Soal dan Pembahasan - Matriks 4. Determinan dan Invers Matriks 3x3

Berikut ini adalah Kumpulan Soal Determinan dan Invers Matriks 3x3 dan Pembahasan. Silakan dimanfaatkan sebaik mungkin.

Tata Cara Belajar:
Cobalah mengerjakan soal-soal yang tersedia secara mandiri. Setelah itu cocokkanlah jawaban kamu dengan pembahasan yang telah disediakan, dengan cara klik "Lihat/Tutup".

Soal No. 1

Supaya matriks $\left( \begin{matrix} 2 & a & 2 \\ 3 & a & a \\ 3 & 1 & 3 \\ \end{matrix} \right)$ tidak mempunyai invers, maka nilai $a$ yang memenuhi adalah ….
A. $\frac{2}{3}$ atau 3
B. $-\frac{2}{3}$ atau 3
C. $-3$ atau $\frac{2}{3}$
D. $-3$ atau $-\frac{2}{3}$
E. $\frac{3}{2}$ atau 3

Penyelesaian: Lihat/Tutup

$\left| \begin{matrix} 2 & a & 2 \\ 3 & a & a \\ 3 & 1 & 3 \\ \end{matrix} \right|\left. \begin{matrix} {} & 2 & a \\ {} & 3 & a \\ {} & 3 & 1 \\ \end{matrix} \right|=0$
$(6a+3a^2+6)-(6a+2a+9a)=0$
$6a+3a^2+6-17a=0$
$3a^2-11a+6=0$
$(3a-2)(a-3)=0$
$3a-2=0\to a=\frac{2}{3}$
$a-3=0\to a=3$
Jadi, nilai $a$ yang memenuhi adalah $\frac{2}{3}$ atau 3.
Jawaban: A

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Soal No. 2

Bila $A=\left( \begin{matrix} 2 & 2 & -2 \\ 1 & 2 & 3 \\ 2 & 3 & 4 \\ \end{matrix} \right)$, determinan matriks A adalah ….
A. 5
B. 4
C. 3
D. 2
E. 1

Penyelesaian: Lihat/Tutup

$\begin{align}\left| A \right| &= \left| \begin{matrix} 2 & 2 & -2 \\ 1 & 2 & 3 \\ 2 & 3 & 4 \\ \end{matrix} \right|\left. \begin{matrix} {} & 2 & 2 \\ {} & 1 & 2 \\ {} & 2 & 3 \\ \end{matrix} \right| \\ &= (16+12-6)-(-8+18+8) \\ &= 22-18 \\ & \left| A \right| &= 4 \end{align}$
Jawaban: B

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Soal No. 3

Bila $A=\left( \begin{matrix} 2 & 2 & -2 \\ 1 & 2 & 3 \\ 2 & 3 & 4 \\ \end{matrix} \right)$, maka invers A adalah ….
A. $\left( \begin{matrix} -\frac{1}{4} & -\frac{7}{2} & \frac{5}{2} \\ \frac{1}{2} & -3 & 2 \\ \frac{1}{4} & \frac{1}{2} & \frac{1}{2} \\ \end{matrix} \right)$
B. $\left( \begin{matrix} \frac{1}{4} & \frac{1}{2} & \frac{5}{2} \\ \frac{1}{2} & 3 & 2 \\ \frac{1}{4} & \frac{7}{2} & \frac{1}{2} \\ \end{matrix} \right)$
C. $\left( \begin{matrix} -\frac{1}{4} & -\frac{7}{2} & \frac{5}{2} \\ \frac{1}{2} & 3 & -2 \\ -\frac{1}{4} & -\frac{1}{2} & \frac{1}{2} \\ \end{matrix} \right)$
D. $\left( \begin{matrix} \frac{1}{4} & 7 & \frac{5}{2} \\ 2 & 3 & 2 \\ 2 & 1 & -2 \\ \end{matrix} \right)$
E. $\left( \begin{matrix} 4 & -7 & 2 \\ 2 & 3 & 2 \\ 2 & -1 & -1 \\ \end{matrix} \right)$

Penyelesaian: Lihat/Tutup

$A=\left( \begin{matrix} 2 & 2 & -2 \\ 1 & 2 & 3 \\ 2 & 3 & 4 \\ \end{matrix} \right)$
Determinan matriks A:
$\begin{align}\left| A \right| &= \left| \begin{matrix} 2 & 2 & -2 \\ 1 & 2 & 3 \\ 2 & 3 & 4 \\ \end{matrix} \right|\left. \begin{matrix} {} & 2 & 2 \\ {} & 1 & 2 \\ {} & 2 & 3 \\ \end{matrix} \right| \\ &= (16+12-6)-(-8+18+8) \\ &= 22-18 \\ \left| A \right| &= 4 \end{align}$
Minor matriks A:
$M_{11}=\left| \begin{matrix} 2 & 3 \\ 3 & 4 \\ \end{matrix} \right|=8-9=-1$
$M_{12}=\left| \begin{matrix} 1 & 3 \\ 2 & 4 \\ \end{matrix} \right|=4-6=-2$
$M_{13}=\left| \begin{matrix} 1 & 2 \\ 2 & 3 \\ \end{matrix} \right|=3-4=-1$
$M_{21}=\left| \begin{matrix} 2 & -2 \\ 3 & 4 \\ \end{matrix} \right|=8+6=14$
$M_{22}=\left| \begin{matrix} 2 & -2 \\ 2 & 4 \\ \end{matrix} \right|=8+4=12$
$M_{23}=\left| \begin{matrix} 2 & 2 \\ 2 & 3 \\ \end{matrix} \right|=6-4=2$
$M_{31}=\left| \begin{matrix} 2 & -2 \\ 2 & 3 \\ \end{matrix} \right|=6+4=10$
$M_{32}=\left| \begin{matrix} 2 & -2 \\ 1 & 3 \\ \end{matrix} \right|=6+2=8$
$M_{33}=\left| \begin{matrix} 2 & 2 \\ 1 & 2 \\ \end{matrix} \right|=4-2=2$
Kofaktor matriks A:
$kof(A)=\left( \begin{matrix} M_{11} & -M_{12} & M_{13} \\ -M_{21} & M_{22} & -M_{23} \\ M_{31} & -M_{32} & M_{33} \\ \end{matrix} \right)$
$kof(A)=\left( \begin{matrix} -1 & 2 & -1 \\ -14 & 12 & -2 \\ 10 & -8 & 2 \\ \end{matrix} \right)$
Adjoint matriks A:
$Adj(A)=[kof(A)]^T$
$Adj(A)=\left( \begin{matrix} -1 & -14 & 10 \\ 2 & 12 & -8 \\ -1 & -2 & 2 \\ \end{matrix} \right)$
Invers matriks A:
$\begin{align}A^{-1} &= \frac{1}{\det (A)}Adj(A) \\ &= \frac{1}{4}\left( \begin{matrix} -1 & -14 & 10 \\ 2 & 12 & -8 \\ -1 & -2 & 2 \\ \end{matrix} \right) \\ A^{-1} &= \left( \begin{matrix} -\frac{1}{4} & -\frac{7}{2} & \frac{5}{2} \\ \frac{1}{2} & 3 & -2 \\ -\frac{1}{4} & -\frac{1}{2} & \frac{1}{2} \\ \end{matrix} \right) \end{align}$
Jawaban: C

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Soal No. 4

Diketahui matriks $M=\left( \begin{matrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{matrix} \right)$ dengan det(M) = 15. Jika matriks $N=\left( \begin{matrix} -2a & -2b & -2c \\ -2d & -2e & -2f \\ -2g & -2h & -2i \\ \end{matrix} \right)$, maka nilai det(N) = ….
A. $-80$
B. 80
C. $-100$
D. 100
E. $-120$

Penyelesaian: Lihat/Tutup

$N=\left( \begin{matrix} -2a & -2b & -2c \\ -2d & -2e & -2f \\ -2g & -2h & -2i \\ \end{matrix} \right)$
$N=-2\left( \begin{matrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{matrix} \right)$
$N=-2M$
$\begin{align}\det (N) &= \det (-2M) \\ &= (-2)^3.\det (M) \\ &= -8\times 15 \\ \det (N) &= -120 \end{align}$
Jawaban: E

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Soal No. 5

Diketahui matriks $A=\left( \begin{matrix} 1 & 0 & 2 \\ -2 & 1 & -3 \\ 0 & -1 & 2 \\ \end{matrix} \right)$ dan adj(A) menyatakan adjoint matriks A. Nilai dari det(adj(A)) = ….
A. 3
B. 5
C. 6
D. 7
E. 9

Penyelesaian: Lihat/Tutup

$A=\left( \begin{matrix} 1 & 0 & 2 \\ -2 & 1 & -3 \\ 0 & -1 & 2 \\ \end{matrix} \right)$
Minor matriks A:
$M_{11}=\left| \begin{matrix} 1 & -3 \\ -1 & 2 \\ \end{matrix} \right|=2-3=-1$
$M_{12}=\left| \begin{matrix} -2 & -3 \\ 0 & 2 \\ \end{matrix} \right|=-4-0=-4$
$M_{13}=\left| \begin{matrix} -2 & 1 \\ 0 & -1 \\ \end{matrix} \right|=2-0=2$
$M_{21}=\left| \begin{matrix} 0 & 2 \\ -1 & 2 \\ \end{matrix} \right|=0+2=2$
$M_{22}=\left| \begin{matrix} 1 & 2 \\ 0 & 2 \\ \end{matrix} \right|=2-0=2$
$M_{23}=\left| \begin{matrix} 1 & 0 \\ 0 & -1 \\ \end{matrix} \right|=-1-0=-1$
$M_{31}=\left| \begin{matrix} 0 & 2 \\ 1 & -3 \\ \end{matrix} \right|=0-2=-2$
$M_{32}=\left| \begin{matrix} 1 & 2 \\ -2 & -3 \\ \end{matrix} \right|=-3+4=1$
$M_{33}=\left| \begin{matrix} 1 & 0 \\ -2 & 1 \\ \end{matrix} \right|=1-0=1$
Kofaktor matriks A:
$kof(A)=\left( \begin{matrix} M_{11} & -M_{12} & M_{13} \\ -M_{21} & M_{22} & -M_{23} \\ M_{31} & -M_{32} & M_{33} \\ \end{matrix} \right)$
$kof(A)=\left( \begin{matrix} -1 & 4 & 2 \\ -2 & 2 & 1 \\ -2 & -1 & 1 \\ \end{matrix} \right)$
Adjoint matriks A:
$Adj(A)=[kof(A)]^T$
$Adj(A)=\left( \begin{matrix} -1 & -2 & -2 \\ 4 & 2 & -1 \\ 2 & 1 & 1 \\ \end{matrix} \right)$
det(Adj(A) adalah:
= $\left| \begin{matrix} -1 & -2 & -2 \\ 4 & 2 & -1 \\ 2 & 1 & 1 \\ \end{matrix} \right|\left. \begin{matrix} {} & -1 & -2 \\ {} & 4 & 2 \\ {} & 2 & 1 \\ \end{matrix} \right|$
= $(-2+4-8)-(-8+1-8)$
= $-6-(-15)$
= 9
Jawaban: E

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Soal No. 6

Jika $A=\left( \begin{matrix} -1 & 2 & 1 \\ 3 & 1 & 3 \\ 3 & -4 & 5 \\ \end{matrix} \right)$ maka $\det (2A^t)$ = ….
A. $4\det A^t$
B. $8\det A$
C. $16\det A^t$
D. $32\det A$
E. $64\det A^t$

Penyelesaian: Lihat/Tutup

Ingat:
Jika $A_{n\times n}$ maka $\det (kA)=k^n.\det (A)$
$\det A^t=\det A$
$\begin{align}\det (2A^t) &= 2^3\det (A^t) \\ &= 8\det A \end{align}$
Jawaban: B

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Soal No. 7

Jika adjoint matriks $A=\left( \begin{matrix} 2 & 1 & -1 \\ -1 & 2 & 1 \\ 1 & -2 & 1 \\ \end{matrix} \right)$ adalah $\left( \begin{matrix} 4 & 1 & 3 \\ 2 & p & -1 \\ 0 & 5 & 5 \\ \end{matrix} \right)$, maka nilai $p$ = ….
A. $-3$
B. 1
C. 2
D. 3
E. 5

Penyelesaian: Lihat/Tutup

$A=\left( \begin{matrix} 2 & 1 & -1 \\ -1 & 2 & 1 \\ 1 & -2 & 1 \\ \end{matrix} \right)$
$\begin{align}\det A &= \left| \begin{matrix} 2 & 1 & -1 \\ -1 & 2 & 1 \\ 1 & -2 & 1 \\ \end{matrix} \right|\left. \begin{matrix} {} & 2 & 1 \\ {} & -1 & 2 \\ {} & 1 & -2 \\ \end{matrix} \right| \\ &= (4+1-2)-(-2-4-1) \\ &= 3-(-7) \\ \det A &= 10 \end{align}$
$Adj(A)=\left( \begin{matrix} 4 & 1 & 3 \\ 2 & p & -1 \\ 0 & 5 & 5 \\ \end{matrix} \right)$
$A^{-1}=\frac{1}{\det A}.Adj(A)$
$A^{-1}=\frac{1}{10}\left( \begin{matrix} 4 & 1 & 3 \\ 2 & p & -1 \\ 0 & 5 & 5 \\ \end{matrix} \right)$
$A.A^{-1}=I$ maka:
$\left( \begin{matrix}2 & 1 & -1 \\ -1 & 2 & 1 \\ 1 & -2 & 1 \\ \end{matrix} \right).\frac{1}{10}\left( \begin{matrix} 4 & 1 & 3 \\ 2 & p & -1 \\ 0 & 5 & 5 \\ \end{matrix} \right)$ = $\left( \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right)$
$\frac{1}{10}\left( \begin{matrix} 8+2+0 & 2+p-5 & 6-1-5 \\ -4+4+0 & -1+2p+5 & -3-2+5 \\ 4-4+0 & 1-2p+5 & 3+2+5 \\ \end{matrix} \right)$ = $\left( \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right)$
$\frac{1}{10}\left( \begin{matrix} 10 & p-3 & 0 \\ 0 & 2p+4 & 0 \\ 0 & 6-2p & 10 \\ \end{matrix} \right)$ = $\left( \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right)$
$\left( \begin{matrix} 1 & \frac{p-3}{10} & 0 \\ 0 & \frac{2p+4}{10} & 0 \\ 0 & \frac{6-2p}{10} & 1 \\ \end{matrix} \right)$ = $\left( \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right)$
$\begin{align}\frac{p-3}{10} &= 0 \\ p-3 &= 0 \\ p &= 3 \end{align}$
Jawaban: D

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Soal No. 8

Jika $P=\left( \begin{matrix} 3 & 1 & 3 \\ 2 & 1 & 2 \\ 3 & 5 & 4 \\ \end{matrix} \right)$ maka $\det (P^{-1})$ = ….
A. $\frac{1}{5}$
B. $\frac{1}{4}$
C. $\frac{1}{3}$
D. $\frac{1}{2}$
E. 1

Penyelesaian: Lihat/Tutup

$\begin{align}\left| P \right| &= \left| \begin{matrix} 3 & 1 & 3 \\ 2 & 1 & 2 \\ 3 & 5 & 4 \\ \end{matrix} \right|\left. \begin{matrix} {} & 3 & 1 \\ {} & 2 & 1 \\ {} & 3 & 5 \\ \end{matrix} \right| \\ &= (12+6+30)-(9+30+8) \\ &= 48-47 \\ \left| P \right| &= 1 \end{align}$
$\det (P^{-1})=\frac{1}{\det (P)}=1$
Jawaban: E

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Soal No. 9

Jika $A=\left( \begin{matrix} 1 & 2 & 4 \\ 3 & 1 & 6 \\ k & 3 & 2 \\ \end{matrix} \right)$ matriks singular, maka nilai $k$ = ….
A. 8
B. 0
C. 1
D. $-1$
E. 2

Penyelesaian: Lihat/Tutup

$\begin{align}\left| A \right| &= 0 \\ \left| \begin{matrix} 1 & 2 & 4 \\ 3 & 1 & 6 \\ k & 3 & 2 \\ \end{matrix} \right|\left. \begin{matrix} {} & 1 & 2 \\ {} & 3 & 1 \\ {} & k & 3 \\ \end{matrix} \right| &= 0 \\ (2+12k+36)-(4k+18+12) &= 0 \\ 12k+38-4k-30 &= 0 \\ 8k+8 &= 0 \\ 8k &= -8 \\ k &= -1 \end{align}$
Jawaban: D

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Soal No. 10

Jika $A=\left( \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{matrix} \right)$ maka det A = ….
A. 0
B. 1
C. 2
D. 3
E. 4

Penyelesaian: Lihat/Tutup

$\begin{align}\left| A \right| &= \left| \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{matrix} \right|\left. \begin{matrix} {} & 1 & 2 \\ {} & 4 & 5 \\ {} & 7 & 8 \\ \end{matrix} \right| \\ &= (45+84+96)-(105+48+72) \\ &= 225-225 \\ \left| A \right| &= 0 \end{align}$
Jawaban: A

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Soal No. 11

Nilai $\left| \begin{matrix} 3 & 2 & 1 \\ 0 & 4 & 5 \\ 0 & 0 & 2 \\ \end{matrix} \right|$ sama dengan ….
A. 12
B. 24
C. 36
D. 48
E. 52

Penyelesaian: Lihat/Tutup

$\begin{align}\left| \begin{matrix} 3 & 2 & 1 \\ 0 & 4 & 5 \\ 0 & 0 & 2 \\ \end{matrix} \right|=\left| \begin{matrix} 3 & 2 & 1 \\ 0 & 4 & 5 \\ 0 & 0 & 2 \\ \end{matrix} \right|\left. \begin{matrix} {} & 3 & 2 \\ {} & 0 & 4 \\ {} & 0 & 0 \\ \end{matrix} \right| \\ &= (24+0+0)-(0+0+0) \\ &= 24 \end{align}$
Jawaban: B

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Soal No. 12

Agar matriks $A=\left( \begin{matrix} 4 & 5 & 7 \\ 8 & x & 14 \\ 7 & 5 & 2 \\ \end{matrix} \right)$ merupakan matriks singular maka $x$ = ….
A. 0
B. 2
C. 5
D. 10
E. 15

Penyelesaian: Lihat/Tutup

A matriks singular maka determinan A bernilai 0
$\left| A \right|=0$
$\left| \begin{matrix} 4 & 5 & 7 \\ 8 & x & 14 \\ 7 & 5 & 2 \\ \end{matrix} \right|=0$
$\left| \begin{matrix} 4 & 5 & 7 \\ 8 & x & 14 \\ 7 & 5 & 2 \\ \end{matrix} \right|\left. \begin{matrix} {} & 4 & 5 \\ {} & 8 & x \\ {} & 7 & 5 \\ \end{matrix} \right|=0$
(8x + 490 + 280) – (49x + 280 + 80) = 0
–41x +410 = 0
–41x = –410
x = 10
Jawaban: D

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Soal No. 13

$\left| \begin{matrix} \sin x & \cos x & 1 \\ 0 & 1 & 0 \\ 1 & \cos x & \sin x \\ \end{matrix} \right|$ = ….
A. $\cos ^2x$
B. $-\sin ^2x$
C. 1
D. $\sin ^2x$
E. $-\cos ^2x$

Penyelesaian: Lihat/Tutup

$\left| \begin{matrix} \sin x & \cos x & 1 \\ 0 & 1 & 0 \\ 1 & \cos x & \sin x \\ \end{matrix} \right|$
= $\left| \begin{matrix} \sin x & \cos x & 1 \\ 0 & 1 & 0 \\ 1 & \cos x & \sin x \\ \end{matrix} \right|\left. \begin{matrix} {} & \sin x & \cos x \\ {} & 0 & 1 \\ {} & 1 & \cos x \\ \end{matrix} \right|$
= $(\sin ^2x+0+0)-(1+0+0)$
= $\sin ^2x$
Jawaban: D

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Soal No. 14

Matriks A berordo 3x3 dan mempunyai determinan 2, maka determinan dari matriks (2A) adalah ….
A. 16
B. 12
C. 18
D. 36
E. 5

Penyelesaian: Lihat/Tutup

Ingat: Jika matriks $A_{n\times n}$ dan $k$ skalar maka $\left| kA \right|=k^n.\left| A \right|$.
$A_{3\times 3}$ dan $\det A=2$ maka:
$\det (2A)=2^3.\det A=8.2=16$
Jawaban: A

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Soal No. 15

Jika $A=\left( \begin{matrix} 0 & 0 & 0 \\ 15 & -5 & 8 \\ 21 & -8 & 31 \\ \end{matrix} \right)$, maka nilai $\left| -3A \right|$ = ….
A. 9
B. 3
C. 1
D. 0
E. $-3$

Penyelesaian: Lihat/Tutup

$\begin{align} \left| A \right| &= \left| \begin{matrix} 0 & 0 & 0 \\ 15 & -5 & 8 \\ 21 & -8 & 31 \\ \end{matrix} \right|\left. \begin{matrix} {} & 0 & 0 \\ {} & 15 & -5 \\ {} & 21 & -8 \\ \end{matrix} \right| \\ &= (0+0+0)-(0+0+0) \\ \left| A \right| &= 0 \end{align}$
$A_{3\times 3}$ maka $\left| -3A \right|=(-3)^3.\left| A \right|=-27\times 0=0$
Jawaban: D

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Soal No. 16

Jika $A=\left( \begin{matrix} -2 & 5 \\ -1 & 3 \\ \end{matrix} \right)$, $B=\left( \begin{matrix} 12 & 2 & 5 \\ 1 & -8 & 3 \\ 24 & 4 & 10 \\ \end{matrix} \right)$ maka nilai $\left| 3A^{25} \right|-3^{25}\left| B^T \right|$ = ….
A. $-9$
B. $-3$
C. $-1$
D. 0
E. 3

Penyelesaian: Lihat/Tutup

$\begin{align}\left| A \right| &= \left| \begin{matrix} -2 & 5 \\ -1 & 3 \\ \end{matrix} \right| \\ &= -2.3-5(-1) \\ &= -6+5 \\ \left| A \right| &= -1 \end{align}$
$\begin{align}\left| B \right| &= \left| \begin{matrix} 12 & 2 & 5 \\ 1 & -8 & 3 \\ 24 & 4 & 10 \\ \end{matrix} \right|\left. \begin{matrix} {} & 12 & 2 \\ {} & -1 & -8 \\ {} & 24 & 4 \\ \end{matrix} \right| \\ &= (-960+144-20)-(-960+144-20) \\ \left| B \right| &= 0 \end{align}$
$\begin{align}\left| 3A^{25} \right|-3^{25}\left| B^T \right| &= 3^2\left| A \right|^{25}-3^{25}\left| B \right| \\ &= 9.(-1)^{25}-3^{25}.0 \\ &= -9 \end{align}$
Jawaban: A

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Soal No. 17

Invers matriks $A=\left( \begin{matrix} 4 & -1 & 5 \\ -2 & 3 & 1 \\ 3 & -1 & 4 \\ \end{matrix} \right)$ adalah ….
A. $\left( \begin{matrix} 13 & -1 & -16 \\ 11 & 1 & -14 \\ -7 & 1 & 10 \\ \end{matrix} \right)$
B. $\left( \begin{matrix} -13 & 1 & 16 \\ -11 & -1 & 14 \\ 7 & -1 & -10 \\ \end{matrix} \right)$
C. $\left( \begin{matrix} \frac{13}{6} & -\frac{1}{6} & -\frac{16}{6} \\ \frac{11}{6} & \frac{1}{6} & -\frac{14}{6} \\ -\frac{7}{6} & \frac{1}{6} & \frac{10}{6} \\ \end{matrix} \right)$
D. $\left( \begin{matrix} -\frac{13}{6} & \frac{1}{6} & \frac{16}{6} \\ -\frac{13}{6} & -\frac{1}{6} & \frac{14}{6} \\ \frac{7}{6} & -\frac{1}{6} & -\frac{10}{6} \\ \end{matrix} \right)$
E. Bukan A, B, C dan D

Penyelesaian: Lihat/Tutup

$A=\left( \begin{matrix} 4 & -1 & 5 \\ -2 & 3 & 1 \\ 3 & -1 & 4 \\ \end{matrix} \right)$
Determinan matriks A:
$\begin{align}\left| A \right| &= \left| \begin{matrix} 4 & -1 & 5 \\ -2 & 3 & 1 \\ 3 & -1 & 4 \\ \end{matrix} \right|\left. \begin{matrix} {} & 4 & -1 \\ {} & -2 & 3 \\ {} & 3 & -1 \\ \end{matrix} \right| \\ &= (48-3+10)-(45-4+8) \\ &= 55-49 \\ \left| A \right| &= 6 \end{align}$
Minor matriks A:
$M_{11}=\left| \begin{matrix} 3 & 1 \\ -1 & 4 \\ \end{matrix} \right|=12+1=13$
$M_{12}=\left| \begin{matrix} -2 & 1 \\ 3 & 4 \\ \end{matrix} \right|=-8-3=-11$
$M_{13}=\left| \begin{matrix} -2 & 3 \\ 3 & -1 \\ \end{matrix} \right|=2-9=-7$
$M_{21}=\left| \begin{matrix} -1 & 5 \\ -1 & 4 \\ \end{matrix} \right|=-4+5=1$
$M_{22}=\left| \begin{matrix} 4 & 5 \\ 3 & 4 \\ \end{matrix} \right|=16-15=1$
$M_{23}=\left| \begin{matrix} 4 & -1 \\ 3 & -1 \\ \end{matrix} \right|=-4+3=-1$
$M_{31}=\left| \begin{matrix} -1 & 5 \\ 3 & 1 \\ \end{matrix} \right|=-1-15=-16$
$M_{32}=\left| \begin{matrix} 4 & 5 \\ -2 & 1 \\ \end{matrix} \right|=4+10=14$
$M_{33}=\left| \begin{matrix} 4 & -1 \\ -2 & 3 \\ \end{matrix} \right|=12-2=10$
Kofaktor matriks A:
$kof(A)=\left( \begin{matrix} M_{11} & -M_{12} & M_{13} \\ -M_{21} & M_{22} & -M_{23} \\ M_{31} & -M_{32} & M_{33} \\ \end{matrix} \right)$
$kof(A)=\left( \begin{matrix} 13 & 11 & -7 \\ -1 & 1 & 1 \\ -16 & -14 & 10 \\ \end{matrix} \right)$
Adjoint matriks A:
$Adj(A)=[kof(A)]^T$
$Adj(A)=\left( \begin{matrix} 13 & -1 & -16 \\ 11 & 1 & -14 \\ -7 & 1 & 10 \\ \end{matrix} \right)$
Invers matriks A:
$\begin{align}A^{-1} &= \frac{1}{\det A}.Adj(A) \\ &= \frac{1}{6}\left( \begin{matrix} 13 & -1 & -16 \\ 11 & 1 & -14 \\ -7 & 1 & 10 \\ \end{matrix} \right) \\ A^{-1} &= \left( \begin{matrix} \frac{13}{6} & -\frac{1}{6} & -\frac{16}{6} \\ \frac{11}{6} & \frac{1}{6} & -\frac{14}{6} \\ -\frac{7}{6} & \frac{1}{6} & \frac{10}{6} \\ \end{matrix} \right) \end{align}$
Jawaban: C

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Soal No. 18

Matriks $A=\left( \begin{matrix} 1 & 3 & 1 \\ 0 & 3 & 1 \\ 1 & 2 & 1 \\ \end{matrix} \right)$ jumlah elemen-elemen baris pertama dari invers matriks A adalah ….
A. $-2$
B. $-1$
C. 0
D. 1
E. 2

Penyelesaian: Lihat/Tutup

$A=\left( \begin{matrix} 1 & 3 & 1 \\ 0 & 3 & 1 \\ 1 & 2 & 1 \\ \end{matrix} \right)$
Determinan matriks A:
$\begin{align}\left| A \right| &= \left| \begin{matrix} 1 & 3 & 1 \\ 0 & 3 & 1 \\ 1 & 2 & 1 \\ \end{matrix} \right|\left. \begin{matrix} {} & 1 & 3 \\ {} & 0 & 3 \\ {} & 1 & 2 \\ \end{matrix} \right| \\ &= (3+3+0)-(3+2+0) \\ &= 6-5 \\ \left| A \right| &= 1 \end{align}$
Minor matriks A:
$M_{11}=\left| \begin{matrix} 3 & 1 \\ 2 & 1 \\ \end{matrix} \right|=3-2=1$
$M_{12}=\left| \begin{matrix} 0 & 1 \\ 1 & 1 \\ \end{matrix} \right|=0-1=-1$
$M_{13}=\left| \begin{matrix} 0 & 3 \\ 1 & 2 \\ \end{matrix} \right|=0-3=-3$
$M_{21}=\left| \begin{matrix} 3 & 1 \\ 2 & 1 \\ \end{matrix} \right|=3-2=1$
$M_{22}=\left| \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right|=1-1=0$
$M_{23}=\left| \begin{matrix} 1 & 3 \\ 1 & 2 \\ \end{matrix} \right|=2-3=-1$
$M_{31}=\left| \begin{matrix} 3 & 1 \\ 3 & 1 \\ \end{matrix} \right|=3-3=0$
$M_{32}=\left| \begin{matrix} 1 & 1 \\ 0 & 1 \\ \end{matrix} \right|=1-0=1$
$M_{33}=\left| \begin{matrix} 1 & 3 \\ 0 & 3 \\ \end{matrix} \right|=3-0=3$
Kofaktor matriks A:
$kof(A)=\left( \begin{matrix} M_{11} & -M_{12} & M_{13} \\ -M_{21} & M_{22} & -M_{23} \\ M_{31} & -M_{32} & M_{33} \\ \end{matrix} \right)$
$kof(A)=\left( \begin{matrix} 1 & 1 & 3 \\ -1 & 0 & 1 \\ 0 & -1 & 3 \\ \end{matrix} \right)$
Adjoint matriks A:
$Adj(A)=[kof(A)]^T$
$Adj(A)=\left( \begin{matrix} 1 & -1 & 0 \\ 1 & 0 & -1 \\ 3 & 1 & 3 \\ \end{matrix} \right)$
Invers matriks A:
$\begin{align}A^{-1} &= \frac{1}{\det A}.Adj(A) \\ &= \frac{1}{1}\left( \begin{matrix} 1 & -1 & 0 \\ 1 & 0 & -1 \\ 3 & 1 & 3 \\ \end{matrix} \right) \\ A^{-1} &= \left( \begin{matrix} 1 & -1 & 0 \\ 1 & 0 & -1 \\ 3 & 1 & 3 \\ \end{matrix} \right) \end{align}$
Jumlah elemen-elemen baris pertama dari invers matriks A adalah:
1 + (-1) + 0 = 0
Jawaban: C

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Soal No. 19

Matriks $A=\left( \begin{matrix} 1 & 2 & 3 \\ 1 & 3 & 3 \\ 1 & 2 & 4 \\ \end{matrix} \right)$ maka $2.A^{-1}$ adalah ….
A. $\left( \begin{matrix} 6 & -2 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \\ \end{matrix} \right)$
B. $\left( \begin{matrix} 6 & -2 & -3 \\ -2 & 2 & 0 \\ -2 & 0 & 2 \\ \end{matrix} \right)$
C. $\left( \begin{matrix} 12 & -4 & -6 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \\ \end{matrix} \right)$
D. $\left( \begin{matrix} 12 & -4 & -6 \\ -2 & 2 & 0 \\ -2 & 0 & 2 \\ \end{matrix} \right)$
E. $\left( \begin{matrix} 6 & -2 & -3 \\ -2 & 2 & 0 \\ -1 & 0 & 1 \\ \end{matrix} \right)$

Penyelesaian: Lihat/Tutup

$A=\left( \begin{matrix} 1 & 2 & 3 \\ 1 & 3 & 3 \\ 1 & 2 & 4 \\ \end{matrix} \right)$
Determinan matriks A:
$\begin{align}\left| A \right| &= \left| \begin{matrix} 1 & 2 & 3 \\ 1 & 3 & 3 \\ 1 & 2 & 4 \\ \end{matrix} \right|\left. \begin{matrix} {} & 1 & 2 \\ {} & 1 & 3 \\ {} & 1 & 2 \\ \end{matrix} \right| \\ &= (12+6+6)-(9+6+8) \\ &= 24-23 \\ \left| A \right| &= 1 \end{align}$
Minor matriks A:
$M_{11}=\left| \begin{matrix} 3 & 3 \\ 2 & 4 \\ \end{matrix} \right|=12-6=6$
$M_{12}=\left| \begin{matrix} 1 & 3 \\ 1 & 4 \\ \end{matrix} \right|=4-3=1$
$M_{13}=\left| \begin{matrix} 1 & 3 \\ 1 & 2 \\ \end{matrix} \right|=2-3=-1$
$M_{21}=\left| \begin{matrix} 2 & 3 \\ 2 & 4 \\ \end{matrix} \right|=8-6=2$
$M_{22}=\left| \begin{matrix} 1 & 3 \\ 1 & 4 \\ \end{matrix} \right|=4-3=1$
$M_{23}=\left| \begin{matrix} 1 & 2 \\ 1 & 2 \\ \end{matrix} \right|=2-2=0$
$M_{31}=\left| \begin{matrix} 2 & 3 \\ 3 & 3 \\ \end{matrix} \right|=6-9=-3$
$M_{32}=\left| \begin{matrix} 1 & 3 \\ 1 & 3 \\ \end{matrix} \right|=3-3=0$
$M_{33}=\left| \begin{matrix} 1 & 2 \\ 1 & 3 \\ \end{matrix} \right|=3-2=1$
Kofaktor matriks A:
$kof(A)=\left( \begin{matrix} M_{11} & -M_{12} & M_{13} \\ -M_{21} & M_{22} & -M_{23} \\ M_{31} & -M_{32} & M_{33} \\ \end{matrix} \right)$
$kof(A)=\left( \begin{matrix} 6 & -1 & -1 \\ -2 & 1 & 0 \\ -3 & 0 & 1 \\ \end{matrix} \right)$
Adjoint matriks A:
$Adj(A)=[kof(A)]^T$
$Adj(A)=\left( \begin{matrix} 6 & -2 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \\ \end{matrix} \right)$
Invers matriks A:
$\begin{align}A^{-1}=\frac{1}{\det A}Adj(A) \\ &= \frac{1}{1}\left( \begin{matrix} 6 & -2 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \\ \end{matrix} \right) \\ A^{-1} &= \left( \begin{matrix} 6 & -2 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \\ \end{matrix} \right) \end{align}$
$2A^{-1}=2\left( \begin{matrix} 6 & -2 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \\ \end{matrix} \right)$
$2A^{-1}=2\left( \begin{matrix} 12 & -4 & -6 \\ -2 & 2 & 0 \\ -2 & 0 & 2 \\ \end{matrix} \right)$
Jawaban: D

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Soal No. 20

Matriks $A=\left( \begin{matrix} 6 & -2 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \\ \end{matrix} \right)$, maka jumlah kuadrat unsur pada baris ketiga dari invers matriks A adalah ….
A. 21
B. 14
C. 7
D. 49
E. 34

Penyelesaian: Lihat/Tutup

$A=\left( \begin{matrix} 6 & -2 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \\ \end{matrix} \right)$
Determinan matriks A:
$\begin{align} \left| A \right| &= \left| \begin{matrix} 6 & -2 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \\ \end{matrix} \right|\left. \begin{matrix} {} & 6 & -2 \\ {} & -1 & 1 \\ {} & -1 & 0 \\ \end{matrix} \right| \\ &= (6+0+0)-(3+0+2) \\ &= 6-5 \\ \left| A \right| &= 1 \end{align}$
Minor matriks A:
$M_{11}=\left| \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right|=1-0=1$
$M_{12}=\left| \begin{matrix} -1 & 0 \\ -1 & 1 \\ \end{matrix} \right|=-1-0=-1$
$M_{13}=\left| \begin{matrix} -1 & 1 \\ -1 & 0 \\ \end{matrix} \right|=0+1=1$
$M_{21}=\left| \begin{matrix} -2 & -3 \\ 0 & 1 \\ \end{matrix} \right|=-2-0=-2$
$M_{22}=\left| \begin{matrix} 6 & -3 \\ -1 & 1 \\ \end{matrix} \right|=6-3=3$
$M_{23}=\left| \begin{matrix} 6 & -2 \\ -1 & 0 \\ \end{matrix} \right|=0-2=-2$
$M_{31}=\left| \begin{matrix} -2 & -3 \\ 1 & 0 \\ \end{matrix} \right|=0+3=3$
$M_{32}=\left| \begin{matrix} 6 & -3 \\ -1 & 0 \\ \end{matrix} \right|=0-3=-3$
$M_{33}=\left| \begin{matrix} 6 & -2 \\ -1 & 1 \\ \end{matrix} \right|=6-2=4$
Kofaktor matriks A:
$kof(A)=\left( \begin{matrix} M_{11} & -M_{12} & M_{13} \\ -M_{21} & M_{22} & -M_{23} \\ M_{31} & -M_{32} & M_{33} \\ \end{matrix} \right)$
$kof(A)=\left( \begin{matrix} 1 & 1 & 1 \\ 2 & 3 & 2 \\ 3 & 3 & 4 \\ \end{matrix} \right)$
Adjoint matriks A:
$Adj(A)=[kof(A)]^T$
$Adj(A)=\left( \begin{matrix} 1 & 2 & 3 \\ 1 & 3 & 3 \\ 1 & 2 & 4 \\ \end{matrix} \right)$
Invers matriks A:
$\begin{align}A^{-1} &= \frac{1}{\det A}Adj(A) \\ &= \frac{1}{1}\left( \begin{matrix} 1 & 2 & 3 \\ 1 & 3 & 3 \\ 1 & 2 & 4 \\ \end{matrix} \right) \\ A^{-1} &= \left( \begin{matrix} 1 & 2 & 3 \\ 1 & 3 & 3 \\ 1 & 2 & 4 \\ \end{matrix} \right) \end{align}$
Jadi, jumlah kuadrat unsur pada baris ketiga dari invers matriks A adalah:
$1^2+2^2+4^2$ = 1 + 4 + 16 = 21
Jawaban: A

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