Exercise Zone : Matriks [6]

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

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StandarSNBTHOTS


No.

Diketahui sistem persamaan linear dua variabel:
2x − 3y = −4
4x + 2y = 8
Nilai 10x + 10y adalah ....
ALTERNATIF PENYELESAIAN
\(\begin{pmatrix}2&-3\\4&2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}-4\\8\end{pmatrix}\)

\(\begin{aligned} D&=\det\begin{pmatrix}2&-3\\4&2\end{pmatrix}\\&=2\cdot2-(-3)\cdot4\\&=4+12\\&=16\end{aligned}\)

\(\begin{aligned} D_x&=\det\begin{pmatrix}-4&-3\\8&2\end{pmatrix}\\&=-4\cdot2-(-3)\cdot8\\&=-8+24\\&=16\end{aligned}\)

\(\begin{aligned} D_y&=\det\begin{pmatrix}2&-4\\4&8\end{pmatrix}\\&=2\cdot8-(-4)\cdot4\\&=16+16\\&=32\end{aligned}\)

\(\begin{aligned} x&=\dfrac{D_x}D\\[3.7pt]&=\dfrac{16}{16}\\[3.7pt]&=1\end{aligned}\)

\(\begin{aligned} y&=\dfrac{D_y}D\\[3.7pt]&=\dfrac{32}{16}\\[3.7pt]&=2\end{aligned}\)

\(\begin{aligned} 10x+10y&=10(1)+10(2)\\&=10+20\\&=30\end{aligned}\)
Jadi, 10x + 10y= 30.

No.

Jika matriks \(\begin{pmatrix}3a&5\\-1&b\end{pmatrix} = \begin{pmatrix}9&5\\-1&a\end{pmatrix}\) maka tentukan nilai dari 2a + b!
ALTERNATIF PENYELESAIAN
\(\begin{aligned} 3a&=9\\ a&=\dfrac93\\[3.7pt] &=3 \end{aligned}\)

\(\begin{aligned} b&=a\\ &=3 \end{aligned}\)
\(\begin{aligned} 2a+b&=2(3)+3\\ &=6+3\\ &=\boxed{\boxed{9}} \end{aligned}\)
Jadi, 2a + b = 9.

No.

Tentukan hasil perkalian matriks \(\begin{pmatrix}2&-2\\2&3\end{pmatrix}\times\begin{pmatrix}2\\2\end{pmatrix}\)!
ALTERNATIF PENYELESAIAN
\(\begin{aligned} \begin{pmatrix}2&-2\\2&3\end{pmatrix}\times\begin{pmatrix}2\\2\end{pmatrix}&=\begin{pmatrix}2\cdot2+(-2)\cdot2\\2\cdot2+3\cdot2\end{pmatrix}\\ &=\begin{pmatrix}4+(-4)\\4+6\end{pmatrix}\\ &=\boxed{\boxed{\begin{pmatrix}0\\10\end{pmatrix}}} \end{aligned}\)
Jadi, \(\begin{pmatrix}2&-2\\2&3\end{pmatrix}\times\begin{pmatrix}2\\2\end{pmatrix}=\begin{pmatrix}0\\10\end{pmatrix}\).

No.

Jika matriks \(A=\begin{pmatrix}1&2\\2&1\end{pmatrix}\) dan \(B=\begin{pmatrix}-1&3\\0&2\end{pmatrix}\) maka tentukan nilai dari 2A + B!
ALTERNATIF PENYELESAIAN
\(\begin{aligned} 2A+B&=2\begin{pmatrix}1&2\\2&1\end{pmatrix}+\begin{pmatrix}-1&3\\0&2\end{pmatrix}\\ &=\begin{pmatrix}2&4\\4&2\end{pmatrix}+\begin{pmatrix}-1&3\\0&2\end{pmatrix}\\ &=\boxed{\boxed{\begin{pmatrix}1&7\\4&4\end{pmatrix}}} \end{aligned}\)
Jadi, \(2A+B=\begin{pmatrix}1&7\\4&4\end{pmatrix}\).

No.

Jika matriks \(A=\begin{pmatrix}1&1\\1&1\end{pmatrix}\) dan \(B=\begin{pmatrix}-1&3\\0&2\end{pmatrix}\) maka tentukan nilai dari 2A + B!
ALTERNATIF PENYELESAIAN
\(\begin{aligned} 2A+B&=2\begin{pmatrix}1&1\\1&1\end{pmatrix}+\begin{pmatrix}-1&3\\0&2\end{pmatrix}\\ &=\begin{pmatrix}2&2\\2&2\end{pmatrix}+\begin{pmatrix}-1&3\\0&2\end{pmatrix}\\ &=\boxed{\boxed{\begin{pmatrix}1&5\\2&4\end{pmatrix}}} \end{aligned}\)
Jadi, \(2A+B=\begin{pmatrix}1&5\\2&4\end{pmatrix}\).

No.

Tentukan hasil perkalian matriks \(\begin{pmatrix}1&1\\2&3\end{pmatrix}\times\begin{pmatrix}1\\2\end{pmatrix}\)!
ALTERNATIF PENYELESAIAN
\(\begin{aligned} \begin{pmatrix}1&1\\2&3\end{pmatrix}\times\begin{pmatrix}1\\2\end{pmatrix}&=\begin{pmatrix}1\cdot1+1\cdot2\\2\cdot1+3\cdot2\end{pmatrix}\\ &=\begin{pmatrix}1+2\\2+6\end{pmatrix}\\ &=\boxed{\boxed{\begin{pmatrix}3\\8\end{pmatrix}}} \end{aligned}\)
Jadi, \(\begin{pmatrix}1&1\\2&3\end{pmatrix}\times\begin{pmatrix}1\\2\end{pmatrix}=\begin{pmatrix}3\\8\end{pmatrix}\).

No.

Diketahui \(A=\begin{pmatrix}3&1\\2&4\end{pmatrix}\), \({B=\begin{pmatrix}0&1\\-1&2\end{pmatrix}}\) dan X matriks berordo 2×2 yang memenuhi persamaan matriks 2AB + X = 0, maka X sama dengan ....
ALTERNATIF PENYELESAIAN
\(\begin{aligned} 2A-B+X&=0\\ 2\begin{pmatrix}3&1\\2&4\end{pmatrix}-\begin{pmatrix}0&1\\-1&2\end{pmatrix}+X&=\begin{pmatrix}0&0\\0&0\end{pmatrix}\\ \begin{pmatrix}6&2\\4&8\end{pmatrix}-\begin{pmatrix}0&1\\-1&2\end{pmatrix}+X&=\begin{pmatrix}0&0\\0&0\end{pmatrix}\\ \begin{pmatrix}6&1\\5&6\end{pmatrix}+X&=\begin{pmatrix}0&0\\0&0\end{pmatrix}\\ X&=\begin{pmatrix}0&0\\0&0\end{pmatrix}-\begin{pmatrix}6&1\\5&6\end{pmatrix}\\ &=\boxed{\boxed{\begin{pmatrix}-6&-1\\-5&-6\end{pmatrix}}} \end{aligned}\)
Jadi, \(X=\begin{pmatrix}-6&-1\\-5&-6\end{pmatrix}\).

No.

Diketahui matriks \(A=\begin{pmatrix}3&-2\\x&y\end{pmatrix}\), \(B=\begin{pmatrix}4&3\\-2&-1\end{pmatrix}\), dan \(C=\begin{pmatrix}16&11\\18&13\end{pmatrix}\). Jika AB = C, Nilai dari x − 2y = ....
  1. −8
  2. −2
  3. 2
  1. 6
  2. 8
ALTERNATIF PENYELESAIAN
\(\begin{aligned} AB&=C\\ \begin{pmatrix}3&-2\\x&y\end{pmatrix}\begin{pmatrix}4&3\\-2&-1\end{pmatrix}&=\begin{pmatrix}16&11\\18&13\end{pmatrix}\\ \begin{pmatrix}16&11\\4x-2y&3x-y\end{pmatrix}&=\begin{pmatrix}16&11\\18&13\end{pmatrix} \end{aligned}\)

4x − 2y = 18 ⟶ 2xy = 9

\(\begin{aligned} 3x-y&=13\\ 2x-y&=9&-\\\hline x&=4 \end{aligned}\)

\(\begin{aligned} 2x-y&=9\\ 2(4)-y&=9\\ 8-y&=9\\ y&=8-9\\ y&=-1 \end{aligned}\)

\(\begin{aligned} x-2y&=4-2(-1)\\ &=4+2\\ &=\boxed{\boxed{6}} \end{aligned}\)
Jadi, x − 2y = 6.
JAWAB: D

No.

Diketahui matriks \(M=\begin{pmatrix}4&5\\2&2\end{pmatrix}\) dan \(L=\begin{pmatrix}-4&-3\\2&1\end{pmatrix}\). Jika matriks M = KL, matriks K−1 = ....
  1. \(\begin{pmatrix}1&-4\\-1&3\end{pmatrix}\)
  2. \(\begin{pmatrix}-1&-4\\1&3\end{pmatrix}\)
  3. \(\begin{pmatrix}1&4\\-1&-3\end{pmatrix}\)
  1. \(\begin{pmatrix}-3&4\\-1&1\end{pmatrix}\)
  2. \(\begin{pmatrix}-3&4\\1&-1\end{pmatrix}\)
ALTERNATIF PENYELESAIAN
\(\begin{aligned} |M|&=4\cdot2-5\cdot2\\ &=8-10\\ &=-2 \end{aligned}\)

\(\begin{aligned} M&=KL\\ K^{-1}M&=K^{-1}KL\\ K^{-1}M&=L\\ K^{-1}MM^{-1}&=LM^{-1}\\ K^{-1}&=LM^{-1}\\ &=\begin{pmatrix}-4&-3\\2&1\end{pmatrix}\dfrac1{-2}\begin{pmatrix}2&-5\\-2&4\end{pmatrix}\\[4pt] &=-\dfrac12\begin{pmatrix}-4&-3\\2&1\end{pmatrix}\begin{pmatrix}2&-5\\-2&4\end{pmatrix}\\[4pt] &=-\dfrac12\begin{pmatrix}-2&8\\2&-6\end{pmatrix}\\ &=\boxed{\boxed{\begin{pmatrix}1&-4\\-1&3\end{pmatrix}}} \end{aligned}\)
Jadi, \(K^{-1}=\begin{pmatrix}1&-4\\-1&3\end{pmatrix}\).
JAWAB: A

No.

Tentukan nilai a, b, c dan d jika ${\begin{pmatrix}a&2b\\c&-d\end{pmatrix}=\begin{pmatrix}2-a&b+4\\ 3c-1&d+3\end{pmatrix}}$
ALTERNATIF PENYELESAIAN
a = 2 − a
$\begin{aligned} 2a&=2\\ a&=\dfrac22\\ a&=1 \end{aligned}$

$\begin{aligned} 2b&=b+4\\ b&=4 \end{aligned}$

$\begin{aligned} c&=3c-1\\ -2c&=-1\\ c&=\dfrac{-1}{-2}\\ c&=\dfrac12 \end{aligned}$

$\begin{aligned} -d&=d+3\\ -2d&=3\\ d&=\dfrac3{-2}\\ &=-\dfrac32 \end{aligned}$
Jadi, a = 1, b = 4, ${c=\dfrac12}$, ${d=-\dfrac32}$.


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