Exercise Zone : Invers Matriks [2]

Table of Contents

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

Tipe:

StandarSNBTHOTS

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
3. 2

4. 1
5. 0

Matematika Zone Indonesia

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

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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. \(-\dfrac12\)
3. 0

4. \(\dfrac12\)
5. 2

Matematika Zone Indonesia

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

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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

4. 1
5. 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

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No.

Jika matriks \(A\cdot B=\begin{pmatrix}4&6\\-7&-14\end{pmatrix}\) dan det A = 7, maka det(7BA⁻¹) adalah....

1. −16
2. −15
3. −14

4. −13
5. −12

Matematika Zone Indonesia

ALTERNATIF PENYELESAIAN

\(\begin{aligned} A\cdot B&=\begin{pmatrix}4&6\\-7&-14\end{pmatrix}\\[4pt] |AB|&=(4)(-14)-(6)(-7)\\ |A||B|&=-56+42\\ 7|B|&=-14\\ |B|&=-2 \end{aligned}\)

\(\begin{aligned} \left|7BA^{-1}\right|&=7^2|B|\left|A^{-1}\right|\\ &=49(-2)\left(\dfrac1{|A|}\right)\\ &=-98\left(\dfrac17\right)\\ &=\boxed{\boxed{-14}} \end{aligned}\)

Jadi, det(7BA⁻¹) = −14.
JAWAB: C

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No.

Diketahui matriks A berordo 2×2 dan \(B=\begin{pmatrix}-3&5\\-1&2\end{pmatrix}\) dan \(C=\begin{pmatrix}4&5\\2&3\end{pmatrix}\). Jika A memenuhi B ⋅ A = C, maka det (2A⁻¹) adalah

1. −2
2. −1
3. \(-\dfrac12\)

4. $\dfrac12$
5. 2

ALTERNATIF PENYELESAIAN

\(\begin{aligned} \det B&=(-3)(2)-(5)(-1)\\ &=-6-(-5)\\ &=-1 \end{aligned}\)

\(\begin{aligned} \det C&=(4)(3)-(2)(5)\\ &=12-10\\ &=2 \end{aligned}\)

\(\begin{aligned} B\cdot A&=C\\ A&=B^{-1}\cdot C\\ \det A&=\det\left(B^{-1}\cdot C\right)\\ &=\dfrac1{\det B}\cdot\det C\\ &=\dfrac1{-1}\cdot2\\ &=-2 \end{aligned}\)

\(\begin{aligned} \det\left(2A^{-1}\right)&=2^2\det A^{-1}\\ &=4\cdot\dfrac1{\det A}\\ &=4\cdot\dfrac1{-2}\\ &=\boxed{\boxed{-2}} \end{aligned}\)

Jadi, det (2A⁻¹) = −2.
JAWAB: A

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