SNBT Zone : Invers Matriks

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.

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

ALTERNATIF PENYELESAIAN

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

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

Jadi, x + y = 31.
JAWAB: B

---

No.

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 2×2 yang memiliki invers dan matriks AC maupun matriks BC tidak memiliki invers, maka nilai 4a² + 9b² =

1. 72
2. 74
3. 76

4. 78
5. 80

ALTERNATIF PENYELESAIAN

|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, 4a² + 9b² = 72.
JAWAB: A

---

No.

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 3det (A) = det (A⁻¹) − 2 adalah

1. \(-\dfrac{29}3\)
2. \(-\dfrac{20}3\)
3. \(\dfrac{20}3\)

4. \(\dfrac{29}3\)
5. \(\dfrac{32}3\)

ALTERNATIF PENYELESAIAN

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

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.

Diketahui matriks A berordo 2×2 dan \({A=\begin{pmatrix}-2&-5\\1&3\end{pmatrix},}\) \({C=\begin{pmatrix}4&6\\3&5\end{pmatrix}.}\) Jika B memenuhi A⋅B = C, maka det (2B⁻¹) adalah

1. 1
2. 2
3. −2

4. −1
5. −4

SKMP Februari 2021

ALTERNATIF PENYELESAIAN

det A = (−2)(3) − (−5)(1) = −1
det C = (4)(5) − (6)(3) = 2

\(\begin{aligned} A\cdot B&=C\\ B&=A^{-1}\cdot C \end{aligned}\)

\(\begin{aligned} \det\left(2B^{-1}\right)&=2^2\det\left(B^{-1}\right)\\ &=\dfrac4{\det B}\\ &=\dfrac4{\det \left(A^{-1}\cdot C\right)}\\ &=\dfrac4{\det A^{-1}\cdot\det C}\\ &=\dfrac{4\det A}{\det C}\\ &=\dfrac{4(-1)}2\\ &=\boxed{\boxed{-2}} \end{aligned}\)

Jadi, det (2B⁻¹) = −2.
JAWAB: C

---