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Exercise Zone : Matriks [3]

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Idham Muqoddas
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Berikut ini adalah kumpulan soal mengenai Matriks. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.

Tipe:

StandarSNBTHOTS


No.

Diketahui matriks ${A=\begin{pmatrix}1&2\\3&7\end{pmatrix}}$, ${B=\begin{pmatrix}a&b\\2&3\end{pmatrix}}$, dan ${A^T\cdot B^T=\begin{pmatrix}13&11\\30&25\end{pmatrix}}$ maka nilai 12a + 9b =
  1. 44
  2. 46
  3. 48
  1. 50
  2. 52
SKMP Juli 2020
ALTERNATIF PENYELESAIAN
$\begin{aligned} A^T\cdot B^T&=\begin{pmatrix}13&11\\30&25\end{pmatrix}\\ \begin{pmatrix}1&3\\2&7\end{pmatrix}\begin{pmatrix}a&2\\b&3\end{pmatrix}&=\begin{pmatrix}13&11\\30&25\end{pmatrix}\\ \begin{pmatrix}a+3b&11\\2a+7b&25\end{pmatrix}&=\begin{pmatrix}13&11\\30&25\end{pmatrix} \end{aligned}$

$\begin{aligned} 2a+7b&=30\\ a+3b&=13\qquad&{\color{red}\times2} \end{aligned}$

$\begin{aligned} 2a+7b&=30\\ 2a+2b&=26\qquad&{\color{red}-}\\\hline b&=4 \end{aligned}$
$\begin{aligned} a+3b&=13\\ a+3(4)&=13\\ a+12&=13\\ a&=1 \end{aligned}$

$\begin{aligned} 12a+9b&=12(1)+9(4)\\ &=12+36\\ &=\boxed{\boxed{48}} \end{aligned}$
Jadi, 12a + 9b = 48.
JAWAB: C

No.

Diketahui matriks $A=\begin{pmatrix}2&-1\\1&4\end{pmatrix}$, $B=\begin{pmatrix}x+y&2\\3&y\end{pmatrix}$, dan $C=\begin{pmatrix}7&2\\3&1\end{pmatrix}$. apabila B − A = C' dan C' = transpose matriks C, maka nilai x⋅y = ....
  1. 10
  2. 15
  1. 20
  2. 25
  1. 30
Big Book Matematika SMA
ALTERNATIF PENYELESAIAN
\(\begin{aligned} B-A&=C'\\ \begin{pmatrix}x+y&2\\3&y\end{pmatrix}-\begin{pmatrix}2&-1\\1&4\end{pmatrix}&=\begin{pmatrix}7&3\\2&1\end{pmatrix}\\[4pt] \begin{pmatrix}x+y-2&3\\2&y-4\end{pmatrix}&=\begin{pmatrix}7&3\\2&1\end{pmatrix} \end{aligned}\)

\(\begin{aligned} y-4&=1\\ y&=5 \end{aligned}\)

\(\begin{aligned} x+y-2&=7\\ x+5-2&=7\\ x+3&=7\\ x&=4 \end{aligned}\)

x⋅y = 4⋅5 = 20
Jadi, x⋅y = 20.
JAWAB: C

No.

Nyatakan matriks $\begin{pmatrix}6&3\\0&8\end{pmatrix}$ sebagai kombinasi linear dari matriks berikut:
$\begin{pmatrix}1&2\\-1&3\end{pmatrix}$, $\begin{pmatrix}0&1\\2&4\end{pmatrix}$, dan $\begin{pmatrix}4&-2\\0&-2\end{pmatrix}$.
Grup Telegram
ALTERNATIF PENYELESAIAN
$\begin{aligned} \begin{pmatrix}6&3\\0&8\end{pmatrix}&=k_1\begin{pmatrix}1&2\\-1&3\end{pmatrix}+k_2\begin{pmatrix}0&1\\2&4\end{pmatrix}+k_3\begin{pmatrix}4&-2\\0&-2\end{pmatrix}\\ &=\begin{pmatrix}k_1+4k_3&2k_1+k_2-2k_3\\-k_1+2k_2&3k_1+4k_2-2k_3\end{pmatrix} \end{aligned}$

$\begin{aligned} k_1&&&+&4k_3&=6\\ 2k_1&+&k_2&-&2k_3&=3\\ -k_1&+&2k_2&&&=0\\ 3k_1&+&4k_2&-&2k_3&=8 \end{aligned}$
$\begin{aligned} \begin{pmatrix}1&0&4&6\\2&1&-2&3\\-1&2&0&0\\3&4&-2&8\end{pmatrix}&\begin{aligned}-2b_1+b_2\\b_1+b_3\\-3b_1+b_3\end{aligned}\\ \begin{pmatrix}1&0&4&6\\0&1&-10&-9\\0&2&4&6\\0&4&-14&-10\end{pmatrix}&\begin{aligned}\dfrac12b_3\\\dfrac12b_4\end{aligned}\\ \begin{pmatrix}1&0&4&6\\0&1&-10&-9\\0&1&2&3\\0&2&-7&-5\end{pmatrix}&\begin{aligned}-b_2+b_3\\-2b_2+b_4\end{aligned}\\ \begin{pmatrix}1&0&4&6\\0&1&-10&-9\\0&0&12&12\\0&0&13&13\end{pmatrix}&\begin{aligned}\dfrac1{12}b_3\\\dfrac1{13}b_4\end{aligned}\\ \begin{pmatrix}1&0&0&2\\0&1&0&1\\0&0&1&1\\0&0&0&0\end{pmatrix} \end{aligned}$
k1 = 2, k2 = 1, k3 = 1
Jadi, $\begin{pmatrix}6&3\\0&8\end{pmatrix}=2\begin{pmatrix}1&2\\-1&3\end{pmatrix}+\begin{pmatrix}0&1\\2&4\end{pmatrix}+\begin{pmatrix}4&-2\\0&-2\end{pmatrix}$.

No.

Diketahui $A=\begin{pmatrix}-3&2\\1&0\end{pmatrix}$, $B=\begin{pmatrix}4&-1\\2&5\end{pmatrix}$ dan $C=\begin{pmatrix}-2&-2\\3&3\end{pmatrix}$, maka A − (B − C) = ...
  1. $\begin{pmatrix}1&5\\0&1\end{pmatrix}$
  2. $\begin{pmatrix}4&3\\-3&2\end{pmatrix}$
  3. $\begin{pmatrix}2&1\\3&-2\end{pmatrix}$
  1. $\begin{pmatrix}-9&1\\2&-2\end{pmatrix}$
  2. $\begin{pmatrix}-3&1\\-3&2\end{pmatrix}$
Grup Telegram
ALTERNATIF PENYELESAIAN
$\begin{aligned} A-(B-C)&=\begin{pmatrix}-3&2\\1&0\end{pmatrix}-\left(\begin{pmatrix}4&-1\\2&5\end{pmatrix}-\begin{pmatrix}-2&-2\\3&3\end{pmatrix}\right)\\ &=\begin{pmatrix}-3&2\\1&0\end{pmatrix}-\begin{pmatrix}6&1\\-1&2\end{pmatrix}\\ &=\boxed{\boxed{\begin{pmatrix}-9&1\\2&-2\end{pmatrix}}} \end{aligned}$
Jadi, $A-(B-C)=\begin{pmatrix}-9&1\\2&-2\end{pmatrix}$.
JAWAB: D

No.

Diketahui $P=\begin{pmatrix}-3&1\\4&-2\end{pmatrix}$ dan $Q=\begin{pmatrix}3&4\\1&2\end{pmatrix}$. Hitunglah (PQ)−1
Grup Telegram
ALTERNATIF PENYELESAIAN
$\begin{aligned} PQ&=\begin{pmatrix}-3&1\\4&-2\end{pmatrix}\begin{pmatrix}3&4\\1&2\end{pmatrix}\\ &=\begin{pmatrix}-8&-10\\10&12\end{pmatrix}\\ \left(PQ\right)^{-1}&=\dfrac1{(-8)(12)-(-10)(10)}\begin{pmatrix}12&10\\-10&-8\end{pmatrix}\\ &=\dfrac1{-96+100}\begin{pmatrix}12&10\\-10&-8\end{pmatrix}\\ &=\dfrac14\begin{pmatrix}12&10\\-10&-8\end{pmatrix}\\ &=\boxed{\boxed{\begin{pmatrix}3&\dfrac52\\-\dfrac52&-2\end{pmatrix}}} \end{aligned}$
Jadi, $\left(PQ\right)^{-1}=\begin{pmatrix}3&\dfrac52\\-\dfrac52&-2\end{pmatrix}$.

No.

X adalah matriks persegi berordo 2×2 yang memenuhi $X\begin{pmatrix}1&2\\3&4\end{pmatrix}=\begin{pmatrix}4&8\\5&8\end{pmatrix}$. Matriks X adalah
  1. $\begin{pmatrix}3&2\\-2&1\end{pmatrix}$
  2. $\begin{pmatrix}3&2\\2&1\end{pmatrix}$
  3. $\begin{pmatrix}-4&0\\-1&-2\end{pmatrix}$
  1. $\begin{pmatrix}4&0\\2&1\end{pmatrix}$
  2. $\begin{pmatrix}4&0\\-1&2\end{pmatrix}$
Grup Telegram
ALTERNATIF PENYELESAIAN
$\begin{aligned} X\begin{pmatrix}1&2\\3&4\end{pmatrix}&=\begin{pmatrix}4&8\\5&8\end{pmatrix}\\ X&=\begin{pmatrix}4&8\\5&8\end{pmatrix}\begin{pmatrix}1&2\\3&4\end{pmatrix}^{-1}\\ &=\begin{pmatrix}4&8\\5&8\end{pmatrix}\cdot\dfrac1{1\cdot4-2\cdot3}\begin{pmatrix}4&-2\\-3&1\end{pmatrix}\\ &=\begin{pmatrix}4&8\\5&8\end{pmatrix}\cdot\dfrac1{-2}\begin{pmatrix}4&-2\\-3&1\end{pmatrix}\\ &=\dfrac1{-2}\begin{pmatrix}4&8\\5&8\end{pmatrix}\begin{pmatrix}4&-2\\-3&1\end{pmatrix}\\ &=\dfrac1{-2}\begin{pmatrix}-8&0\\-4&-2\end{pmatrix}\\ &=\boxed{\boxed{\begin{pmatrix}4&0\\2&1\end{pmatrix}}} \end{aligned}$
Jadi, $X=\begin{pmatrix}4&0\\2&1\end{pmatrix}$
JAWAB: D

No.

Diberikan sistem persamaan linear berikut:
3x − 4y − 3z = 12
−2x + 7y − 6z = 9
5x + 8y − z = −10
Nyatakanlah matriks koefisien sistem persamaan linear tersebut
Grup Telegram
ALTERNATIF PENYELESAIAN
$\begin{pmatrix}3&4&-3\\-2&7&-6\\5&8&-1\end{pmatrix}$
Jadi, matriks koefisien sistem persamaan linear tersebut adalah $\begin{pmatrix}3&4&-3\\-2&7&-6\\5&8&-1\end{pmatrix}$.

No.

Diketahui matriks $A=\begin{pmatrix}5&2\\0&3\end{pmatrix}$ dan $B=\begin{pmatrix}-3&1\\17&0\end{pmatrix}$, Jika AT transpose matriks A dan AX = B + AT, maka determinan matriks X adalah ....
  1. $\dfrac{15}{13}$
  2. $\dfrac{13}{15}$
  3. $-\dfrac{13}{15}$
  1. −1
  2. 1
Grup Telegram
ALTERNATIF PENYELESAIAN
|A|= 5⋅3 − 2⋅0 = 15

$A^T=\begin{pmatrix}5&0\\2&3\end{pmatrix}$

$B+A^T=\begin{pmatrix}-3&1\\17&0\end{pmatrix}+\begin{pmatrix}5&0\\2&3\end{pmatrix}=\begin{pmatrix}2&1\\19&3\end{pmatrix}$

|B + AT| = 2⋅3 − 1⋅19 = −13

$\begin{aligned} AX&=B+A^T\\ |A||X|&=\left|B+A^T\right|\\ 15|X|&=-13\\ |X|&=\boxed{\boxed{-\dfrac{13}{15}}} \end{aligned}$
Jadi, determinan matriks X adalah $-\dfrac{13}{15}$.
JAWAB: C

No.

Untuk persamaan ${2\begin{pmatrix}x&3y\\3&y\end{pmatrix}+\begin{pmatrix}3&x-6\\1&x\end{pmatrix}=\begin{pmatrix}11&10\\7&8\end{pmatrix}}$, harga x + y adalah
  1. −2
  2. 2
  3. 4
  1. 6
  2. 7
Grup Telegram
ALTERNATIF PENYELESAIAN
$\begin{aligned} 2\begin{pmatrix}x&3y\\3&y\end{pmatrix}+\begin{pmatrix}3&x-6\\1&x\end{pmatrix}&=\begin{pmatrix}11&10\\7&8\end{pmatrix}\\ \begin{pmatrix}2x&6y\\6&2y\end{pmatrix}+\begin{pmatrix}3&x-6\\1&x\end{pmatrix}&=\begin{pmatrix}11&10\\7&8\end{pmatrix}\\ \begin{pmatrix}2x+3&6y+x-6\\7&2y+x\end{pmatrix}&=\begin{pmatrix}11&10\\7&8\end{pmatrix} \end{aligned}$

$\begin{aligned} 2x+3&=11\\ 2x&=8\\ x&=4 \end{aligned}$

$\begin{aligned} 2y+x&=8\\ 2y+4&=8\\ 2y&=4\\ y&=2 \end{aligned}$

$\begin{aligned} x+y&=2+4\\ &=\boxed{\boxed{6}} \end{aligned}$
Jadi, harga x + y adalah 6.
JAWAB: D

No.

Jika $\begin{pmatrix}5&4\\1&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}4\\6&1\end{pmatrix}$, nilai 3x + y adalah ....
  1. 5
  2. 6
  3. 7
  1. 8
  2. 9
$\begin{aligned} \begin{pmatrix}5&4\\1&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}&=\begin{pmatrix}6\\1\end{pmatrix}\\ \begin{pmatrix}5x+4y\\x+y\end{pmatrix}&=\begin{pmatrix}6\\1\end{pmatrix} \end{aligned}$
$\begin{aligned} 5x+4y&=6\\ x+y&=1\qquad&{\color{red}\times4} \end{aligned}$

$\begin{aligned} 5x+4y&=6\\ 4x+4y&=4\qquad&{\color{red}-}\\\hline x&=2 \end{aligned}$

$\begin{aligned} x+y&=1\\ 2+y&=1\\ y&=1-2\\ &=-1\end{aligned}$

$\begin{aligned} 3x+y&=3(2)+(-1)\\ &=6-1\\ &=\boxed{\boxed{5}}\end{aligned}$
ALTERNATIF PENYELESAIAN
Jadi, 3x + y = 5.
JAWAB: A


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