Exercise Zone : Matriks [5]

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

Tipe:

StandarSNBTHOTS


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 xy = ....
  1. $\dfrac52$
  2. $\dfrac{15}2$
  1. $\dfrac{19}2$
  2. $\dfrac{22}2$
  1. $\dfrac{23}2$
ALTERNATIF PENYELESAIAN
\(\begin{aligned} \begin{pmatrix}5&-2\\9&-4\end{pmatrix}\begin{pmatrix}2&-1\\x&x+y\end{pmatrix}&=\begin{pmatrix}1&0\\0&1\end{pmatrix}\\[4pt] \begin{pmatrix}10-2x&-5-2x-2y\\18-4x&-9-4x-4y\end{pmatrix}&=\begin{pmatrix}1&0\\0&1\end{pmatrix} \end{aligned}\)

\(\begin{aligned} 10-2x&=1\\ -2x&=-9\\ x&=\dfrac92 \end{aligned}\)

\(\begin{aligned} -5-2x-2y&=0\\ -5-2\left(\dfrac92\right)-2y&=0\\[4pt] -5-9-2y&=0\\ -14-2y&=0\\ -2y&=14\\ y&=-7 \end{aligned}\)

\(\begin{aligned} x-y&=\dfrac92-(-7)\\[4pt] &=\dfrac92+\dfrac{14}2\\ &=\color{blue}\boxed{\boxed{\color{black}\dfrac{23}2}} \end{aligned}\)
Jadi, $x-y=\dfrac{23}2$.
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.

\(\begin{pmatrix}x+1&6\\4&y+2\end{pmatrix}=\begin{pmatrix}5&6\\4&9\end{pmatrix}\). Nilai x + y = ....
ALTERNATIF PENYELESAIAN
\(\begin{aligned} x+1&=5\\ x&=5-1\\ &=4 \end{aligned} \)

\(\begin{aligned} y+2&=9\\ y&=9-2\\ &=7 \end{aligned} \)
\(\begin{aligned} x+y&=4+7\\ y&=11 \end{aligned} \)
Jadi, x + y = 11.

No.

Diketahui matriks \({C=\begin{pmatrix}2&3\\4&m\end{pmatrix}}\). Jika det C = −2 maka nilai m adalah ....
ALTERNATIF 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 \({Q=\begin{pmatrix}1&3\\5&7\end{pmatrix}}\). Jika det P = det Q maka nilai p adalah ....
ALTERNATIF 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 \({R=\begin{pmatrix}4&2\\3&5\end{pmatrix}}\). det R2 = ....
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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