Exercise Zone : Matriks [5]
Table of Contents
Tipe:
No.
Perhatikan matriks berikut,\(\begin{pmatrix}5&2&7\\1&5&9\\4&0&2\end{pmatrix}\)
elemen a23 adalah ....
ALTERNATIF PENYELESAIAN
a23 adalah elemen di baris 2 kolom 3 yaitu 9.
Jadi, elemen a23 adalah 9.
No.
Diketahui persamaan matriks $\begin{pmatrix}5&-2\\9&-4\end{pmatrix}\begin{pmatrix}2&-1\\x&x+y\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}$. Nilai- $\dfrac52$
- $\dfrac{15}2$
- $\dfrac{19}2$
- $\dfrac{22}2$
- $\dfrac{23}2$
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\begin{pmatrix}5&-2\\9&-4\end{pmatrix}\begin{pmatrix}2&-1\\x&x+y\end{pmatrix}&=\begin{pmatrix}1&0\\0&1\end{pmatrix}\\[4pt]
\begin{pmatrix}10-2x&-5-2x-2y\\18-4x&-9-4x-4y\end{pmatrix}&=\begin{pmatrix}1&0\\0&1\end{pmatrix}
\end{aligned}\)
\(\begin{aligned} 10-2x&=1\\ -2x&=-9\\ x&=\dfrac92 \end{aligned}\)
\(\begin{aligned} -5-2x-2y&=0\\ -5-2\left(\dfrac92\right)-2y&=0\\[4pt] -5-9-2y&=0\\ -14-2y&=0\\ -2y&=14\\ y&=-7 \end{aligned}\)
\(\begin{aligned} x-y&=\dfrac92-(-7)\\[4pt] &=\dfrac92+\dfrac{14}2\\ &=\color{blue}\boxed{\boxed{\color{black}\dfrac{23}2}} \end{aligned}\)
\(\begin{aligned} 10-2x&=1\\ -2x&=-9\\ x&=\dfrac92 \end{aligned}\)
\(\begin{aligned} -5-2x-2y&=0\\ -5-2\left(\dfrac92\right)-2y&=0\\[4pt] -5-9-2y&=0\\ -14-2y&=0\\ -2y&=14\\ y&=-7 \end{aligned}\)
\(\begin{aligned} x-y&=\dfrac92-(-7)\\[4pt] &=\dfrac92+\dfrac{14}2\\ &=\color{blue}\boxed{\boxed{\color{black}\dfrac{23}2}} \end{aligned}\)
Jadi, $x-y=\dfrac{23}2$.
JAWAB: E
JAWAB: E
No.
\(\begin{pmatrix}1&6&2\\4&2&5\end{pmatrix}+\begin{pmatrix}0&5&9\\1&3&2\end{pmatrix}=\) ....ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\begin{pmatrix}1&6&2\\4&2&5\end{pmatrix}+\begin{pmatrix}0&5&9\\1&3&2\end{pmatrix}&=\begin{pmatrix}1+0&6+5&2+9\\4+1&2+3&5+2\end{pmatrix}\\
&=\begin{pmatrix}1&11&11\\5&5&7\end{pmatrix}
\end{aligned}
\)
Jadi, \(\begin{pmatrix}1&6&2\\4&2&5\end{pmatrix}+\begin{pmatrix}0&5&9\\1&3&2\end{pmatrix}=\begin{pmatrix}1&11&11\\5&5&7\end{pmatrix}\).
No.
\(\begin{pmatrix}1&6&2\\4&2&5\end{pmatrix}-\begin{pmatrix}0&5&9\\1&3&2\end{pmatrix}=\) ....ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\begin{pmatrix}1&6&2\\4&2&5\end{pmatrix}+\begin{pmatrix}0&5&9\\1&3&2\end{pmatrix}&=\begin{pmatrix}1-0&6-5&2-9\\4-1&2-3&5-2\end{pmatrix}\\
&=\begin{pmatrix}1&1&-7\\3&-1&3\end{pmatrix}
\end{aligned}
\)
Jadi, \(\begin{pmatrix}1&6&2\\4&2&5\end{pmatrix}+\begin{pmatrix}0&5&9\\1&3&2\end{pmatrix}=\begin{pmatrix}1&1&-7\\3&-1&3\end{pmatrix}\).
No.
\(10\begin{pmatrix}1&6&2\\4&2&5\end{pmatrix}=\) ....ALTERNATIF PENYELESAIAN
\(\begin{aligned}
10\begin{pmatrix}1&6&2\\4&2&5\end{pmatrix}&=\begin{pmatrix}10\cdot1&10\cdot6&10\cdot2\\10\cdot4&10\cdot2&10\cdot5\end{pmatrix}\\
&=\begin{pmatrix}10&60&20\\40&20&50\end{pmatrix}
\end{aligned}
\)
Jadi, \(10\begin{pmatrix}1&6&2\\4&2&5\end{pmatrix}=\begin{pmatrix}10&60&20\\40&20&50\end{pmatrix}\).
No.
Determinan matriks \(\begin{pmatrix}1&6\\4&2\end{pmatrix}\) adalah ....ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\det&=1\cdot2-6\cdot4\\
&=2-24\\
&=-22
\end{aligned}
\)
Jadi, determinan matriks \(\begin{pmatrix}1&6\\4&2\end{pmatrix}\) adalah −22.
No.
\(\begin{pmatrix}x+1&6\\4&y+2\end{pmatrix}=\begin{pmatrix}5&6\\4&9\end{pmatrix}\). NilaiALTERNATIF PENYELESAIAN
\(\begin{aligned}
x+1&=5\\
x&=5-1\\
&=4
\end{aligned}
\)
\(\begin{aligned} y+2&=9\\ y&=9-2\\ &=7 \end{aligned} \)
\(\begin{aligned} y+2&=9\\ y&=9-2\\ &=7 \end{aligned} \)
\(\begin{aligned}
x+y&=4+7\\
y&=11
\end{aligned}
\)
Jadi, x + y = 11.
No.
Diketahui matriks \({C=\begin{pmatrix}2&3\\4&m\end{pmatrix}}\). JikaALTERNATIF PENYELESAIAN
\(\begin{aligned}
\det C&=-2\\
2m-12&=-2\\
2m&=-2+12\\
2m&=10\\
m&=\dfrac{10}2\\
m&=5
\end{aligned}\)
Jadi, m = 5.
No.
Diketahui matriks \({P=\begin{pmatrix}p&4\\6&2\end{pmatrix}}\) dan \({Q=\begin{pmatrix}1&3\\5&7\end{pmatrix}}\). JikaALTERNATIF PENYELESAIAN
\(\begin{aligned}
\det P&=\det Q\\
2p-24&=-8\\
2p&=-8+24\\
2p&=16\\
p&=\dfrac{16}2\\
p&=8
\end{aligned}\)
Jadi, p = 8.
No.
Diketahui matriks \({R=\begin{pmatrix}4&2\\3&5\end{pmatrix}}\).ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\det R&=4\cdot5-2\dot3\\
&=12\end{aligned}\)
\(\begin{aligned}\det R^2&=(\det R)^2\\
&=(12)^2\\
&=144
\end{aligned}\)
Jadi, det R2 = 144.
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