Exercise Zone : Matriks [6]
Table of Contents

Tipe:
No.
Diketahui sistem persamaan linear dua variabel:2x − 3y = −4
4x + 2y = 8
Nilai
ALTERNATIF PENYELESAIAN
\(\begin{pmatrix}2&-3\\4&2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}-4\\8\end{pmatrix}\)
\(\begin{aligned} D&=\det\begin{pmatrix}2&-3\\4&2\end{pmatrix}\\&=2\cdot2-(-3)\cdot4\\&=4+12\\&=16\end{aligned}\)
\(\begin{aligned} D_x&=\det\begin{pmatrix}-4&-3\\8&2\end{pmatrix}\\&=-4\cdot2-(-3)\cdot8\\&=-8+24\\&=16\end{aligned}\)
\(\begin{aligned} D_y&=\det\begin{pmatrix}2&-4\\4&8\end{pmatrix}\\&=2\cdot8-(-4)\cdot4\\&=16+16\\&=32\end{aligned}\)
\(\begin{aligned} D&=\det\begin{pmatrix}2&-3\\4&2\end{pmatrix}\\&=2\cdot2-(-3)\cdot4\\&=4+12\\&=16\end{aligned}\)
\(\begin{aligned} D_x&=\det\begin{pmatrix}-4&-3\\8&2\end{pmatrix}\\&=-4\cdot2-(-3)\cdot8\\&=-8+24\\&=16\end{aligned}\)
\(\begin{aligned} D_y&=\det\begin{pmatrix}2&-4\\4&8\end{pmatrix}\\&=2\cdot8-(-4)\cdot4\\&=16+16\\&=32\end{aligned}\)
\(\begin{aligned}
x&=\dfrac{D_x}D\\[3.7pt]&=\dfrac{16}{16}\\[3.7pt]&=1\end{aligned}\)
\(\begin{aligned} y&=\dfrac{D_y}D\\[3.7pt]&=\dfrac{32}{16}\\[3.7pt]&=2\end{aligned}\)
\(\begin{aligned} 10x+10y&=10(1)+10(2)\\&=10+20\\&=30\end{aligned}\)
\(\begin{aligned} y&=\dfrac{D_y}D\\[3.7pt]&=\dfrac{32}{16}\\[3.7pt]&=2\end{aligned}\)
\(\begin{aligned} 10x+10y&=10(1)+10(2)\\&=10+20\\&=30\end{aligned}\)
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 dariALTERNATIF PENYELESAIAN
\(\begin{aligned}
3a&=9\\
a&=\dfrac93\\[3.7pt]
&=3
\end{aligned}\)
\(\begin{aligned} b&=a\\ &=3 \end{aligned}\)
\(\begin{aligned} b&=a\\ &=3 \end{aligned}\)
\(\begin{aligned}
2a+b&=2(3)+3\\
&=6+3\\
&=\boxed{\boxed{9}}
\end{aligned}\)
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 dariALTERNATIF 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 dariALTERNATIF 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 matriksALTERNATIF 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- −8
- −2
- 2
- 6
- 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 ⟶ 2x − y = 9
\(\begin{aligned} 3x-y&=13\\ 2x-y&=9&-\\\hline x&=4 \end{aligned}\)
\(\begin{aligned} 2x-y&=9\\ 2(4)-y&=9\\ 8-y&=9\\ y&=8-9\\ y&=-1 \end{aligned}\)
\(\begin{aligned} x-2y&=4-2(-1)\\ &=4+2\\ &=\boxed{\boxed{6}} \end{aligned}\)
\(\begin{aligned} 3x-y&=13\\ 2x-y&=9&-\\\hline x&=4 \end{aligned}\)
\(\begin{aligned} 2x-y&=9\\ 2(4)-y&=9\\ 8-y&=9\\ y&=8-9\\ y&=-1 \end{aligned}\)
\(\begin{aligned} x-2y&=4-2(-1)\\ &=4+2\\ &=\boxed{\boxed{6}} \end{aligned}\)
Jadi, x − 2y = 6 .
JAWAB: D
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- \(\begin{pmatrix}1&-4\\-1&3\end{pmatrix}\)
- \(\begin{pmatrix}-1&-4\\1&3\end{pmatrix}\)
- \(\begin{pmatrix}1&4\\-1&-3\end{pmatrix}\)
- \(\begin{pmatrix}-3&4\\-1&1\end{pmatrix}\)
- \(\begin{pmatrix}-3&4\\1&-1\end{pmatrix}\)
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
|M|&=4\cdot2-5\cdot2\\
&=8-10\\
&=-2
\end{aligned}\)
\(\begin{aligned} M&=KL\\ K^{-1}M&=K^{-1}KL\\ K^{-1}M&=L\\ K^{-1}MM^{-1}&=LM^{-1}\\ K^{-1}&=LM^{-1}\\ &=\begin{pmatrix}-4&-3\\2&1\end{pmatrix}\dfrac1{-2}\begin{pmatrix}2&-5\\-2&4\end{pmatrix}\\[4pt] &=-\dfrac12\begin{pmatrix}-4&-3\\2&1\end{pmatrix}\begin{pmatrix}2&-5\\-2&4\end{pmatrix}\\[4pt] &=-\dfrac12\begin{pmatrix}-2&8\\2&-6\end{pmatrix}\\ &=\boxed{\boxed{\begin{pmatrix}1&-4\\-1&3\end{pmatrix}}} \end{aligned}\)
\(\begin{aligned} M&=KL\\ K^{-1}M&=K^{-1}KL\\ K^{-1}M&=L\\ K^{-1}MM^{-1}&=LM^{-1}\\ K^{-1}&=LM^{-1}\\ &=\begin{pmatrix}-4&-3\\2&1\end{pmatrix}\dfrac1{-2}\begin{pmatrix}2&-5\\-2&4\end{pmatrix}\\[4pt] &=-\dfrac12\begin{pmatrix}-4&-3\\2&1\end{pmatrix}\begin{pmatrix}2&-5\\-2&4\end{pmatrix}\\[4pt] &=-\dfrac12\begin{pmatrix}-2&8\\2&-6\end{pmatrix}\\ &=\boxed{\boxed{\begin{pmatrix}1&-4\\-1&3\end{pmatrix}}} \end{aligned}\)
Jadi, \(K^{-1}=\begin{pmatrix}1&-4\\-1&3\end{pmatrix}\).
JAWAB: A
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
$\begin{aligned} 2a&=2\\ a&=\dfrac22\\ a&=1 \end{aligned}$
$\begin{aligned} 2b&=b+4\\ b&=4 \end{aligned}$
$\begin{aligned} c&=3c-1\\ -2c&=-1\\ c&=\dfrac{-1}{-2}\\ c&=\dfrac12 \end{aligned}$
$\begin{aligned} -d&=d+3\\ -2d&=3\\ d&=\dfrac3{-2}\\ &=-\dfrac32 \end{aligned}$
$\begin{aligned} 2a&=2\\ a&=\dfrac22\\ a&=1 \end{aligned}$
$\begin{aligned} 2b&=b+4\\ b&=4 \end{aligned}$
$\begin{aligned} c&=3c-1\\ -2c&=-1\\ c&=\dfrac{-1}{-2}\\ c&=\dfrac12 \end{aligned}$
$\begin{aligned} -d&=d+3\\ -2d&=3\\ d&=\dfrac3{-2}\\ &=-\dfrac32 \end{aligned}$
Jadi, a = 1, b = 4, ${c=\dfrac12}$, ${d=-\dfrac32}$.
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