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Exercise Zone : Matriks [2]

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Idham Muqoddas
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Berikut ini adalah kumpulan soal mengenai Matriks tipe standar. Mau tanya soal? Gabung aja ke grup Facebook https://web.facebook.com/groups/matematikazoneid/ atau Telegram https://t.me/matematikazoneidgrup.

Tipe:

StandarSNBTHOTS



No.

Diketahui persamaan matriks \({\begin{pmatrix}3x&3y\\6&18\end{pmatrix}-2\begin{pmatrix}x&6\\-1&y+1\end{pmatrix}=\begin{pmatrix}2&2x-y\\8&8\end{pmatrix}}\). Nilai dari x − y =
  1. −2
  2. 0
  1. 2
  2. 4
  1. 6
Grup Facebook
ALTERNATIF PENYELESAIAN
\(\begin{aligned} \begin{pmatrix}3x&3y\\6&18\end{pmatrix}-2\begin{pmatrix}x&6\\-1&y+1\end{pmatrix}&=\begin{pmatrix}2&2x-y\\8&8\end{pmatrix}\\ \begin{pmatrix}3x&3y\\6&18\end{pmatrix}-\begin{pmatrix}2x&12\\-2&2y+2\end{pmatrix}&=\begin{pmatrix}2&2x-y\\8&8\end{pmatrix}\\ \begin{pmatrix}x&3y-12\\8&-2y+16\end{pmatrix}&=\begin{pmatrix}2&2x-y\\8&8\end{pmatrix}\\ \end{aligned}\)
x = 2
\(\begin{aligned} -2y+16&=8\\ -2y&=-8\\ y&=\boxed{4} \end{aligned}\)

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

No.

Diberikan matriks \({P = \begin{pmatrix}3&-1\\5&2\end{pmatrix}}\) dan \({Q = \begin{pmatrix}3r&2\\r&p+1\end{pmatrix}}\) dengan r ≠ 0 dan p ≠ 0. Supaya matriks PQ tidak mempunyai invers, maka nilai 3p + 2 = ....
  1. 4
  2. 3
  1. 2
  2. 1
  1. 0
Grup Facebook
ALTERNATIF PENYELESAIAN
Tidak punya invers artinya det = 0.

\(\begin{aligned} \left|PQ\right|&=0\\ |P||Q|&=0 \end{aligned}\)
Karena |P| ≠ 0 maka
\(\begin{aligned} |Q|&=0\\ 3r(p+1)-2r&=0\\ 3pr+3r-2r&=0\\ 3pr+r&=0\\ r(3p+1)&=0\\ 3p+1&=0\\ 3p+2&=\boxed{\boxed{1}} \end{aligned}\)
Jadi, 3p + 2 = 1.
JAWAB: D

No.

Jika diketahui matriks A memenuhi persamaan \({\begin{pmatrix}5&1\\7&2\end{pmatrix}A=\begin{pmatrix}3&-2\\-3&1\end{pmatrix}\begin{pmatrix}3&4\\1&2\end{pmatrix}}\), maka determinan dari A−1 adalah
  1. −2
  2. $-\dfrac12$
  1. 0
  2. $\dfrac12$
  1. 2
Grup Facebook
ALTERNATIF PENYELESAIAN
\(\begin{aligned} \begin{pmatrix}5&1\\7&2\end{pmatrix}A&=\begin{pmatrix}3&-2\\-3&1\end{pmatrix}\begin{pmatrix}3&4\\1&2\end{pmatrix}\\ \begin{vmatrix}5&1\\7&2\end{vmatrix}|A|&=\begin{vmatrix}3&-2\\-3&1\end{vmatrix}\begin{vmatrix}3&4\\1&2\end{vmatrix}\\ (5\cdot2-1\cdot7)|A|&=(3\cdot1-(-2)\cdot(-3))(3\cdot2-4\cdot1)\\ (10-7)|A|&=(3-6)(6-4)\\ 3|A|&=(-3)(2)\\ 3|A|&=-6\\ |A|&=-2 \end{aligned}\)

\(\begin{aligned} \left|A^{-1}\right|&=\dfrac1{|A|}\\ &=\boxed{\boxed{-\dfrac12}} \end{aligned}\)
Jadi, determinan dari A−1 adalah \(-\dfrac12\).
JAWAB: B

No.

Diketahui matriks \({P=\begin{pmatrix}2&-3&6\\5&0&-2\\1&4&-4\end{pmatrix}}\). Nilai (a12 − a31) dari transpose P adalah
  1. −4
  2. −2
  1. −1
  2. 1
  1. 11
Grup Facebook
ALTERNATIF PENYELESAIAN
\(P^T=\begin{pmatrix}2&5&1\\-3&0&4\\6&-2&-4\end{pmatrix}\)

\(\begin{aligned} a_{12}-a_{31}&=5-6\\ &=\boxed{\boxed{-1}} \end{aligned}\)
Jadi, nilai (a12 − a31) dari transpose P adalah −1.
JAWAB: C

No.

Matriks X2×2 yang memenuhi persamaan AXA−1 = B jika matriks \({A=\begin{pmatrix}2&-2\\1&3\end{pmatrix}}\) dan \({B=\begin{pmatrix}3&2\\-1&-2\end{pmatrix}}\) adalah ....
  1. \(\begin{pmatrix}3&2\\-1&-2\end{pmatrix}\)
  2. \(\begin{pmatrix}2&-1\\2&1\end{pmatrix}\)
  3. \(\begin{pmatrix}3&2\\-1&2\end{pmatrix}\)
  1. \(\begin{pmatrix}2&-1\\-2&-1\end{pmatrix}\)
  2. \(\begin{pmatrix}2&-1\\-2&1\end{pmatrix}\)
Brainly.co.id
ALTERNATIF PENYELESAIAN
\(\begin{aligned} A^{-1}&=\dfrac1{(2)(3)-(-2)(1)}\begin{pmatrix}3&2\\-1&2\end{pmatrix}\\ &=\dfrac18\begin{pmatrix}3&2\\-1&2\end{pmatrix} \end{aligned}\)

\(\begin{aligned} AXA^{-1}&=B\\ AX&=BA\\ x&=A^{-1}BA\\ &=\dfrac18\begin{pmatrix}3&2\\-1&2\end{pmatrix}\begin{pmatrix}3&2\\-1&-2\end{pmatrix}\begin{pmatrix}2&-2\\1&3\end{pmatrix}\\ &=\dfrac18\begin{pmatrix}7&2\\-5&-6\end{pmatrix}\begin{pmatrix}2&-2\\1&3\end{pmatrix}\\ &=\dfrac18\begin{pmatrix}16&-8\\-16&-8\end{pmatrix}\\ &=\boxed{\boxed{\begin{pmatrix}2&-1\\-2&-1\end{pmatrix}}} \end{aligned}\)
Jadi, \(X=\begin{pmatrix}2&-1\\-2&-1\end{pmatrix}\).
JAWAB: D

No.

Diketahui matriks \(A=\begin{pmatrix}2&-5\\-5&12\end{pmatrix}\) dan \(B=\begin{pmatrix}1&-2\\-1&1\end{pmatrix}\). Tentukan (3AB)−1
Grup Telegram
ALTERNATIF PENYELESAIAN
\(\begin{aligned} 3AB&=3\begin{pmatrix}2&-5\\-5&12\end{pmatrix}\begin{pmatrix}1&-2\\-1&1\end{pmatrix}\\ &=3\begin{pmatrix}(2)(1)+(-5)(-1)&(2)(-2)+(-5)(1)\\(-5)(1)+(12)(-1)&(-5)(-2)+(12)(1)\end{pmatrix}\\ &=3\begin{pmatrix}2+5&-4-5\\-5-12&10+12\end{pmatrix}\\ &=3\begin{pmatrix}7&-9\\-17&22\end{pmatrix}\\ &=\boxed{\boxed{\begin{pmatrix}21&-27\\-54&66\end{pmatrix}}} \end{aligned}\)
Jadi, \((3AB)^{-1}=\begin{pmatrix}21&-27\\-54&66\end{pmatrix}\).

No.

Diketahui matriks \(A=\begin{pmatrix}1&0\\2&3\end{pmatrix}\), \(B=\begin{pmatrix}-1&-3\\2&0\end{pmatrix}\), dan memenuhi persamaan AX + 2B = I, dengan I adalah matriks identitas. Maka nilai determinan matriks X adalah
  1. 6
  2. 7
  3. 8
  1. 9
  2. 10
SKMP Agustus 2020
ALTERNATIF PENYELESAIAN
\(\begin{aligned} |A|&=(1)(3)-(0)(2)\\ &=3-0\\ &=3 \end{aligned}\)

\(\begin{aligned} AX+2B&=I\\ AX&=I-2B\\ &=\begin{pmatrix}1&0\\0&1\end{pmatrix}-2\begin{pmatrix}-1&-3\\2&0\end{pmatrix}\\ &=\begin{pmatrix}1&0\\0&1\end{pmatrix}-\begin{pmatrix}-2&-6\\4&0\end{pmatrix}\\ &=\begin{pmatrix}3&6\\-4&1\end{pmatrix}\\ |AX|&=\begin{vmatrix}3&6\\-4&1\end{vmatrix}\\ |A||X|&=(3)(1)-(6)(-4)\\ 3|X|&=3+24\\ &=27\\ |X|&=\boxed{\boxed{9}} \end{aligned}\)
Jadi, nilai determinan matriks X adalah 9.
JAWAB: D

No.

Jika diketahui matriks A memenuhi persamaan \({\begin{pmatrix}2&1\\4&5\end{pmatrix}A=\begin{pmatrix}3&1\\3&2\end{pmatrix}\begin{pmatrix}2&5\\1&3\end{pmatrix}}\), maka determinan dari A−1 adalah
  1. −2
  2. \(-\dfrac12\)
  3. 0
  1. 1
  2. 2
SKMP Juli 2020
ALTERNATIF PENYELESAIAN
\(\begin{aligned} \begin{pmatrix}2&1\\4&5\end{pmatrix}A&=\begin{pmatrix}3&1\\3&2\end{pmatrix}\begin{pmatrix}2&5\\1&3\end{pmatrix}\\ \begin{vmatrix}2&1\\4&5\end{vmatrix}|A|&=\begin{vmatrix}3&1\\3&2\end{vmatrix}\begin{vmatrix}2&5\\1&3\end{vmatrix}\\ 6|A|&=(3)(1)\\ 6|A|&=3\\ |A|&=\dfrac36\\ &=\dfrac12 \end{aligned}\)

\(\begin{aligned} \left|A^{-1}\right|&=\dfrac1{|A|}\\ &=\dfrac1{\dfrac12}\\ &=\boxed{\boxed{2}} \end{aligned}\)
Jadi, determinan dari A−1 adalah 2.
JAWAB: E

No.

Jika matriks \(A=\begin{pmatrix}2&1\\3&-4\end{pmatrix}\), \(B=\begin{pmatrix}-4&-1\\-3&2\end{pmatrix}\), dan \(C=\begin{pmatrix}-11&0\\0&-11\end{pmatrix}\), maka (A × B) − C sama dengan
  1. \(\begin{pmatrix}1&1\\1&1\end{pmatrix}\)
  2. \(\begin{pmatrix}1&0\\0&1\end{pmatrix}\)
  3. \(\begin{pmatrix}0&1\\1&0\end{pmatrix}\)
  1. \(\begin{pmatrix}-1&-1\\-1&-1\end{pmatrix}\)
  2. \(\begin{pmatrix}0&0\\0&0\end{pmatrix}\)
Grup Telegram
ALTERNATIF PENYELESAIAN
\(\begin{aligned} (A\times B)-C&=\begin{pmatrix}2&1\\3&-4\end{pmatrix}\begin{pmatrix}-4&-1\\-3&2\end{pmatrix}-\begin{pmatrix}-11&0\\0&-11\end{pmatrix}\\ &=\begin{pmatrix}-11&0\\0&-11\end{pmatrix}-\begin{pmatrix}-11&0\\0&-11\end{pmatrix}\\ &=\boxed{\boxed{\begin{pmatrix}0&0\\0&0\end{pmatrix}}} \end{aligned}\)
Jadi, \((A\times B)-C=\begin{pmatrix}0&0\\0&0\end{pmatrix}\).
JAWAB: E

No.

Jika diketahui \(\begin{pmatrix}x+3\\4-y\end{pmatrix}=\begin{pmatrix}7\\5\end{pmatrix}\), Nilai dari x + y = ....
Grup Telegram
ALTERNATIF PENYELESAIAN
\(\begin{aligned} x+3&=7\\ x&=4 \end{aligned}\)

\(\begin{aligned} 4-y&=5\\ -y&=1\\ y&=-1 \end{aligned}\)
\(\begin{aligned} x+y&=4+(-1)\\ &=\boxed{\boxed{3}} \end{aligned}\)
Jadi, x + y = 3.


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