x) > 0)$ ) adalah sebagai berikut:

$\begin{array}{ll} a > 1 & 0 < a < 1 \\ \begin{aligned} ^{a} \log f (x) \leq^{a} \log g (x) \\ \Rightarrow f (x) \leq g (x) \end{aligned} & \begin{aligned} ^{a} \log f (x) \leq^{a} \log g (x) \\ \Rightarrow f (x) \geq g (x) \end{aligned} \\ \begin{aligned} ^{a} \log f (x) <^{a} \log g (x) \\ \Rightarrow f (x) < g (x) \end{aligned} & \begin{aligned} ^{a} \log f (x) <^{a} \log g (x) \\ \Rightarrow f (x) > g (x) \end{aligned} \\ \begin{aligned} ^{a} \log f (x) \geq^{a} \log g (x) \\ \Rightarrow f (x) \geq g (x) \end{aligned} & \begin{aligned} ^{a} \log f (x) \geq^{a} \log g (x) \\ \Rightarrow f (x) \leq g (x) \end{aligned} \\ \begin{aligned} ^{a} \log f (x) >^{a} \log g (x) \\ \Rightarrow f (x) > g (x) \end{aligned} & \begin{aligned} ^{a} \log f (x) >^{a} \log g (x) \\ \Rightarrow f (x) < g (x) \end{aligned} \end{array}$

Contoh Soal 6 Matriks

$\begin{array}{ll} 26. & (UM UGM 2004) \\ Nilai-nilai x agar matriks \\ (\begin{array}{c} 5 x & 5 \\ 4 & x \end{array}) \\ tidak memiliki invers adalah . . . . \\ \begin{array}{llll} a . & 4 atau 5 \\ b . & - 2 atau 2 \\ c . & - 4 atau 5 \\ d . & - 6 atau 4 \\ e . & 0 \end{array} \\ Jawab : b \\ \begin{aligned} supaya matriks (\begin{array}{c} 5 x & 5 \\ 4 & x \end{array}) \\ tidak memiliki invers, maka \\ determinan matriks (\begin{array}{c} 5 x & 5 \\ 4 & x \end{array}) = 0 \\ Sehingga \\ | \begin{array}{c} 5 x & 5 \\ 4 & x \end{array} | = 0 \\ \Leftrightarrow 5 x^{2} - 20 = 0 \\ \Leftrightarrow x^{2} = 4 \\ \Leftrightarrow x = \pm 2 \end{aligned} \end{array}$

$\begin{array}{ll} 27. & (UM UGM 2005) \\ Matriks (\begin{array}{c} x & 1 \\ - 2 & 1 - x \end{array}) \\ tidak memiliki invers untuk \\ nilai x = . . . . \\ \begin{array}{llll} a . & - 1 atau - 2 \\ b . & - 1 atau 0 \\ c . & - 1 atau 1 \\ d . & - 1 atau 2 \\ e . & 1 atau 2 \end{array} \\ Jawab : d \\ Mirip dengan pembahasan no. 26 \\ \begin{aligned} Nilai | \begin{array}{c} x & 1 \\ - 2 & 1 - x \end{array} | = 0 \\ \Leftrightarrow x - x^{2} - (- 2) = 0 \\ \Leftrightarrow 2 + x - x^{2} = 0 \\ \Leftrightarrow x^{2} - x - 2 = 0 \\ \Leftrightarrow (x - 2) (x + 1) = 0 \\ \Leftrightarrow x = 2 atau x = - 1 \end{aligned} \end{array}$

$\begin{array}{ll} 28. & (Mat Das SIMAK UI 2014) \\ Jika matriks A adalah invers \\ dari matriks \frac{1}{3} . (\begin{array}{c} - 1 & - 3 \\ 4 & 5 \end{array}) dan \\ A (\begin{array}{c} x \\ y \end{array}) = (\begin{array}{c} 1 \\ 3 \end{array}) maka nilai 2 x + y adalah . . . . \\ \begin{array}{llllllll} a . & - \frac{10}{3} \\ b . & - \frac{1}{3} \\ c . & 1 \\ d . & \frac{9}{7} \\ e . & \frac{20}{3} \end{array} \\ Jawab : b \\ \begin{aligned} Misalkan diketahui matriks \\ B = \frac{1}{3} . (\begin{array}{c} - 1 & - 3 \\ 4 & 5 \end{array}), \\ maka A = {(\frac{1}{3} . (\begin{array}{c} - 1 & - 3 \\ 4 & 5 \end{array}))}^{- 1} \\ selanjutnya A (\begin{array}{c} x \\ y \end{array}) = (\begin{array}{c} 1 \\ 3 \end{array}) \\ (\begin{array}{c} x \\ y \end{array}) = A^{- 1} (\begin{array}{c} 1 \\ 3 \end{array}), \\ ingat bahwa {(A^{- 1})}^{- 1} = A \\ (\begin{array}{c} x \\ y \end{array}) = {({(\frac{1}{3} . (\begin{array}{c} - 1 & - 3 \\ 4 & 5 \end{array}))}^{- 1})}^{- 1} (\begin{array}{c} 1 \\ 3 \end{array}) \\ (\begin{array}{c} x \\ y \end{array}) = \frac{1}{3} . (\begin{array}{c} - 1 & - 3 \\ 4 & 5 \end{array}) (\begin{array}{c} 1 \\ 3 \end{array}) \\ (\begin{array}{c} x \\ y \end{array}) = \frac{1}{3} (\begin{array}{c} - 1 - 9 \\ 4 + 15 \end{array}) = (\begin{array}{c} - \frac{10}{3} \\ \frac{19}{3} \end{array}) \\ 2 x + y = 2 (- \frac{10}{3}) + \frac{19}{3} \\ = \frac{- 20 + 19}{3} = - \frac{1}{3} \end{aligned} \end{array}$

Contoh Soal 5 Matriks

$\begin{array}{ll} 21. & (SPMB 2003) \\ Diketahu matriks A = (\begin{array}{c} a & b \\ c & d \end{array}) . \\ Jika A^{t} = A^{- 1} dengan A^{t} \\ adalah transpose matriks A, \\ maka nilai a d - b c = . . . . \\ \begin{array}{lllllll} a . & - 1 atau - \sqrt{2} \\ b . & 1 atau \sqrt{2} \\ c . & - \sqrt{2} atau - \sqrt{2} \\ d . & - 1 atau 1 \\ e . & 1 atau - \sqrt{2} \end{array} \\ Jawab : d \\ \begin{aligned} Diketahui matriks A = (\begin{array}{c} a & b \\ c & d \end{array}) \\ dan A^{t} = A^{- 1}, maka \\ A^{t} = A^{- 1} \\ {(\begin{array}{c} a & b \\ c & d \end{array})}^{t} = \frac{1}{a d - b c} \times Adjoin Matriks A \\ (\begin{array}{c} a & c \\ b & d \end{array}) = \frac{1}{a d - b c} (\begin{array}{c} d & - b \\ - c & a \end{array}), \\ didapatkan hubungan \\ c = \frac{- b}{a d - b c} . . . . . . . . . . . . . . . (1) \\ b = \frac{- c}{a d - b c} . . . . . . . . . . . . . . . (2) \\ Persamaan (2) disubstitusikan ke persamaan (1) \\ c = \frac{- \frac{- c}{a d - b c}}{a d - b c} \\ 1 = (a d - b c)^{2} \\ (a d - b c) = - 1 atau 1 \end{aligned} \end{array}$

$\begin{array}{ll} 22. & Diketahu matriks H \\ yang memenuhi persamaan \\ H (\begin{array}{c} 3 & 2 \\ 1 & 4 \end{array}) = (\begin{array}{c} 7 & 8 \\ 4 & 6 \end{array}), \\ maka nilai dari d e t H adalah . . . . \\ \begin{array}{llllllll} a . & - 3 \\ b . & - 2 \\ c . & - 1 \\ d . & 1 \\ e . & 2 \end{array} \\ Jawab : d \\ \begin{array}{c} Alternatif 1 \\ \begin{aligned} H.A & = B \\ {H.A.A}^{- 1} & = {B.A}^{- 1} \\ H & = {B.A}^{- 1} \\ = (\begin{array}{c} 7 & 8 \\ 4 & 6 \end{array}) . \frac{1}{| \begin{array}{c} 3 & 2 \\ 1 & 4 \end{array} |} (\begin{array}{c} 4 & - 2 \\ - 1 & 3 \end{array}) \\ = (\begin{array}{c} 7 & 8 \\ 4 & 6 \end{array}) . \frac{1}{12 - 2} (\begin{array}{c} 4 & - 2 \\ - 1 & 3 \end{array}) \\ = \frac{1}{10} (\begin{array}{c} 28 + (- 8) & (- 14) + 24 \\ 16 + (- 6) & (- 8) + 18 \end{array}) \\ H & = \frac{1}{10} (\begin{array}{c} 20 & 10 \\ 10 & 10 \end{array}) = (\begin{array}{c} 2 & 1 \\ 1 & 1 \end{array}) \\ d e t H & = | \begin{array}{c} 2 & 1 \\ 1 & 1 \end{array} | = 2.1 - 1.1 = 2 - 1 = 1 \end{aligned} \\ Alternatif 2 \\ \begin{aligned} H.A & = B {\begin{cases} d e t H & = | H | \\ d e t A & = | A | = | \begin{array}{c} 3 & 4 \\ 2 & 1 \end{array} | \\ = 12 - 2 = 10 \\ d e t B & = | B | = | \begin{array}{c} 7 & 8 \\ 4 & 6 \end{array} | \\ = 42 - 32 = 10 \end{cases} \\ | H | . | A | & = | B | \\ | H | & = \frac{| B |}{| A |} \\ = \frac{10}{10} \\ = 1 \end{aligned} \end{array} \end{array}$

$\begin{array}{ll} 23. & (UM UGM 2006) \\ Apabila x dan y memenuhi \\ persamaan matriks \\ (\begin{array}{c} 1 & - 2 \\ - 1 & 3 \end{array}) (\begin{array}{c} x \\ y \end{array}) = (\begin{array}{c} - 1 \\ 2 \end{array}), \\ maka x + y = . . . . \\ \begin{array}{llll} a . & 1 \\ b . & 2 \\ c . & 3 \\ d . & 4 \\ e . & 5 \end{array} \\ Jawab : b \\ \begin{aligned} (\begin{array}{c} 1 & - 2 \\ - 1 & 3 \end{array}) (\begin{array}{c} x \\ y \end{array}) & = (\begin{array}{c} - 1 \\ 2 \end{array}) \\ A . X & = B \\ A^{- 1} . A . X & = A^{- 1} . B \\ A^{0} . X & = A^{- 1} . B \\ X & = A^{- 1} . B \\ (\begin{array}{c} x \\ y \end{array}) & = {(\begin{array}{c} 1 & - 2 \\ - 1 & 3 \end{array})}^{- 1} (\begin{array}{c} - 1 \\ 2 \end{array}) \\ (\begin{array}{c} x \\ y \end{array}) = \frac{1}{| \begin{array}{c} 1 & - 2 \\ - 1 & 3 \end{array} |} & (\begin{array}{c} 3 & 2 \\ 1 & 1 \end{array}) (\begin{array}{c} - 1 \\ 2 \end{array}) \\ (\begin{array}{c} x \\ y \end{array}) = \frac{1}{3 - 2} & (\begin{array}{c} 3. (- 1) + 2.2 \\ 1. (- 1) + 1.2 \end{array}) \\ (\begin{array}{c} x \\ y \end{array}) & = (\begin{array}{c} 1 \\ 1 \end{array}) \\ x + y & = 1 + 1 = 2 \end{aligned} \end{array}$

$\begin{array}{ll} 24. & (KSM Matematika Kabupten 2019) \\ Matriks A dengan entri bulat dan \\ berukuran 2x2, dikalikan dengan matriks \\ (\begin{array}{c} 1 & 2 \\ 2 & 2 \end{array}) dari kanan menghasilkan matriks \\ yang semua entrinya bilangan prima . \\ Jika determinan dari matriks A juga \\ bilangan prima, maka nilai minimum dari \\ d e t A adalah . . . . \\ \begin{array}{llll} a . & 2 \\ b . & 3 \\ c . & 5 \\ d . & 7 \end{array} \\ Jawab : a \\ \begin{aligned} (\begin{array}{c} 1 & 2 \\ 2 & 2 \end{array}) & \times A_{2 \times 2} = (\begin{array}{c} α & β \\ γ & δ \end{array}) \\ | \begin{array}{c} 1 & 2 \\ 2 & 2 \end{array} | & \times | A_{2 \times 2} | = | \begin{array}{c} α & β \\ γ & δ \end{array} | \\ | A_{2 \times 2} | = \frac{| \begin{array}{c} α & β \\ γ & δ \end{array} |}{| \begin{array}{c} 1 & 2 \\ 2 & 2 \end{array} |} \\ | A_{2 \times 2} | = \frac{(α δ - β γ)}{- 2} \\ | A_{2 \times 2} | = \frac{(β γ - α δ)}{2} \\ Karena & | A_{2 \times 2} | bilangan prima \\ akan m & engakibatkan (β γ - α δ) \\ harus h & abis dibagi 2, oleh karenanya \\ menyeb & abkan (β γ - α δ) berupa bilangan \\ genap. & Dan karena (β γ - α δ) genap, \\ maka p & astilah | A_{2 \times 2} | juga bernilai genap \\ sehingg & a nilai | A_{2 \times 2} | pastilah 2 \end{aligned} \end{array}$

$\begin{array}{ll} 25. & (UM UGM 2005) Jika \\ (\begin{array}{c} x & y \end{array}) (\begin{array}{c} \sin α & \cos α \\ - \cos α & \sin α \end{array}) = (\begin{array}{c} \sin A & \cos A \end{array}) \\ dan A suatu konstanta, maka x + y = . . . . \\ \begin{array}{llll} a . & - 2 \\ b . & - 1 \\ c . & 0 \\ d . & 1 \\ e . & 2 \end{array} \\ Jawab : d \\ \begin{aligned} (\begin{array}{c} x & y \end{array}) (\begin{array}{c} \sin α & \cos α \\ - \cos α & \sin α \end{array}) = (\begin{array}{c} \sin A & \cos A \end{array}) \\ (\begin{array}{c} x \sin α - y \cos α & x \cos α + y \sin α \end{array}) = (\begin{array}{c} \sin A & \cos A \end{array}) \\ {\begin{cases} \sin A & = x \sin α - y \cos α = \sqrt{x^{2} + y^{2}} \cos (α - \tan^{- 1} \frac{x}{- y}) \\ \cos A & = x \cos α + y \sin α = \sqrt{x^{2} + y^{2}} \cos (α - \tan^{- 1} \frac{y}{x}) \end{cases} \\ Supaya \cos A = \sqrt{x^{2} + y^{2}} \cos (α - \tan^{- 1} \frac{y}{x}), maka \\ {\begin{cases} \sqrt{x^{2} + y^{2}} & = 1 \\ \tan^{- 1} \frac{y}{x} & = 0 \Rightarrow {\begin{cases} y & = 0 \\ x & = 1 \end{cases} \end{cases} \\ Sehingga x + y = 1 + 0 = 1 \end{aligned} \end{array}$

Contoh Soal 4 Matriks

$\begin{array}{ll} 16. & Determinan untuk matriks (\begin{array}{c} 2 & - 5 \\ 3 & - 1 \end{array}) = . . . . \\ \begin{array}{llllllll} a . & - 17 \\ b . & - 13 \\ c . & 11 \\ d . & 13 \\ e . & 17 \end{array} \\ Jawab : d \\ \begin{aligned} Determinan & dari matriks (\begin{array}{c} 2 & - 5 \\ 3 & - 1 \end{array}) \\ = | \begin{array}{c} 2 & - 5 \\ 3 & - 1 \end{array} | = 2 (- 1) - 3 (- 5) \\ = - 2 + 15 \\ = 13 \end{aligned} \end{array}$

$\begin{array}{ll} 17. & Determinan untuk matriks \\ (\begin{array}{c} 2 & - 1 & - 1 \\ 1 & 4 & - 1 \\ 1 & - 2 & 3 \end{array}) = . . . . \\ \begin{array}{llllllll} a . & 10 \\ b . & 18 \\ c . & 22 \\ d . & 30 \\ e . & 36 \end{array} \\ Jawab : a \\ \begin{aligned} Determinan dari matriks \\ (\begin{array}{c} 2 & - 1 & - 1 \\ 1 & 4 & - 1 \\ 1 & - 2 & 3 \end{array}) \\ = | \begin{array}{c} 2 & - 1 & - 1 \\ 1 & 4 & - 1 \\ 1 & - 2 & 3 \end{array} | \\ = + (2.4 .3) + (- 1. - 1.1) + (- 1.1 . - 2) \\ - (1.4 . - 1) - (- 2. - 1.2) - (3.1 . - 1) \\ = 24 + 1 + 2 + 4 - 24 + 3 \\ = 10 \end{aligned} \end{array}$

$\begin{array}{ll} 18. & Jika diketahu matriks \\ A = (\begin{array}{c} x + 3 & - 2 \\ - 16 & 2 x - 6 \end{array}), \\ maka nilai dari x supaya matriks \\ A tidak memiliki invers adalah . . . . \\ \begin{array}{llllllll} a . & 1 \\ b . & 2 \\ c . & 3 \\ d . & 4 \\ e . & 5 \end{array} \\ Jawab : e \\ \begin{aligned} Invers & dari matriks A adalah A^{- 1} . \\ A^{- 1} & = \frac{1}{d e t A} \times A d j o i n A . \\ Karen & a tidak memiliki invers, \\ maka & d e t A = 0, sehingga \\ d e t A & = | \begin{array}{c} x + 3 & - 2 \\ - 16 & 2 x - 6 \end{array} | = 0 \\ \Leftrightarrow (x + 3) (2 x - 6) - (- 16. - 2) = 0 \\ (masing-masing ruas dibagi 2) \\ \Leftrightarrow (x + 3) (x - 3) - 16 = 0 \\ \Leftrightarrow x^{2} - 9 - 16 = 0 \\ \Leftrightarrow x^{2} - 25 = 0 \\ \Leftrightarrow (x + 5) (x - 5) = 0 \\ \Leftrightarrow x + 5 = 0 atau x - 5 = 0 \\ \Leftrightarrow x = - 5 atau x = 5 \end{aligned} \end{array}$

$\begin{array}{ll} 19. & Jika | \begin{array}{c} 5^{2 x} & - 5 \\ 1 & 1 \end{array} | = {6.5}^{x} \\ maka 5^{2 x} adalah . . . . \\ \begin{array}{llll} a . & 625 atau 1 \\ b . & 25 atau 1 \\ c . & 25 atau 0 \\ d . & 5 atau 1 \\ e . & 5 atau 0 \end{array} \\ Jawab : b \\ \begin{aligned} 5^{2 x} + 5 = {6.5}^{x} \\ 5^{2 x} - {6.5}^{x} + 5 = 0 \\ (5^{x} - 1) (5^{x} - 5) = 0 \\ 5^{x} - 1 = 0 atau 5^{x} - 5 = 0 \\ 5^{x} = 1 atau 5^{x} = 5 \\ 5^{x} = 5^{0} atau 5^{x} = 5^{1} \\ x = 0 atau x = 1 \\ maka \\ 5^{2 x} = {\begin{cases} 5^{2.1} & = 5^{2} = 25 \\ 5^{2.0} & = 5^{0} = 1 \end{cases} \end{aligned} \end{array}$

$\begin{array}{ll} 20. & Diketahu determinan suatu \\ matriks adalah | \begin{array}{c} x & 1 & 2 \\ x & 1 & x \\ 5 & - 3 & 7 \end{array} | = 0. \\ Jika p dan q adalah akar-akar \\ yang memenuhi persamaan tersebut \\ maka nilai dari p + q adalah . . . . \\ \begin{array}{llllllll} a . & - 3 \\ b . & - \frac{1}{3} \\ c . & - 1 \\ d . & \frac{1}{3} \\ e . & 3 \end{array} \\ Jawab : d \\ \begin{aligned} Diketahui ba & hwa : \\ | \begin{array}{c} x & 1 & 2 \\ x & 1 & x \\ 5 & - 3 & 7 \end{array} | & = 0 \\ + (x .1 .7) + & (1. x .5) + (2. x . - 3) \\ - (5.1 .2) & - (- 3. x . x) - (7. x .1) = 0 \\ 7 x + 5 x - 6 x & - 10 + 3 x^{2} - 7 x = 0 \\ 3 x^{2} - x - 10 & = 0 {\begin{cases} p & salah satu akar \\ q & salah satu akar yang lain, \end{cases} \\ dengan & {\begin{cases} a & = 3 \\ b & = - 1 \\ c & = - 10 \end{cases} . \\ maka p + q & = - \frac{b}{a} = - \frac{- 1}{3} \\ = \frac{1}{3} \end{aligned} \end{array}$

Lanjutan 4 Materi Matriks (Matematika Wajib Kelas XI)

$F. Invers Matriks ordo 2x2$

Perhatikanlah kembali materi sebelumnya berkaitan determinan matriks 2x2, yaitu

$\begin{matrix} \begin{aligned} Jika matriks A = (\begin{array}{c} a & b \\ c & d \end{array}) \\ maka determinan matriks A \\ ditentukan dengan \\ d e t A = | \begin{array}{c} a & b \\ c & d \end{array} | = a d - b c \end{aligned} \end{matrix}$

Jika $d e t A$ bernilai tidak sama dengan nol, maka invers matriks ordo 2x2 yang selanjutnya dilambangkan dengan $A^{- 1}$ dapat ditentukan dengan formula:

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

$\begin{aligned} Sebagai CONTOH \\ Diketahui sebuah matrik ordo 2 x 2, yaitu : \\ A = (\begin{array}{c} - 3 & 2 \\ - 1 & 4 \end{array}), maka A^{^{- 1}} adalah : \\ A^{^{- 1}} = \frac{1}{| \begin{array}{c} - 3 & 2 \\ - 1 & 4 \end{array} |} (\begin{array}{c} 4 & - 2 \\ 1 & - 3 \end{array}) \\ = \frac{1}{12 - (- 2)} (\begin{array}{c} 4 & - 2 \\ 1 & - 3 \end{array}) \\ = \frac{1}{14} (\begin{array}{c} 4 & - 2 \\ 1 & - 3 \end{array}) \\ = (\begin{array}{c} \frac{4}{14} & \frac{- 2}{14} \\ \frac{1}{14} & \frac{- 3}{14} \end{array}) \\ = (\begin{array}{c} \frac{2}{7} & - \frac{1}{7} \\ \frac{1}{14} & - \frac{3}{14} \end{array}) \end{aligned}$

$CONTOH SOAL$

$\begin{array}{ll} 1. & Tentukanlah invers matriks berikut \\ a . B = (\begin{array}{c} 5 & - 3 \\ 4 & - 2 \end{array}) \\ b . C = (\begin{array}{c} - 3 & - 5 \\ 6 & 9 \end{array}) \\ c . P = (\begin{array}{c} - 1 & 2 \\ - 3 & 6 \end{array}) \\ d . Q = (\begin{array}{c} 6 & 9 \\ 2 & 3 \end{array}) \\ Jawab : yang dibahas poin a saja \\ B^{- 1} = \frac{1}{d e t B} (\begin{array}{c} - 2 & 3 \\ - 4 & 5 \end{array}) \\ = \frac{1}{- 10 - (- 12)} (\begin{array}{c} - 2 & 3 \\ - 4 & 5 \end{array}) = \frac{1}{2} (\begin{array}{c} - 2 & 3 \\ - 4 & 5 \end{array}) \\ = (\begin{array}{c} - 1 & \frac{3}{2} \\ - 2 & \frac{5}{2} \end{array}) \\ b. Silahkan dicoba sendiri \\ c. Silahkan dicoba sendiri \\ d. Silahkan dicoba sendiri \end{array}$

$\begin{array}{ll} 2. & Diketahui matriks \\ E = (\begin{array}{c} 4 & 2 \\ 5 & 3 \end{array}) \\ a . Tentukanlah \\ (i) E^{- 1} (iii) {(E^{- 1})}^{t} \\ (i) E^{t} (iv) {(E^{t})}^{- 1} \\ b . Dengan menggunakan hasil-hasil \\ pada a. apakah {(E^{- 1})}^{t} = {(E^{t})}^{- 1} \\ Jawab : \\ \begin{aligned} a . (i) E^{- 1} & = \frac{1}{2} (\begin{array}{c} 3 & - 2 \\ - 5 & 4 \end{array}) = (\begin{array}{c} \frac{3}{2} & - 1 \\ - \frac{5}{2} & 2 \end{array}) \\ (ii) E^{t} & = (\begin{array}{c} 4 & 5 \\ 2 & 3 \end{array}) \\ (iii) & {(E^{- 1})}^{t} = (\begin{array}{c} \frac{3}{2} & - \frac{5}{2} \\ - 1 & 2 \end{array}) \\ (iv) & {(E^{t})}^{- 1} = \frac{1}{2} (\begin{array}{c} 3 & - 5 \\ - 2 & 4 \end{array}) = (\begin{array}{c} \frac{3}{2} & - \frac{5}{2} \\ - 1 & 2 \end{array}) \\ b . Dari ha & sil yang didapat dapat disimpulkan \\ {(E^{- 1})}^{t} = {(E^{t})}^{- 1} \end{aligned} \end{array}$

DAFTAR PUSTAKA

Wirodikromo, S. 2003. Matematika 2000 untuk SMU Jilid 2 Kelas 1 Semester 2. Jakarta: ERLANGGA

Lanjutan 3 Materi Matriks (Matematika Wajib Kelas XI)

$E. Determinan Matriks$

$1. Ordo 2x2$

Misalkan A adalah matriks persegi berordo 2x2 dan dituliskan dengan $A = (\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix})$ dengan $a_{11} dan a_{22}$ sebagai elemen dari diagonal utama dan $a_{12} dan a_{21}$ adalah elemen yang menempati diagonal samping, perhatikan lagi matriks A berikut:

$A = (\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix})$

maka determinan dari matriks A yang berordo 2x2 adalah perkalian elemen pada diagonal utama dikurangi dengan hasil kali perkalian diagonal samping dan di tuliskan dengan det A atau tanda |...|. Sehingga dari pengertian tersebut kita dapat menuliskan bahwa determinan dari matriks A dalah:

$det . A = (\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix})$ sama dengan

$det . A = | \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix} | = a_{11} \times a_{22} - a_{12} \times a_{21}$

$\begin{aligned} Sebagai CONTOH \\ Diketahui sebuah matrik ordo 2 x 2, yaitu : \\ A = (\begin{array}{c} - 3 & 2 \\ - 1 & 4 \end{array}), maka d e t A adalah : \\ d e t A = | \begin{array}{c} - 3 & 2 \\ - 1 & 4 \end{array} | = (- 3) \times (- 4) - (- 1) \times (2) \\ = 12 - (- 2) = 12 + 2 = 14 \end{aligned}$

$2. Ordo 3x3$

Ada dua buah cara minimal dalam menentukan determinan matriks ordo 3x3, yaitu:

cara menjabarkan mengikuti baris atau kolom(ekspansi kofaktor)
aturan Sarrus

Adapun penjelasan lebih lanjut adalah sebagai berikut

\begin{aligned} Misalkan diberikan matriks ordo 3 x 3 \\ A = (\begin{array}{c} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}) \end{aligned}

2.1 Menjabarkan mengikuti baris atau kolom

\begin{aligned} det A & = a_{11} | \begin{array}{c} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array} | - a_{12} | \begin{array}{c} a_{21} & a_{23} \\ a_{31} & a_{33} \end{array} | + a_{13} | \begin{array}{c} a_{21} & a_{22} \\ a_{31} & a_{32} \end{array} | \\ Catatan : \\ tanda a_{i j} = positif jika i + j genap \\ tanda a_{i j} = negatif jika i + j ganjil \end{aligned}

Anda juga bisa menjabarkan mengikuti baris yang lain termasuk juga menjabarkan mengikuti kolom. Sehingga total cara menjabarkan ini, karena ada 3 baris dan 3 kolom total akan ada sebanyak 6 cara menentukan determinan dari matriks A tersebut.

2.2 Aturan Sarrus

\begin{aligned} det A & = 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{aligned}

\begin{aligned} Sebagai CONTOH MENJABARKAN \\ Diketahui sebuah matrik ordo 3 x 3, yaitu : \\ A = (\begin{array}{c} 1 & 2 & 3 \\ 1 & 3 & 4 \\ 1 & 4 & 3 \end{array}), maka d e t A dengan \\ menjabarkan baris pertama adalah : \\ d e t A = 1 | \begin{array}{c} 3 & 4 \\ 4 & 3 \end{array} | - 2 | \begin{array}{c} 1 & 4 \\ 1 & 3 \end{array} | + 3 | \begin{array}{c} 1 & 3 \\ 1 & 4 \end{array} | \\ = (9 - 16) - 2 (3 - 4) + 3 (4 - 3) \\ = - 7 + 2 + 3 \\ = - 2 \end{aligned}

\begin{aligned} Dan berikut CONTOH aturan SARRUS \\ Diketahui sebuah matrik ordo 3 x 3, yaitu : \\ B = (\begin{array}{c} 2 & 1 & 3 \\ 3 & 1 & 4 \\ 4 & 1 & 3 \end{array}), maka d e t B dengan \\ metode SARRUS adalah : \\ d e t B = (2.1 .3) + (1.4 .4) + (3.1 .3) \\ - (4.1 .3) - (1.4 .2) - (3.1 .3) \\ = 6 + 16 + 9 - 12 - 8 - 9 = 2 \end{aligned}

CONTOH SOAL

\begin{array}{ll} 1. & Diketahui matriks-matriks persegi berikut \\ a . (\begin{array}{c} 2 & 3 \\ 6 & 7 \end{array}) c . (\begin{array}{c} - 2 & - 3 \\ 6 & 7 \end{array}) \\ b . (\begin{array}{c} 0 & 4 \\ - 3 & 6 \end{array}) d . (\begin{array}{c} \sqrt{3} & 3 \sqrt{3} \\ \sqrt{2} & - 2 \sqrt{2} \end{array}) \\ Tentukanlah determinan dari \\ matriks-matriks persegi di atas \\ Jawab : \\ \begin{array}{ll} \begin{aligned} a . & | \begin{array}{c} 2 & 3 \\ 6 & 7 \end{array} | \\ = (2) . (7) - (3) . (6) \\ = 14 - 18 \\ = - 4 \end{aligned} & \begin{aligned} b . & | \begin{array}{c} 0 & 4 \\ - 3 & 6 \end{array} | \\ = (0) . (6) - (4) . (- 3) \\ = 0 - (- 12) \\ = 12 \end{aligned} \\ \begin{aligned} c . & | \begin{array}{c} - 2 & - 3 \\ 6 & 7 \end{array} | \\ = (- 2) . (7) - (- 3) . (6) \\ = (- 14) - (- 18) \\ = - 14 + 18 \\ = 4 \end{aligned} & \begin{aligned} d . & | \begin{array}{c} \sqrt{3} & 3 \sqrt{3} \\ \sqrt{2} & - 2 \sqrt{2} \end{array} | \\ = (\sqrt{3}) . (- 2 \sqrt{2}) \\ - (3 \sqrt{3}) . (\sqrt{2}) \\ = - 2 \sqrt{6} - 3 \sqrt{6} \\ = - 5 \sqrt{6} \end{aligned} \end{array} \end{array}

\begin{array}{ll} 2. & Tentukanlah nilai x yang memenuhi persamaan \\ | \begin{array}{c} 1 - x & 3 \\ 2 & 3 - x \end{array} | = 2 \\ Jawab : \\ \begin{aligned} | \begin{array}{c} 1 - x & 3 \\ 2 & 3 - x \end{array} | & = 2 \\ (1 - x) (3 - x) - (3) (2) & = 2 \\ 3 - x - 3 x + x^{2} - 6 & = 2 \\ x^{2} - 4 x - 3 & = 2 \\ x^{2} - 4 x - 5 & = 0 \\ (x - 5) (x + 1) & = 0 \\ x - 5 = 0 atau x + 1 & = 0 \\ x = 5 atau x = - 1 \end{aligned} \end{array}

\begin{array}{ll} 3. & Diketahui matriks-matriks persegi berikut \\ (i) . (\begin{array}{c} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 4 \end{array}) (iii) . (\begin{array}{c} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{array}) \\ (ii) . (\begin{array}{c} - 1 & - 2 & - 3 \\ - 2 & 6 & 0 \\ - 3 & 0 & 6 \end{array}) (iv) . (\begin{array}{c} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{array}) \\ Tentukanlah determinan matriks-matriks \\ di atas dengan cara \\ a . S a r r u s \\ b . Menjabarkan baris pertama \\ c . Menjabarkan baris kedua \\ d . Menjabarkan baris ketiga \\ e . Menjabarkan kolom pertama \\ f . Menjabarkan kolom kedua \\ g . Menjabarkan kolom ketiga \end{array}

. \begin{aligned} Jawab : \\ \begin{array}{ll} (i) . (\begin{array}{c} 1 & 1 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 4 \end{array}) & (ii) . (\begin{array}{c} - 1 & - 2 & - 3 \\ - 2 & 6 & 0 \\ - 3 & 0 & 6 \end{array}) \\ (iii) . (\begin{array}{c} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{array}) & (iv) . (\begin{array}{c} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{array}) \\ \begin{aligned} (i) . & | \begin{array}{c} 1 & 1 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 4 \end{array} | \\ = (1) (4) (4) + \\ (1) (5) (3) + \\ (3) (2) (5) + \\ - (3) (4) (3) \\ - (5) (5) (1) \\ - (4) (2) (1) \\ = 16 + 15 + 30 \\ - 36 - 25 - 8 \\ = - 8 \end{aligned} & \begin{aligned} (ii) . & | \begin{array}{c} - 1 & - 2 & - 3 \\ - 2 & 6 & 0 \\ - 3 & 0 & 6 \end{array} | \\ = (- 1) (6) (6) + \\ (- 2) (0) (- 3) + \\ (- 3) (- 2) (0) + \\ - (- 3) (6) (- 3) \\ - (0) (0) (- 1) \\ - (6) (- 2) (- 2) \\ = - 36 + 0 + 0 \\ - 54 - 0 - 24 \\ = - 114 \end{aligned} \\ \begin{aligned} (iii) . & | \begin{array}{c} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{array} | \\ = (1) (5) (9) + \\ (2) (6) (7) + \\ (3) (4) (8) + \\ - (7) (5) (3) \\ - (8) (6) (1) \\ - (9) (4) (2) \\ = 45 + 84 + 96 \\ - 105 - 48 - 72 \\ = 0 \end{aligned} & \begin{aligned} (iv) . & | \begin{array}{c} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{array} | \\ = (2) (2) (2) + \\ (1) (1) (1) + \\ (1) (1) (1) + \\ - (1) (2) (1) \\ - (1) (1) (2) \\ - (2) (1) (1) \\ = 8 + 1 + 1 \\ - 2 - 2 - 2 \\ = 4 \end{aligned} \end{array} \\ yang belum dibahas silahkan dibuat latihan \end{aligned}

DAFTAR PUSTAKA

Wirodikromo, S. 2003. Matematika 2000 untuk SMU Jilid 2 Kelas 1 Semester 2. Jakarta: ERLANGGA

Contoh Soal 3 Matriks

$\begin{array}{ll} 11. & Diketahui matriks \\ A = (\begin{array}{c} ^{a} \log 6 & 2 \\ 1 & a + 3 b \end{array}), \\ B = (\begin{array}{c} 0 & - 5 \\ - 6 & 3 a - 5 b \end{array}), dan \\ C = (\begin{array}{c} ^{a} \log 2 & - \frac{1}{2} \\ - 2 (b + c) & 3 \end{array}), \\ serta I adalah matriks identitas . \\ Jika 2 A + B - 2 C = 2 I, \\ maka nilai 4 a + b + c adalah . . . . \\ \begin{array}{llllllll} a . & 1 \\ b . & 5 \\ c . & 7 \\ d . & 11 \\ e . & 13 \end{array} \\ Jawab : e \\ \begin{aligned} 2 A + B - 2 C = 2 I \\ 2 (\begin{array}{c} ^{a} \log 6 & 2 \\ 1 & a + 3 b \end{array}) + (\begin{array}{c} 0 & - 5 \\ - 6 & 3 a - 5 b \end{array}) \\ - 2 (\begin{array}{c} ^{a} \log 2 & - \frac{1}{2} \\ - 2 (b + c) & 3 \end{array}) = 2 (\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}) \\ (\begin{array}{c} 2.^{a} \log 6 - 2.^{2} \log 2 & 2.2 - 5 - 2 (- \frac{1}{2}) \\ 2.1 - 6 - 2 (- 2 (b + c)) & 2 (a + 3 b) + 3 a - 5 b - 2.3 \end{array}) = (\begin{array}{c} 2 & 0 \\ 0 & 2 \end{array}) \\ {\begin{cases} 2 & = 2.^{a} \log 6 - 2.^{2} \log 2 \\ 0 & = 2.1 - 6 - 2 (- 2 (b + c)) \\ 2 & = 2 (a + 3 b) + 3 a - 5 b - 2.3 \end{cases} \\ \begin{array}{cc} dari persamaan (1) & dari persamaan (2) \\ \begin{aligned} 2.^{a} \log 6 - 2.^{2} \log 2 & = 2 \\ ^{a} \log 6^{2} -^{2} \log 2^{2} & = 2 \\ ^{a} \log \frac{6^{2}}{2^{2}} & = 2 \\ ^{a} \log 9 & = 2 \\ 9 & = a^{2} \\ 3 & = a \\ 12 & = 4 a \end{aligned} & \begin{aligned} 2.1 - 6 - 2 (- 2 (b + c)) & = 0 \\ 2 - 6 + 4 (b + c) & = 0 \\ 4 (b + c) & = 4 \\ b + c & = 1 \\ sehingga diperoleh & , \\ 4 a + b + c = 12 + 1 \\ = 13 \end{aligned} \end{array} \end{aligned} \end{array}$

$\begin{array}{ll} 12. & Jika (\begin{array}{c} - 4 x & 2 y \\ y & x \end{array}) (\begin{array}{c} 2 \\ - 3 \end{array}) = (\begin{array}{c} 2 \\ - 12 \end{array}), \\ maka nilai x y = . . . . \\ \begin{array}{llllllll} a . & - 6 \\ b . & - 3 \\ c . & 2 \\ d . & 3 \\ e . & 6 \end{array} \\ Jawab : a \\ \begin{array}{cc} \begin{aligned} (\begin{array}{c} - 4 x & 2 y \\ y & x \end{array}) (\begin{array}{c} 2 \\ - 3 \end{array}) & = (\begin{array}{c} 2 \\ - 12 \end{array}) \\ (\begin{array}{c} - 4 x .2 + 2 y . - 3 \\ y .2 + x . - 3 \end{array}) & = (\begin{array}{c} 2 \\ - 12 \end{array}) \\ (\begin{array}{c} - 8 x - 6 y \\ - 3 x + 2 y \end{array}) & = (\begin{array}{c} 2 \\ - 12 \end{array}) \\ SPLDV \end{aligned} & \begin{aligned} - 8 x - 6 y & = 2 (\times 1) \\ - 3 x + 2 y & = - 12 (\times 3) \\ menjadi \\ - 8 x - 6 y & = 2 \\ - 9 x + 6 y & = - 36_{+} \\ - - - - & - - - \\ - 17 x & = - 34 \\ x & = 2 \end{aligned} \\ \begin{aligned} - 8 x - 6 y & = 2 \\ - 8 (2) - 6 y & = 2 \\ - 16 - 6 y & = 2 \\ - 6 y & = 2 + 16 \\ - 6 y & = 18 \\ y & = - 3 \\ sehingga \\ x y & = 2. (- 3) = - 6 \end{aligned} \end{array} \end{array}$

$\begin{array}{ll} 13. & Diketahui N = (\begin{array}{c} - 2 & 3 \\ - 1 & 4 \end{array}) \\ dan M = (\begin{array}{c} - 1 & 3 \\ - 1 & 5 \end{array}) . \\ Jika N^{2} = p N - q M, \\ : maka nilai p - q = . . . . \\ \begin{array}{llllllll} a . & 2 \\ b . & 3 \\ c . & 4 \\ d . & 5 \\ e . & 6 \end{array} \\ Jawab : a \\ \begin{aligned} N^{2} = p N - q M \\ (\begin{array}{c} - 2 & 3 \\ - 1 & 4 \end{array}) \times (\begin{array}{c} - 2 & 3 \\ - 1 & 4 \end{array}) = p (\begin{array}{c} - 2 & 3 \\ - 1 & 4 \end{array}) - q (\begin{array}{c} - 1 & 3 \\ - 1 & 5 \end{array}) \\ (\begin{array}{c} - 2. - 2 + 3. - 1 & - 2.3 + 3.4 \\ - 1. - 2 + 4. - 1 & - 1.3 + 4.4 \end{array}) = (\begin{array}{c} - 2 p + q & 3 p - 3 q \\ - p + q & 4 p - 5 q \end{array}) \\ (\begin{array}{c} 4 - 3 & - 6 + 12 \\ 2 - 4 & - 3 + 16 \end{array}) = (\begin{array}{c} - 2 p + q & 3 p - 3 q \\ - p + q & 4 p - 5 q \end{array}) \\ (\begin{array}{c} 1 & 6 \\ - 2 & 13 \end{array}) = (\begin{array}{c} - 2 p + q & 3 p - 3 q \\ - p + q & 4 p - 5 q \end{array}) \\ \begin{array}{cc} \begin{aligned} - 2 p + q & = 1 \\ - p + q & = - 2_{-} \\ - - - - & - - - \\ - p & = 3 \\ p & = - 3 \end{aligned} & \begin{aligned} - p + q & = - 2 \\ - (- 3) + q & = - 2 \\ q & = - 2 - 3 \\ q & = - 5 \\ sehingga & didapatkan \\ p - q & = - 3 - (- 5) \\ = 2 \end{aligned} \end{array} \end{aligned} \end{array}$

$\begin{array}{ll} 14. & Diketahui matriks Z = (\begin{array}{c} - 2 & 6 \\ - 3 & 5 \end{array}) \\ dan f (x) = x^{2} - x . \\ Jika f (Z) = (\begin{array}{c} - 3 p - 8 q & 12 \\ - 6 & - 2 (p + q) \end{array}), \\ maka nilai p^{2} - q^{2} = . . . . \\ \begin{array}{llllllll} a . & 5 \\ b . & 7 \\ c . & 9 \\ d . & 12 \\ e . & 15 \end{array} \\ Jawab : b \\ \begin{aligned} f (Z) & = (\begin{array}{c} - 3 p - 8 q & 12 \\ - 6 & - 2 (p + q) \end{array}) \\ Z^{2} - Z & = (\begin{array}{c} - 3 p - 8 q & 12 \\ - 6 & - 2 (p + q) \end{array}) \\ (\begin{array}{c} - 2 & 6 \\ - 3 & 5 \end{array}) \times (\begin{array}{c} - 2 & 6 \\ - 3 & 5 \end{array}) - (\begin{array}{c} - 2 & 6 \\ - 3 & 5 \end{array}) & = (\begin{array}{c} - 3 p - 8 q & 12 \\ - 6 & - 2 (p + q) \end{array}) \\ (\begin{array}{c} 4 - 18 & - 12 + 30 \\ 6 - 15 & - 18 + 25 \end{array}) - (\begin{array}{c} - 2 & 6 \\ - 3 & 5 \end{array}) & = (\begin{array}{c} - 3 p - 8 q & 12 \\ - 6 & - 2 (p + q) \end{array}) \\ (\begin{array}{c} - 12 & 12 \\ - 6 & 2 \end{array}) & = (\begin{array}{c} - 3 p - 8 q & 12 \\ - 6 & - 2 (p + q) \end{array}) \end{aligned} \\ \begin{aligned} Sehingga \\ - 12 & = - 3 p - 8 q . . . . . . . . . . . . . . . . . (1) \\ - 1 & = p + q . . . . . . . . . . . . . . . . . . . . . . (2) \\ persamaan & (2) ke persamaan (1) \\ - 12 & = - 3 p - 3 q - 5 q = - 3 (p + q) - 5 q \\ - 12 & = - 3 (- 1) - 5 q \\ - 12 & = 3 - 5 q \\ 5 q & = 3 + 12 \\ q & = 3 . . . . . . . . . . . . . . . . . . . . . . . . (3) \\ persamaan & (3) ke persamaan (2) \\ p + q & = - 1 \\ p & = - 1 - q = - 1 - 3 = - 4 \\ p^{2} - q^{2} & = (- 4)^{2} - 3^{2} = 16 - 9 \\ = 7 \end{aligned} \end{array}$

$\begin{array}{ll} 15. & (SBMPTN 2013) \\ Jika A = (\begin{array}{c} 2 & - 1 & 1 \\ a & b & c \end{array}), \\ B = (\begin{array}{c} - 2 & 1 \\ 1 & - 1 \\ 0 & 2 \end{array}) dan \\ A B = (\begin{array}{c} - 5 & 5 \\ 3 & - 3 \end{array}) \\ maka nilai 2 c - a = . . . . \\ \begin{array}{llllllll} a . & 0 \\ b . & 2 \\ c . & 4 \\ d . & 5 \\ e . & 6 \end{array} \\ Jawab : a \\ \begin{aligned} A B = (\begin{array}{c} - 5 & 5 \\ 3 & - 3 \end{array}) \\ (\begin{array}{c} 2 & - 1 & 1 \\ a & b & c \end{array}) (\begin{array}{c} - 2 & 1 \\ 1 & - 1 \\ 0 & 2 \end{array}) = (\begin{array}{c} - 5 & 5 \\ 3 & - 3 \end{array}) \\ (\begin{array}{c} - 5 & 5 \\ - 2 a + b & a - b + 2 c \end{array}) = (\begin{array}{c} - 5 & 5 \\ 3 & - 3 \end{array}) \\ \begin{array}{lllll} - 2 a + b & = 3 \\ a - b + 2 c & = - 3 & + \\ 2 c - a & = 0 \end{array} \end{aligned} \end{array}$

DAFTAR PUSTAKA

Budhi, W.S. 2018. Bupena Matematika SMA/MA Kelas XI Kelompok Wajib. Jakarta: ERLANGGA
Kanginan, M., Terzalgi, Z. 2014. Matematika untuk SMA-MA/SMK Kelas XI (Wajib). Bandung: SEWU.
Sharma, S. N. 2017. Jelajah Matematika 2 SMA Kelas XI Program Wajib. Jakarta: YUDHISTIRA.
Suparmin, S. Malau, A. 2014. Mainstream Matematika Dasar & Matematika IPA untuk Siswa SMA/MA Kelompok IPA. Bandung: YRAMA WIDYA.

Contoh Soal 2 Matriks

$\begin{array}{ll} 6. & Diketahui matriks \\ M = (\begin{array}{c} - 6 & 9 & - 15 \\ 3 & - 6 & 12 \end{array}) \\ dan N = (\begin{array}{c} 2 & - 3 & 5 \\ - 1 & 2 & - 4 \end{array}) . \\ Nilai k yang memenuhi jika \\ M = k N adalah . . . . \\ \begin{array}{llllllll} a . & - \frac{1}{3} \\ b . & \frac{1}{3} \\ c . & - 1 \\ d . & - 3 \\ e . & 3 \end{array} \\ Jawab : d \\ \begin{aligned} Diketahu bahwa \\ M = k N \\ (perkalian suatu matrik dengan skalar) \\ (\begin{array}{c} - 6 & 9 & - 15 \\ 3 & - 6 & 12 \end{array}) \\ = (\begin{array}{c} - 3. 2 & - 3. - 3 & - 3. 5 \\ - 3. - 1 & - 3. 2 & - 3. - 4 \end{array}) \\ = - 3 (\begin{array}{c} 2 & - 3 & 5 \\ - 1 & 2 & - 4 \end{array}) \\ = k (\begin{array}{c} 2 & - 3 & 5 \\ - 1 & 2 & - 4 \end{array}) \\ sehingga dari kesamaan tersebut \\ maka k = - 3 \end{aligned} \end{array}$

$\begin{array}{ll} 7. & Hasil dari (\begin{array}{c} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array}) \times (\begin{array}{c} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{array}) \\ adalah . . . \\ \begin{array}{llllllll} a . & (\begin{array}{c} 22 & 28 \\ 49 & 64 \end{array}) \\ b . & (\begin{array}{c} 22 & 49 \\ 28 & 64 \end{array}) \\ c . & (\begin{array}{c} 64 & 28 \\ 49 & 22 \end{array}) \\ d . & (\begin{array}{c} 2 & 8 & 18 \\ 4 & 15 & 30 \end{array}) \\ e . & (\begin{array}{c} 1 & 4 & 6 \\ 4 & 15 & 30 \end{array}) \end{array} \\ Jawab : a \\ \begin{aligned} {(\begin{array}{c} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array})}_{2 \times 3} \times {(\begin{array}{c} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{array})}_{3 \times 2} \\ = {(\begin{array}{c} 1.1 + 2.3 + 3.5 & 1.2 + 2.4 + 3.6 \\ 4.1 + 5.3 + 6.5 & 4.2 + 5.4 + 6.6 \end{array})}_{2 \times 2} \\ = {(\begin{array}{c} 1 + 6 + 15 & 2 + 8 + 18 \\ 4 + 15 + 30 & 8 + 20 + 36 \end{array})}_{2 \times 2} \\ = {(\begin{array}{c} 22 & 28 \\ 49 & 64 \end{array})}_{2 \times 2} \end{aligned} \end{array}$

$\begin{array}{ll} 8. & Jika diketahui matriks \\ A = (\begin{array}{c} 0 & 1 \\ 3 & 2 \end{array}) . \\ maka hasil dari A^{3} adalah . . . . \\ \begin{array}{llllllll} a . & (\begin{array}{c} 5 & 8 \\ 20 & 22 \end{array}) \\ b . & (\begin{array}{c} 6 & 7 \\ 21 & 20 \end{array}) \\ c . & (\begin{array}{c} 6 & 7 \\ 20 & 22 \end{array}) \\ d . & (\begin{array}{c} 7 & 8 \\ 20 & 23 \end{array}) \\ e . & (\begin{array}{c} 7 & 9 \\ 20 & 23 \end{array}) \end{array} \\ Jawab : b \\ \begin{aligned} Dike & tahui bahwa \\ A & = (\begin{array}{c} 0 & 1 \\ 3 & 2 \end{array}) \\ mak & a \\ A^{2} & = A \times A \\ = (\begin{array}{c} 0 & 1 \\ 3 & 2 \end{array}) \times (\begin{array}{c} 0 & 1 \\ 3 & 2 \end{array}) \\ = (\begin{array}{c} 0 + 3 & 0 + 2 \\ 0 + 6 & 3 + 4 \end{array}) \\ = (\begin{array}{c} 3 & 2 \\ 6 & 7 \end{array}) \\ A^{3} & = A^{2} \times A \\ = (\begin{array}{c} 3 & 2 \\ 6 & 7 \end{array}) \times (\begin{array}{c} 0 & 1 \\ 3 & 2 \end{array}) \\ = (\begin{array}{c} 0 + 6 & 3 + 4 \\ 0 + 21 & 6 + 14 \end{array}) \\ = (\begin{array}{c} 6 & 7 \\ 21 & 20 \end{array}) \end{aligned} \end{array}$

$\begin{array}{ll} 9. & (SBMPTN Mat IPA 2014) \\ Jika A adalah matriks yang berordo \\ 2 \times 2 dan memenuhi \\ (\begin{array}{c} x & 1 \end{array}) \times A \times (\begin{array}{c} x \\ 1 \end{array}) = x^{2} - 5 x + 8, \\ maka matriks A yang mungkin adalah . . . . \\ \begin{array}{llllllll} a . & (\begin{array}{c} 1 & - 5 \\ 8 & 0 \end{array}) \\ b . & (\begin{array}{c} 1 & 5 \\ 8 & 0 \end{array}) \\ c . & (\begin{array}{c} 1 & 8 \\ - 5 & 0 \end{array}) \\ d . & (\begin{array}{c} 1 & 3 \\ - 8 & 8 \end{array}) \\ e . & (\begin{array}{c} 1 & - 3 \\ 8 & - 8 \end{array}) \end{array} \\ Jawab : d \\ \begin{aligned} (\begin{array}{c} x & 1 \end{array}) \times A \times (\begin{array}{c} x \\ 1 \end{array}) & = x^{2} - 5 x + 8 \\ (\begin{array}{c} x & 1 \end{array}) \times (\begin{array}{c} p & q \\ r & s \end{array}) \times (\begin{array}{c} x \\ 1 \end{array}) & = x^{2} - 5 x + 8 \\ (\begin{array}{c} x p + r & x q + s \end{array}) \times (\begin{array}{c} x \\ 1 \end{array}) & = x^{2} - 5 x + 8 \\ (\begin{array}{c} x^{2} p + x r + x q + s \end{array}) & = x^{2} - 5 x + 8 \\ p x^{2} + (q + r) x + s & = x^{2} - 5 x + 8 \end{aligned} \\ \begin{aligned} {\begin{cases} p & = 1 \\ q + r & = - 5 \\ s & = 8 \end{cases} \Rightarrow (\begin{array}{c} 1 & . . . \\ . . . & 8 \end{array}) \\ Sehingga yang paling mungkin \\ adalah (\begin{array}{c} 1 & 3 \\ - 8 & 8 \end{array}) \end{aligned} \end{array}$

$\begin{array}{ll} 10. & Diketahui \\ (\begin{array}{c} ^{x} \log a & \log (2 a - 6) \\ \log (b - 2) & 1 \end{array}) = (\begin{array}{c} \log b & 1 \\ \log a & 1 \end{array}) \\ maka nilai x adalah . . . . \\ \begin{array}{llllllll} a . & 1 \\ b . & 2 \\ c . & 4 \\ d . & 6 \\ e . & 8 \end{array} \\ Jawab : e \\ \begin{aligned} (\begin{array}{c} ^{x} \log a & \log (2 a - 6) \\ \log (b - 2) & 1 \end{array}) = (\begin{array}{c} \log b & 1 \\ \log a & 1 \end{array}) \\ maka \\ {\begin{cases} ^{x} \log a & = \log b . . . . . . . . . (1) \\ \log (2 a - 6) & = 1 . . . . . . . . . . . . . . (2) \\ \log (b - 2) & = \log a . . . . . . . . . (3) \end{cases} \\ Sehingga dari persamaan (2) \\ akan didapatkan \\ \log (2 a - 6) = 1 = \log 10 \\ (2 a - 6) = 10 \\ a = 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . (4) \\ persamaan (4) ke persamaan (3), \\ maka \\ \log (b - 2) = \log a \\ b - 2 = a = 8 \\ b = 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (5) \\ Selanjutnya dari persamaan (5) \\ akan diperoleh \\ ^{x} \log a = \log b \\ ^{x} \log 8 = \log 10 = 1 \\ x^{1} = 8 \\ \Leftrightarrow x = 8 \end{aligned} \end{array}$

Contoh Soal 1 Matriks

$\begin{array}{l} 1. & Diketahui matriks \\ A = (\begin{array}{c} 2020 & - 4 & - 3 & 2 \\ 2020 & - 6 & - 7 & 1 \\ 2020 & 4 & - 3 & 0 \\ 2020 & 6 & - 7 & 8 \end{array}) \\ Ordo dari matriks A adalah . . . . \\ \begin{array}{llllllll} a . & 3 \times 2 \\ b . & 3 \times 3 \\ c . & 3 \times 4 \\ d . & 4 \times 3 \\ e . & 4 \times 4 \end{array} \\ Jawab : e \\ Cukup jelas \\ Karena matriknya mengandung \\ 4 baris \times 4 kolom \end{array}$

$\begin{array}{ll} 2. & Diketahui matriks \\ B = (\begin{array}{c} 1 & 2 & 3 & 2020 \\ 5 & 13 & 7 & 2019 \\ 11 & 14 & 3 & - 2018 \\ - 15 & 6 & 17 & 2017 \end{array}) \\ Jika b_{i j} menunjukkan elemen \\ yang terletak pada baris ke - i \\ dan kolom ke - j pada matriks B \\ di atas, maka b_{43} = . . . . \\ \begin{array}{llllllll} a . & 3 \\ b . & 9 \\ c . & - 1 \\ d . & 3 \\ e . & 17 \end{array} \\ Jawab : e \\ \begin{aligned} Perh & atikan bahwa \\ B_{4 \times 4} & = (\begin{array}{c} b_{11} & b_{12} & b_{13} & b_{14} \\ b_{21} & b_{22} & b_{23} & b_{24} \\ b_{31} & b_{32} & b_{33} & b_{34} \\ b_{41} & b_{42} & b_{43} & b_{44} \end{array}) \\ = (\begin{array}{c} 1 & 2 & 3 & 2020 \\ 5 & 13 & 7 & 2019 \\ 11 & 14 & 3 & - 2018 \\ - 15 & 6 & 17 & 2017 \end{array}) \\ sehi & ngga entri b_{43} = 17 \end{aligned} \end{array}$

$\begin{array}{ll} 3. & Diketahui matriks C adalah matriks \\ berordo 3 \times 3. Jika c_{i j} = 4 j - 5 i, \\ maka matriks C tersebut adalah . . . . \\ \begin{array}{llllllll} a . & (\begin{array}{c} - 1 & 3 & 7 \\ - 6 & - 2 & 2 \\ - 11 & - 7 & - 3 \end{array}) \\ b . & (\begin{array}{c} - 1 & 7 & 3 \\ - 6 & 2 & - 2 \\ - 7 & - 11 & - 3 \end{array}) \\ c . & (\begin{array}{c} - 1 & - 7 & - 11 \\ - 6 & 7 & 3 \\ - 2 & 2 & - 3 \end{array}) \\ d . & (\begin{array}{c} - 1 & - 6 & - 11 \\ 3 & - 2 & 2 \\ 7 & 2 & - 3 \end{array}) \\ e . & (\begin{array}{c} - 1 & - 2 & - 3 \\ 3 & - 6 & - 11 \\ 7 & - 7 & 2 \end{array}) \end{array} \\ Jawab : a \\ \begin{aligned} Dike & tahui bahwa c_{i j} = 4 j - 5 i, \\ mak & a \\ C_{3 \times 3} & = (\begin{array}{c} c_{11} & c_{12} & c_{13} \\ c_{21} & c_{22} & c_{23} \\ c_{31} & c_{32} & c_{33} \end{array}) \\ = (\begin{array}{c} 4.1 - 5.1 & 4.2 - 5.1 & 4.3 - 5.1 \\ 4.1 - 5.2 & 4.2 - 5.2 & 4.3 - 5.2 \\ 4.1 - 5.3 & 4.2 - 5.3 & 4.3 - 5.3 \end{array}) \\ = (\begin{array}{c} 4 - 5 & 8 - 5 & 12 - 5 \\ 4 - 10 & 8 - 10 & 12 - 10 \\ 4 - 15 & 8 - 15 & 12 - 15 \end{array}) \\ = (\begin{array}{c} - 1 & 3 & 7 \\ - 6 & - 2 & 2 \\ - 11 & - 7 & - 3 \end{array}) \end{aligned} \end{array}$

$\begin{array}{ll} 4. & Jika diketahui matriks \\ X = (\begin{array}{c} - 7 & 9 & - 1 \\ 4 & - 6 & 15 \end{array}) . \\ maka transpose matriks X adalah . . . . \\ \begin{array}{llllllll} a . & X^{t} = (\begin{array}{c} 4 & - 6 & 15 \\ - 7 & 9 & - 1 \end{array}) \\ b . & X^{t} = (\begin{array}{c} 15 & - 6 & 4 \\ - 1 & 9 & - 7 \end{array}) \\ c . & X^{t} = (\begin{array}{c} - 7 & 4 \\ 9 & - 6 \\ - 1 & 15 \end{array}) \\ d . & X^{t} = (\begin{array}{c} 4 & - 7 \\ - 6 & 9 \\ 15 & - 1 \end{array}) \\ e . & X^{t} = (\begin{array}{c} 15 & - 1 \\ - 6 & 9 \\ 4 & - 7 \end{array}) \end{array} \\ Jawab : c \\ \begin{aligned} Dike & tahui bahwa \\ X_{2 \times 3} & = (\begin{array}{c} x_{11} & x_{12} & x_{13} \\ x_{21} & x_{22} & x_{23} \end{array}) \\ = (\begin{array}{c} - 7 & 9 & - 1 \\ 4 & - 6 & 15 \end{array}) \\ maka \\ X_{3 \times 2}^{t} & = (\begin{array}{c} x_{11} & x_{21} \\ x_{12} & x_{22} \\ x_{13} & x_{23} \end{array}) = (\begin{array}{c} - 7 & 4 \\ 9 & - 6 \\ - 1 & 15 \end{array}) \\ adal & ah sebuah matriks baru \\ deng & an ordo 3 \times 2 \end{aligned} \end{array}$

$\begin{array}{ll} 5. & Diketahui matriks P = (\begin{array}{c} a & 4 \\ 2 b & 3 c \end{array}) \\ dan Q = (\begin{array}{c} 2 c - 3 b & 2 a + 1 \\ a & b + 7 \end{array}) . \\ Nilai c yang memenuhi jika \\ P = 2 Q^{t} adalah . . . . \\ \begin{array}{llllllll} a . & - 2 \\ b . & 3 \\ c . & 5 \\ d . & 8 \\ e . & 10 \end{array} \\ Jawab : d \\ \begin{aligned} P & = 2 Q^{t} \\ (\begin{array}{c} a & 4 \\ 2 b & 3 c \end{array}) & = 2 {(\begin{array}{c} 2 c - 3 b & 2 a + 1 \\ a & b + 7 \end{array})}^{t} \\ (\begin{array}{c} a & 4 \\ 2 b & 3 c \end{array}) & = 2 (\begin{array}{c} 2 c - 3 b & a \\ 2 a + 1 & b + 7 \end{array}) \\ (\begin{array}{c} a & 4 \\ 2 b & 3 c \end{array}) & = (\begin{array}{c} 4 c - 6 b & 2 a \\ 4 a + 2 & 2 b + 14 \end{array}) \\ (kesamaan 2 buah matriks) \\ akibat & nya \\ {\begin{cases} a & = 4 c - 6 b . . . . . . . . . . . . . . . . . . (1) \\ 4 & = 2 a . . . . . . . . . . . . . . . . . . . . . . . . (2) \\ 2 b & = 4 a + 2 . . . . . . . . . . . . . . . . . . . . . . (3) \\ 3 c & = 2 b + 14 . . . . . . . . . . . . . . . . . . . . . . (4) \end{cases} \\ dari & persamaan (2) \\ 2 a = 4 \Rightarrow a = 2 . . . . (5) \\ pers & amaan (5) hasilnya \\ disu & bstitusikan ke persamaan (3), \\ yait & u \\ 2 b & = 4 a + 2 \Rightarrow 2 b = 4 (2) + 2 = 10 \\ b & = 5 . . . . . . . . . . . . . . . . . . . . . (6) \\ pers & amaan (6) hasilnya disbstitusikan \\ ke p & ersamaan \\ (4), & dan akan mendapatkan \\ 3 c & = 2 b + 14 \Rightarrow 3 c = 2 (5) + 14 = 24 \\ c & = 8 \end{aligned} \end{array}$

Lanjutan 2 Materi Matriks (Matematika Wajib Kelas XI)

$C. Tarnspose dan Kesamaan Dua Buah Matriks$

$\begin{array}{cl} 1. & Transpose Matriks \\ \begin{aligned} Membentuk matriks baru dari matriks \\ dengan cara mengubah baris matriks ke - i \\ menjadi kolom ke - i, pada matriks baru \\ dan demikian pula untuk kolomnya . Jika \\ matriks pertama adalah A maka matriks \\ transposenya adalah A^{'} atau A^{t} \end{aligned} \\ 2. & Kesamaan Duan Buah Matriks \\ \begin{aligned} Misalkan matriks A = (a_{i j}) dan B = (b_{i j}) \\ adalah dua buah matriks berordo sama, \\ maka matriks A dikatakan sama dengan matriks B \\ jika elemen-elemen yang seletak sama pada \\ kedua matriks tersebut bernilai sama \end{aligned} \end{array}$

$\begin{array}{l} Berikut contoh transpose \\ A = (\begin{array}{c} 1 & 2 \\ - 7 & 0 \\ 5 & 4 \end{array}) \Rightarrow A^{t} = (\begin{array}{c} 1 & - 7 & 5 \\ 2 & 0 & 4 \end{array}) \\ DAn berikut contoh kesamaan dua matriks \\ A = (\begin{array}{c} 1 & 2 \\ - 7 & 0 \\ 5 & 4 \end{array}), D = (\begin{array}{c} 1 & 2 \\ - 7 & 0 \\ 5 & 4 \end{array}), \Rightarrow A = D \end{array}$

$D. Operasi Matriks$

$\begin{array}{clll} No & Operasi & Ketentuan & Contoh \\ Matriks \\ 1 & Penjumlahan & & ordo sama & A = (\begin{array}{c} 1 \\ 2 \end{array}), B = (\begin{array}{c} 8 \\ 9 \end{array}), maka \\ 2 & Pengurangan & ordo sama & A + B = (\begin{array}{c} 1 + 8 \\ 2 + 9 \end{array}) = (\begin{array}{c} 9 \\ 11 \end{array}) \\ 3 & Perkalian & Dengan & k (\begin{array}{c} p & q \\ r & s \end{array}) = (\begin{array}{c} k p & k q \\ k r & k s \end{array}) \\ Skalar & mengalikan \\ ke setiap elemen \\ 4 & Perkalian & \begin{aligned} Dua matriks \\ dapat dikalikan \\ jika \\ banyaknya kolom \\ matriks pertama \\ sama dengan \\ banyaknya baris \\ matriks kedua \end{aligned} & \begin{aligned} E = & (\begin{array}{c} 1 & 2 \\ 3 & - 1 \end{array}), F = (\begin{array}{c} 5 \\ 0 \end{array}), \\ maka \\ E & \times F \\ = {(\begin{array}{c} 1 & 2 \\ 3 & - 1 \end{array})}_{2 \times 2} \times {(\begin{array}{c} 5 \\ 0 \end{array})}_{2 \times 1} \\ syarat memenuhi yaitu : \\ kolom matriks 1 \\ = baris matriks 2 \\ dan hasilnya adalah \\ matriks baru \\ dengan ordo \\ banyak baris matriks 1 \\ kali banyak \\ kolom matriks 2 \\ Dan aturan perkaliannya \\ adalah \\ elemen baris matriks 1 kali \\ elemen kolom matriks 2 \\ sehingga \\ = {(\begin{array}{c} 1 (5) + 2 (0) \\ 3 (5) + - 1 (0) \end{array})}_{2 \times 1} \\ = (\begin{array}{c} 5 + 0 \\ 15 - 0 \end{array}) = {(\begin{array}{c} 5 \\ 15 \end{array})}_{2 \times 1} \end{aligned} \end{array}$

$CONTOH SOAL$

$\begin{array}{ll} ∙ & Penjumlahan \\ \begin{aligned} (\begin{array}{c} 1 & 2 \\ 3 & - 4 \end{array}) + (\begin{array}{c} 5 & 6 \\ 7 & - 8 \end{array}) \\ = (\begin{array}{c} 1 + 5 & 2 + 6 \\ 3 + 7 & (- 4) + (- 8) \end{array}) \\ = (\begin{array}{c} 6 & 8 \\ 10 & - 12 \end{array}) \end{aligned} \\ ∙ & Lawan suatu matriks \\ \begin{aligned} Jika A = (\begin{array}{c} 1 & 2 \\ 3 & - 4 \end{array}), \\ maka lawan matriks A adalah -A, \\ Sehingga - A = (\begin{array}{c} - 1 & - 2 \\ - 3 & 4 \end{array}) \end{aligned} \\ ∙ & Pengurangan \\ \begin{aligned} (\begin{array}{c} 1 & 2 \\ 3 & - 4 \end{array}) - (\begin{array}{c} 5 & 6 \\ 7 & - 8 \end{array}) \\ = (\begin{array}{c} 1 - 5 & 2 - 6 \\ 3 - 7 & (- 4) - (- 8) \end{array}) \\ = (\begin{array}{c} - 4 & - 4 \\ - 4 & 4 \end{array}) \end{aligned} \\ ∙ & Perkalian \\ \begin{aligned} (1) Perkalian suatu matriks dengan skalar k \\ 2 \times (\begin{array}{c} 1 & 2 \\ 3 & - 4 \end{array}) = (\begin{array}{c} 2 \times 1 & 2 \times 2 \\ 2 \times 3 & 2 \times (- 4) \end{array}) \\ = (\begin{array}{c} 2 & 4 \\ 6 & - 8 \end{array}) \end{aligned} \\ \begin{aligned} (2) Perkalian antara dua buah matriks \\ (\begin{array}{c} 1 & 2 \\ 3 & - 4 \end{array}) \times (\begin{array}{c} 5 & 6 \\ 7 & - 8 \end{array}) \\ perhatikan syarat memenuhi \\ = (\begin{array}{c} 1 \times 5 + 2 \times 7 & 1 \times 6 + 2 \times (- 8) \\ 3 \times 5 + (- 4) \times 7 & 3 \times 6 + (- 4) \times (- 8) \end{array}) \\ = (\begin{array}{c} 5 + 14 & 6 + (- 16) \\ 15 + (- 28) & 18 + 32 \end{array}) \\ = (\begin{array}{c} 19 & - 10 \\ - 13 & 50 \end{array}) \end{aligned} \end{array}$

Lanjutan 1 Materi Matriks (Matematika Wajib Kelas XI)

$B. Jenis-Jenis Matriks$

$\begin{array}{lllll} No & Jenis & Keterangan & Contoh & Ordo \\ 1. & Matriks & Matriks yang elemen & (\begin{array}{c} 1 & 3 & - 5 \end{array}) & 1 \times 3 \\ Baris & penyusunnya satu baris saja \\ 2. & Matriks & Matriks yang elemen & (\begin{array}{c} 5 \\ - 5 \\ 2 \end{array}) & 3 \times 1 \\ Kolom & penyusunnya tepat satu kolom saja \\ 3. & Matriks & Matriks yang semua elemennya & (\begin{array}{c} 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}) & 2 \times 3 \\ Nol & adalah bilangan nol \\ 4. & Matriks & Matriks yang jumlah & (\begin{array}{c} 2 & 8 \\ 6 & 1 \end{array}) & 2 \times 2 \\ Persegi & baris dan kolomnya sama \end{array}$

$\begin{array}{lllll} No & Jenis & Keterangan & Contoh & Ordo \\ 5. & Matriks & Matriks Persegi yang semua & (\begin{array}{c} 1 & 0 \\ 0 & 7 \end{array}) & 2 \times 2 \\ Diagonal & elemennya nol kecuali pada & (\begin{array}{c} 1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 6 \end{array}) & 3 \times 3 \\ diagonal utama \\ 6. & Matriks & Matriks yang elemen semuanya & (\begin{array}{c} 5 \\ - 5 \\ 2 \end{array}) & 3 \times 1 \\ Identitas & nol kecuali pada diagonal \\ utama berupa angka 1 & (\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}) & 2 \times 2 \end{array}$

$\begin{array}{lllll} No & Jenis & Keterangan & Contoh & Ordo \\ 7. & Matriks & matriks persegi yang semua elemen & (\begin{array}{c} 5 & 8 \\ 0 & 7 \end{array}) & 2 \times 3 \\ Segitiga & di bawah diagonal utama berupa & (\begin{array}{c} 1 & 2 & 3 \\ 0 & 3 & 5 \\ 0 & 0 & 6 \end{array}) & 3 \times 3 \\ atas & bilangan nol \\ 8. & Matriks & Matriks persegi yang semua elemen & (\begin{array}{c} 1 & 0 & 0 \\ 4 & 2 & 0 \\ 5 & 6 & 3 \end{array}) & 3 \times 3 \\ segitiga & di bawah diagonal utama berupa \\ bawah & angka nol & (\begin{array}{c} 1 & 0 \\ 6 & 2 \end{array}) & 2 \times 2 \end{array}$

$\begin{array}{lllll} No & Jenis & Keterangan & Contoh & Ordo \\ 9. & Matriks & Suatu matriks disebut sebagai matriks & (\begin{array}{c} 5 & 0 \\ 8 & 7 \end{array}) & 2 \times 2 \\ Simetris & simetris jika dan hanya jika elemen-elemen \\ utama & yang letaknya simetris terhadap diagonal \\ atau bernilai sama \end{array}$