Matriks (2) : Soal dan Pembahasan (DISCONTINUED)

Berikut ini adalah kumpulan soal mengenai matriks. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.

• 1
• 2
• 3

---

No. 11
Jika matriks {A=\begin{pmatrix}a&-3\\1&1\end{pmatrix}} merupakan matriks yang mempunyai invers, maka hasil kali semua nilai a yang mungkin sehingga {3\det(A)=\det\left(A^{-1}\right)-2} adalah

2. 
   
3. 
   

5. 
   

\begin{aligned}
\det(A)&=a\cdot1-(-3)\cdot1\\
&=a+3
\end{aligned}

CARA 1

\begin{aligned}
A^{-1}&=\dfrac1{\det(A)}\begin{pmatrix}1&3\\-1&a\end{pmatrix}\\
&=\dfrac1{a+3}\begin{pmatrix}1&3\\-1&a\end{pmatrix}\\
&=\begin{pmatrix}\dfrac1{a+3}&\dfrac3{a+3}\\-\dfrac1{a+3}&\dfrac{a}{a+3}\end{pmatrix}\\
\det\left(A^{-1}\right)&=\dfrac1{a+3}\cdot\dfrac{a}{a+3}-\dfrac3{a+3}\cdot\left(-\dfrac1{a+3}\right)\\
&=\dfrac{a}{(a+3)^2}+\dfrac3{(a+3)^2}\\
&=\dfrac{a+3}{(a+3)^2}\\
&=\dfrac1{a+3}
\end{aligned}

\begin{aligned}
3\det(A)&=\det\left(A^{-1}\right)-2\\
3(a+3)&=\dfrac1{a+3}-2\qquad\color{red}{\times (a+3)}\\
3(a+3)^2&=1-2(a+3)\\
3\left(a^2+6a+9\right)&=1-2a-6\\
3a^2+18a+27&=-2a-5\\
3a^2+18a+27+2a+5&=0\\
3a^2+20a+32&=0
\end{aligned}

a_1a_2=\dfrac{32}3

CARA 2

\begin{aligned}
3\det(A)&=\det\left(A^{-1}\right)-2\\
3\det(A)&=\dfrac1{\det(A)}-2&\color{red}{\times\det(A)}\\
3\left(\det(A)\right)^2&=1-2\det(A)\\
3\left(\det(A)\right)^2+2\det(A)-1&=0\\
\left(3\det(A)-1\right)\left(\det(A)+1\right)&=0\\
(3(a+3)-1)(a+3+1)&=0\\
(3a+8)(a+4)&=0
\end{aligned}

\begin{aligned}
a_1a_2&=\left(-\dfrac83\right)(-4)\\
&=\dfrac{32}3
\end{aligned}

Jadi, hasil kali semua nilai a yang mungkin adalah \dfrac{32}3.
JAWAB: E

---

No. 12
Diketahui persamaan matriks:
{2\begin{pmatrix}x&2\\6&6\end{pmatrix}+\begin{pmatrix}2&4\\-2&0\end{pmatrix}=\begin{pmatrix}1&3\\2&4\end{pmatrix}\begin{pmatrix}-1&2\\3&y\end{pmatrix}}
Nilai x-y=

1. -2
2. 2
   
3. 0
   

4. 1
5. -1
   

\begin{aligned}
2\begin{pmatrix}x&2\\6&6\end{pmatrix}+\begin{pmatrix}2&4\\-2&0\end{pmatrix}&=\begin{pmatrix}1&3\\2&4\end{pmatrix}\begin{pmatrix}-1&2\\3&y\end{pmatrix}\\
\begin{pmatrix}2x&4\\12&12\end{pmatrix}+\begin{pmatrix}2&4\\-2&0\end{pmatrix}&=\begin{pmatrix}8&3y+2\\10&4y+4\end{pmatrix}\\
\begin{pmatrix}2x+2&8\\10&12\end{pmatrix}&=\begin{pmatrix}8&3y+2\\10&4y+4\end{pmatrix}
\end{aligned}

\begin{aligned}
2x+2&=8\\
2x&=6\\
x&=3
\end{aligned}

\begin{aligned}
3y+2&=8\\
3y&=6\\
y&=2
\end{aligned}

\begin{aligned}
x-y&=3-2\\
&=\boxed{\boxed{1}}
\end{aligned}

Jadi, x-y=1.
JAWAB: D

---

No. 13

Misalkan A^T adalah transpose matriks A dan {I=\begin{pmatrix}1&0\\0&1\end{pmatrix}}. Jika {A=\begin{pmatrix}8&0\\a&b\end{pmatrix}} sehingga {5A=3A^T+16I}, maka nilai {10a+4b} adalah

1. 32
2. 34
   
3. 36
   

4. 35
5. 40
   

{A^T=\begin{pmatrix}8&a\\0&b\end{pmatrix}}

\begin{aligned}
5A&=3A^T+16I\\
5\begin{pmatrix}8&0\\a&b\end{pmatrix}&=3\begin{pmatrix}8&a\\0&b\end{pmatrix}+16\begin{pmatrix}1&0\\0&1\end{pmatrix}\\
\begin{pmatrix}40&0\\5a&5b\end{pmatrix}&=\begin{pmatrix}24&3a\\0&3b\end{pmatrix}+\begin{pmatrix}16&0\\0&16\end{pmatrix}\\
\begin{pmatrix}40&0\\5a&5b\end{pmatrix}&=\begin{pmatrix}40&3a\\0&3b+16\end{pmatrix}
\end{aligned}

\begin{aligned}
0&=3a\\
a&=0
\end{aligned}

\begin{aligned}
5b&=3b+16\\
2b&=16\\
4b&=32
\end{aligned}

\begin{aligned}
10a+4b&=10(0)+32\\
&=\boxed{\boxed{32}}
\end{aligned}

Jadi, 10a+4b=32.
JAWAB: A

---

No. 14

Diketahui matriks A berordo 2\times2 dan {B=\begin{pmatrix}-3&5\\-1&2\end{pmatrix}} dan {C=\begin{pmatrix}4&5\\2&3\end{pmatrix}}. Jika A memenuhi {B\cdot A=C}, maka \det\left(2A^{-1}\right) adalah

1. -2
2. -1
3. -\dfrac12

4. \dfrac12
5. 2

\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\left(2A^{-1}\right)=-2.
JAWAB: A

---

No. 15

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. 4
   
3. -2
   

4. 0
5. 6
   

\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=\boxed{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: C

---

No. 16

Diberikan matriks {P = \begin{pmatrix}3&-1\\5&2\end{pmatrix}} dan {Q = \begin{pmatrix}3r&2\\r&p+1\end{pmatrix}} dengan {r\neq0} dan {p\neq0}. Supaya matriks PQ tidak mempunyai invers, maka nilai {3p + 2 =}

1. 4
2. 3
   
3. 2
   

4. 1
5. 0
   

Tidak punya invers artinya det = 0.
\begin{aligned}
\left|PQ\right|&=0\\
|P||Q|&=0
\end{aligned}
Karena |P|\neq0 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. 17

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

\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. 18

Diketahui matriks {A=\begin{pmatrix}3&a\\b&2\end{pmatrix}} dan {B=\begin{pmatrix}a&b\\3&2\end{pmatrix}}. Jika C adalah matriks berukuran 2\times2 yang memiliki invers dan matriks AC maupun matriks BC tidak memiliki invers, maka nilai {4a^2+9b^2=}

1. 72
2. 74
   
3. 76
   

4. 78
5. 80
   

|A|=6-ab
|B|=2a-3b

\begin{aligned}
AC&=0\\
|A||C|&=0\\
(6-ab)|C|&=0\\
6-ab&=0\\
ab&=6
\end{aligned}

\begin{aligned}
BC&=0\\
|B||C|&=0\\
(2a-3b)|C|&=0\\
2a-3b&=0
\end{aligned}

\begin{aligned}
4a^2+9b^2&=(2a)^2+(-3b)^2\\
&=(2a+(-3b))^2-2(2a)(-3b)\\
&=(2a-3b)^2+12ab\\
&=0^2+12(6)\\
&=\boxed{\boxed{72}}
\end{aligned}

Jadi, 4a^2+9b^2=72.
JAWAB: A

---

No. 19

Jika {\begin{pmatrix}1&2\\1&3\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}a\\b\end{pmatrix}} dan {\begin{pmatrix}5&2\\3&1\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}=\begin{pmatrix}1\\4\end{pmatrix}}, maka nilai {x+y=}

1. 30
2. 31
   
3. 32
   

4. 33
5. 34
   

\begin{aligned}
\begin{pmatrix}5&2\\3&1\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}&=\begin{pmatrix}1\\4\end{pmatrix}\\
\begin{pmatrix}5&2\\3&1\end{pmatrix}\begin{pmatrix}1&2\\1&3\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}&=\begin{pmatrix}1\\4\end{pmatrix}\\
\begin{pmatrix}7&16\\4&9\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}&=\begin{pmatrix}1\\4\end{pmatrix}\\
\begin{pmatrix}x\\y\end{pmatrix}&=\begin{pmatrix}7&16\\4&9\end{pmatrix}^{-1}\begin{pmatrix}1\\4\end{pmatrix}\\
&=\dfrac1{(7)(9)-(16)(4)}\begin{pmatrix}9&-16\\-4&7\end{pmatrix}\begin{pmatrix}1\\4\end{pmatrix}\\
&=\dfrac1{(-1}\begin{pmatrix}-55\\24\end{pmatrix}\\
&=\begin{pmatrix}55\\-24\end{pmatrix}
\end{aligned}
x=55
y=-24

\begin{aligned}
x+y&=55+(-24)\\
&=\boxed{\boxed{31}}
\end{aligned}

Jadi, x+y=31.
JAWAB: B

---

No. 20

Diketahui matriks A berukuran {3\times3} dan memenuhi {A\begin{pmatrix}3\\2\\1\end{pmatrix}=\begin{pmatrix}2\\4\\2\end{pmatrix}} dan {A\begin{pmatrix}3\\1\\2\end{pmatrix}=\begin{pmatrix}2\\2\\2\end{pmatrix}}, matriks {A\begin{pmatrix}6\\4\\2\end{pmatrix}=}

1. \begin{pmatrix}4\\8\\4\end{pmatrix}
2. \begin{pmatrix}8\\4\\8\end{pmatrix}
3. \begin{pmatrix}12\\8\\4\end{pmatrix}

4. \begin{pmatrix}4\\8\\12\end{pmatrix}
5. \begin{pmatrix}12\\8\\12\end{pmatrix}

CARA BIASA

Misal A=\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}

\begin{aligned}
A\begin{pmatrix}3\\2\\1\end{pmatrix}&=\begin{pmatrix}2\\4\\2\end{pmatrix}\\
\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}\begin{pmatrix}3\\2\\1\end{pmatrix}&=\begin{pmatrix}2\\4\\2\end{pmatrix}\\
\begin{pmatrix}3a+2b+c\\3d+2e+f\\3g+2h+i\end{pmatrix}&=\begin{pmatrix}2\\4\\2\end{pmatrix}
\end{aligned}

3a+2b+c=2
3d+2e+f=4
3g+2h+i=2

\begin{aligned}
A\begin{pmatrix}6\\4\\2\end{pmatrix}&=\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}\begin{pmatrix}6\\4\\2\end{pmatrix}\\
&=\begin{pmatrix}6a+4b+2c\\6d+4e+2f\\6g+4h+2i\end{pmatrix}\\
&=\begin{pmatrix}2(3a+2b+c)\\2(3d+2e+f)\\2(3g+2h+i)\end{pmatrix}\\
&=\begin{pmatrix}2(2)\\2(4)\\2(2)\end{pmatrix}\\
&=\boxed{\boxed{\begin{pmatrix}4\\8\\4\end{pmatrix}}}
\end{aligned}

CARA CEPAT

\begin{aligned}
A\begin{pmatrix}6\\4\\2\end{pmatrix}&=A\cdot2\begin{pmatrix}3\\2\\1\end{pmatrix}\\
&=2A\begin{pmatrix}3\\2\\1\end{pmatrix}\\
&=2\begin{pmatrix}2\\4\\2\end{pmatrix}\\
&=\boxed{\boxed{\begin{pmatrix}4\\8\\4\end{pmatrix}}}
\end{aligned}

Jadi, A\begin{pmatrix}6\\4\\2\end{pmatrix}=\begin{pmatrix}4\\8\\4\end{pmatrix}.
JAWAB: A

• 1
• 2
• 3