Exercise Zone : Matriks [4]
Table of Contents
Tipe:
No.
Transpos dari matriks $A=\begin{pmatrix}1&-2&3\\-4&5&-6\end{pmatrix}$ adalahALTERNATIF PENYELESAIAN
$A^{T}=\begin{pmatrix}1&-4\\-2&5\\3&-6\end{pmatrix}$
Jadi, $A^{T}=\begin{pmatrix}1&-4\\-2&5\\3&-6\end{pmatrix}$.
No.
Determinan dari matriks $\begin{pmatrix}2&-3&4\\-5&6&-7\\8&-9&10\end{pmatrix}$ adalah ....ALTERNATIF PENYELESAIAN
$\begin{aligned}
\det&=\begin{vmatrix}2&-3&4\\-5&6&-7\\8&-9&10\end{vmatrix}\\
&=2\begin{vmatrix}6&-7\\-9&10\end{vmatrix}-(-3)\begin{vmatrix}-5&-7\\8&10\end{vmatrix}+4\begin{vmatrix}-5&6\\8&-9\end{vmatrix}\\
&=2(6\cdot10-(-7)\cdot(-9))+3((-5)\cdot10-(-7)\cdot8)+4((-5)\cdot(-9)-6\cdot8)\\
&=2(60-63)+3(-50+56)+4(45-48)\\
&=2(-3)+3(-6)+4(-3)\\
&=-6-18-12\\
&=\boxed{\boxed{-36}}
\end{aligned}$
Jadi, determinan dari matriks $\begin{pmatrix}2&-3&4\\-5&6&-7\\8&-9&10\end{pmatrix}$ adalah −36.
No.
Tentukan nilai p dan q jika $\begin{pmatrix}2p&4\\-3&-\dfrac12q\end{pmatrix}=\begin{pmatrix}-4&4\\-3&5\end{pmatrix}$ALTERNATIF PENYELESAIAN
2p = −4
$\begin{aligned} p&=\dfrac{-4}2\\ &=\boxed{\boxed{-2}}\end{aligned}$
$\begin{aligned}-\dfrac12q&=5\\ q&=5\cdot(-2)\\ &=\boxed{\boxed{-10}}\end{aligned}$
$\begin{aligned} p&=\dfrac{-4}2\\ &=\boxed{\boxed{-2}}\end{aligned}$
$\begin{aligned}-\dfrac12q&=5\\ q&=5\cdot(-2)\\ &=\boxed{\boxed{-10}}\end{aligned}$
Jadi, p = −2 dan q = −10.
No.
Tentukan hasil kali dari kedua matriks berikut.$\begin{pmatrix}-1&-6\\7&5\end{pmatrix}\begin{pmatrix}-9\\5\end{pmatrix}$
ALTERNATIF PENYELESAIAN
$\begin{aligned}
\begin{pmatrix}-1&-6\\7&5\end{pmatrix}\begin{pmatrix}-9\\5\end{pmatrix}&=\begin{pmatrix}(-1)(-9)+(-6)(5)\\ (7)(-9)+(5)(5)\end{pmatrix}\\
&=\begin{pmatrix}9+(-30)\\-63+25\end{pmatrix}\\
&=\boxed{\boxed{\begin{pmatrix}-21\\-38\end{pmatrix}}}
\end{aligned}$
Jadi, $\begin{pmatrix}-1&-6\\7&5\end{pmatrix}\begin{pmatrix}-9\\5\end{pmatrix}=\begin{pmatrix}-21\\-38\end{pmatrix}$.
No.
Tentukan hasil kali dari kedua matriks berikut.$8\begin{pmatrix}4\\6\end{pmatrix}\begin{pmatrix}-5&5\end{pmatrix}$
ALTERNATIF PENYELESAIAN
$\begin{aligned}
8\begin{pmatrix}4\\6\end{pmatrix}\begin{pmatrix}-5&5\end{pmatrix}&=\begin{pmatrix}32\\48\end{pmatrix}\begin{pmatrix}-5&5\end{pmatrix}\\
&=\begin{pmatrix}(32)(-5)&(32)(5)\\ (48)(-5)&(48)(5)\end{pmatrix}\\
&=\boxed{\boxed{\begin{pmatrix}-160&160\\-240&240\end{pmatrix}}}
\end{aligned}$
Jadi, $8\begin{pmatrix}4\\6\end{pmatrix}\begin{pmatrix}-5&5\end{pmatrix}=\begin{pmatrix}-160&160\\-240&240\end{pmatrix}$.
No.
$\begin{pmatrix}-6&-2\\-8&7\end{pmatrix}\cdot\begin{pmatrix}-2&5\\0&5\end{pmatrix}=$ ....ALTERNATIF PENYELESAIAN
$\begin{aligned}
\begin{pmatrix}-6&-2\\-8&7\end{pmatrix}\cdot\begin{pmatrix}-2&5\\0&5\end{pmatrix}&=\begin{pmatrix}(-6)(-2)+(-2)(0)&(-6)(5)+(-2)(5)\\ (-8)(-2)+(7)(0)&(-8)(5)+(7)(5)\end{pmatrix}\\
&=\begin{pmatrix}12+0&-30+(-10)\\16+0&-40+35\end{pmatrix}\\
&=\boxed{\boxed{\begin{pmatrix}12&-40\\16&-5\end{pmatrix}}}
\end{aligned}$
Jadi, $\begin{pmatrix}-6&-2\\-8&7\end{pmatrix}\cdot\begin{pmatrix}-2&5\\0&5\end{pmatrix}=\begin{pmatrix}12&-40\\16&-5\end{pmatrix}$.
No.
Berikut ini adalah tabel banyaknya kendaraan dalam 2 hari.Senin | Selasa | |
Mobil | 2 | 3 |
Motor | 4 | 5 |
ALTERNATIF PENYELESAIAN
$\begin{pmatrix}2&3\\4&5\end{pmatrix}$
Jadi, matriksnya adalah $\begin{pmatrix}2&3\\4&5\end{pmatrix}$.
No.
Jika $\begin{pmatrix}{^x\negmedspace\log a}&\log(2a-2)\\ \log(b-4)&1\end{pmatrix}=\begin{pmatrix}\log b&1\\\log a&1\end{pmatrix}$ maka- 6
- 10
- 1
- 2
- 4
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\log(2a-2)&=1\\
2a-2&=10\\
2a&=12\\
a&=6
\end{aligned}\)
\(\begin{aligned} \log(b-4)&=\log a\\ \log(b-4)&=\log 6\\ b-4&=6\\ b&=10 \end{aligned}\)
\(\begin{aligned} {^x\negmedspace\log a}&=\log b\\ {^x\negmedspace\log 6}&=\log 10\\ {^x\negmedspace\log 6}&=1\\ x&=\color{blue}\boxed{\boxed{\color{black}6}} \end{aligned}\)
\(\begin{aligned} \log(b-4)&=\log a\\ \log(b-4)&=\log 6\\ b-4&=6\\ b&=10 \end{aligned}\)
\(\begin{aligned} {^x\negmedspace\log a}&=\log b\\ {^x\negmedspace\log 6}&=\log 10\\ {^x\negmedspace\log 6}&=1\\ x&=\color{blue}\boxed{\boxed{\color{black}6}} \end{aligned}\)
Jadi, x = 6.
JAWAB: A
JAWAB: A
No.
Jika matriks $M=\begin{pmatrix}1&2\\-3&1\end{pmatrix}$ dan $N=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ maka matriks- $\begin{pmatrix}-1&2\\2&1\end{pmatrix}$
- $\begin{pmatrix}2&1\\-1&2\end{pmatrix}$
- $\begin{pmatrix}2&-1\\1&2\end{pmatrix}$
- $\begin{pmatrix}2&-1\\-1&2\end{pmatrix}$
- $\begin{pmatrix}4&-2\\2&4\end{pmatrix}$
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
2MN-NM&=2\begin{pmatrix}1&2\\-3&1\end{pmatrix}\begin{pmatrix}0&-1\\1&0\end{pmatrix}-\begin{pmatrix}0&-1\\1&0\end{pmatrix}\begin{pmatrix}1&2\\-3&1\end{pmatrix}\\[4pt]
&=\begin{pmatrix}4&-2\\2&4\end{pmatrix}-\begin{pmatrix}2&-1\\1&2\end{pmatrix}\\
&=\color{blue}\boxed{\boxed{\color{black}\begin{pmatrix}2&-1\\1&2\end{pmatrix}}}
\end{aligned}\)
Jadi, $2MN-NM=\begin{pmatrix}2&-1\\1&2\end{pmatrix}$.
JAWAB: C
JAWAB: C
No.
Matriks berikut,$\begin{pmatrix}2&3&4\\5&6&7\end{pmatrix}$
mempunyai ordo ....
ALTERNATIF PENYELESAIAN
banyaknya baris matriks tersebut adalah 2. Banyaknya kolom matriks tersebut adalah 3. Jadi ordonya adalah 2×3 .
Jadi, ordonya adalah 2×3 .
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