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Berikut ini adalah kumpulan soal mengenai Invers Matriks. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.

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StandarSNBTHOTS


No.

Jika \(P=\begin{pmatrix}1&2\\1&3\end{pmatrix}\) dan \(\begin{pmatrix}x&y\\-z&z\end{pmatrix}=2P^{-1}\) dengan P−1 menyatakan invers matriks P, maka x + y = ....
Matematika Zone Id
ALTERNATIF PENYELESAIAN
\begin{aligned} |P|&=(1)(3)-(2)(1)\\ &=1 \end{aligned} \begin{aligned} P^{-1}&=\dfrac1{|P|}\begin{pmatrix}3&-2\\-1&1\end{pmatrix}\\ &=\dfrac11\begin{pmatrix}3&-2\\-1&1\end{pmatrix}\\ &=\begin{pmatrix}3&-2\\-1&1\end{pmatrix} \end{aligned}
\begin{aligned} \begin{pmatrix}x&y\\-z&z\end{pmatrix}&=2P^{-1}\\ \begin{pmatrix}x&y\\-z&z\end{pmatrix}&=2\begin{pmatrix}3&-2\\-1&1\end{pmatrix}\\ \begin{pmatrix}x&y\\-z&z\end{pmatrix}&=\begin{pmatrix}6&-4\\-2&2\end{pmatrix} \end{aligned} x = 6 dan y = −4 \begin{aligned} x+y&=6+(-4)\\ &=2 \end{aligned}
Jadi, x + y = 2.

No.

Jika matriks \(A=\begin{pmatrix}3&-2\\7&-5\end{pmatrix}\), invers matriks A adalah A−1 = ....
Guru Matematika Indonesia
ALTERNATIF PENYELESAIAN
\begin{aligned} \det A&=\begin{vmatrix}3&-2\\7&-5\end{vmatrix}\\ &=3\cdot(-5)-(-2)\cdot7\\ &=-15+14\\ &=-1 \end{aligned}
\begin{aligned} A^{-1}&=\dfrac1{\det A}\begin{pmatrix}-5&2\\-7&3\end{pmatrix}\\ &=\dfrac1{-1}\begin{pmatrix}-5&2\\-7&3\end{pmatrix}\\ &=\boxed{\boxed{\begin{pmatrix}5&-2\\7&-3\end{pmatrix}}}\end{aligned}
Jadi, \(A^{-1}=\begin{pmatrix}5&-2\\7&-3\end{pmatrix}\).

No.

Sebuah matriks diketahui sebagai berikut :\[A=\begin{pmatrix}2&1\\3&2\end{pmatrix}\] Tentukan invers matriks!
Latihan PAS Gajil 2021
ALTERNATIF PENYELESAIAN
\begin{aligned} \det A&=2\cdot2-1\cdot3\\ &=4-3\\ &=1 \end{aligned} \begin{aligned} A^{-1}&=\dfrac1{\det A}\begin{pmatrix}2&-1\\-3&2\end{pmatrix}\\ &=\dfrac1{1}\begin{pmatrix}2&-1\\-3&2\end{pmatrix}\\ &=\boxed{\boxed{\begin{pmatrix}2&-1\\-3&2\end{pmatrix}}} \end{aligned}
Jadi, \(A^{-1}=\begin{pmatrix}2&-1\\-3&2\end{pmatrix}\).

No.

Sebuah matriks diketahui sebagai berikut : \[A=\begin{pmatrix}1&2\\1&3\end{pmatrix}\] Tentukan invers matriks!
Latihan Soal Pra UKK Semester Genap Kelas XI
ALTERNATIF PENYELESAIAN
\begin{aligned} \det A&=1\cdot3-2\cdot1\\ &=3-2\\ &=1 \end{aligned} \begin{aligned} A^{-1}&=\dfrac1{\det A} \begin{pmatrix}3&-2\\-1&1\end{pmatrix}\\ &=\dfrac11\begin{pmatrix}3&-2\\-1&1\end{pmatrix}\\ &=\boxed{\boxed{\begin{pmatrix}3&-2\\-1&1\end{pmatrix}}} \end{aligned}
Jadi, invers matriks A adalah \(A^{-1}=\begin{pmatrix}3&-2\\-1&1\end{pmatrix}\).

No.

Diketahui $P=\begin{pmatrix}-3&1\\4&-2\end{pmatrix}$ dan $Q=\begin{pmatrix}3&4\\1&2\end{pmatrix}$. Hitunglah (PQ)−1
Grup Telegram
ALTERNATIF PENYELESAIAN
$\begin{aligned} PQ&=\begin{pmatrix}-3&1\\4&-2\end{pmatrix}\begin{pmatrix}3&4\\1&2\end{pmatrix}\\ &=\begin{pmatrix}-8&-10\\10&12\end{pmatrix}\\ \left(PQ\right)^{-1}&=\dfrac1{(-8)(12)-(-10)(10)}\begin{pmatrix}12&10\\-10&-8\end{pmatrix}\\ &=\dfrac1{-96+100}\begin{pmatrix}12&10\\-10&-8\end{pmatrix}\\ &=\dfrac14\begin{pmatrix}12&10\\-10&-8\end{pmatrix}\\ &=\boxed{\boxed{\begin{pmatrix}3&\dfrac52\\-\dfrac52&-2\end{pmatrix}}} \end{aligned}$
Jadi, $\left(PQ\right)^{-1}=\begin{pmatrix}3&\dfrac52\\-\dfrac52&-2\end{pmatrix}$.

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.

X adalah matriks persegi berordo 2×2 yang memenuhi $X\begin{pmatrix}1&2\\3&4\end{pmatrix}=\begin{pmatrix}4&8\\5&8\end{pmatrix}$. Matriks X adalah
  1. $\begin{pmatrix}3&2\\-2&1\end{pmatrix}$
  2. $\begin{pmatrix}3&2\\2&1\end{pmatrix}$
  3. $\begin{pmatrix}-4&0\\-1&-2\end{pmatrix}$
  1. $\begin{pmatrix}4&0\\2&1\end{pmatrix}$
  2. $\begin{pmatrix}4&0\\-1&2\end{pmatrix}$
Grup Telegram
ALTERNATIF PENYELESAIAN
$\begin{aligned} X\begin{pmatrix}1&2\\3&4\end{pmatrix}&=\begin{pmatrix}4&8\\5&8\end{pmatrix}\\ X&=\begin{pmatrix}4&8\\5&8\end{pmatrix}\begin{pmatrix}1&2\\3&4\end{pmatrix}^{-1}\\ &=\begin{pmatrix}4&8\\5&8\end{pmatrix}\cdot\dfrac1{1\cdot4-2\cdot3}\begin{pmatrix}4&-2\\-3&1\end{pmatrix}\\ &=\begin{pmatrix}4&8\\5&8\end{pmatrix}\cdot\dfrac1{-2}\begin{pmatrix}4&-2\\-3&1\end{pmatrix}\\ &=\dfrac1{-2}\begin{pmatrix}4&8\\5&8\end{pmatrix}\begin{pmatrix}4&-2\\-3&1\end{pmatrix}\\ &=\dfrac1{-2}\begin{pmatrix}-8&0\\-4&-2\end{pmatrix}\\ &=\boxed{\boxed{\begin{pmatrix}4&0\\2&1\end{pmatrix}}} \end{aligned}$
Jadi, $X=\begin{pmatrix}4&0\\2&1\end{pmatrix}$
JAWAB: D

No.

Matriks A yang memenuhi \(\begin{pmatrix}2&k\\1&0\end{pmatrix}A=\begin{pmatrix}2&4k\\1&0\end{pmatrix}\) adalah ....
Matematika Zone Indonesia
ALTERNATIF PENYELESAIAN
\(\begin{aligned} \begin{pmatrix}2&k\\1&0\end{pmatrix}A&=\begin{pmatrix}2&4k\\1&0\end{pmatrix}\\ A&=\begin{pmatrix}2&k\\1&0\end{pmatrix}^{-1}\begin{pmatrix}2&4k\\1&0\end{pmatrix}\\ &=\dfrac1{2\cdot0-k\cdot1}\begin{pmatrix}0&-k\\-1&2\end{pmatrix}\begin{pmatrix}2&4k\\1&0\end{pmatrix}\\ &=\dfrac1{-k}\begin{pmatrix}-k&0\\0&-4k\end{pmatrix}\\ &=\begin{pmatrix}1&0\\0&4\end{pmatrix} \end{aligned}\)
Jadi, \(A=\begin{pmatrix}1&0\\0&4\end{pmatrix}\).

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.

Diberikan matriks \(P=\begin{pmatrix}2&-1\\4&3\end{pmatrix}\) dan \(Q=\begin{pmatrix}2r&1\\r&p+1\end{pmatrix}\) dengan r ≠ 0 dan p ≠ 0. Matriks PQ tidak mempunyai invers apabila nilai p = ....
Matematika Zone Indonesia
ALTERNATIF PENYELESAIAN
\(\begin{aligned} |PQ|&=0\\ |P||Q|&=0\\ ((2)(3)-(-1)(4))((2r)(p+1)-(1)(r))&=0\\ (10)(2pr+2r-r)&=0\\ 2pr+r&=0\\ r(2p+1)&=0\\ 2p+1&=0\\ 2p&=-1\\ p&=-\dfrac12 \end{aligned}\)
Jadi, \(p=-\dfrac12\).


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