SBMPTN Zone : Matriks

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

Tipe:

Standar SBMPTN HOTS

---

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 {3\det(A)=\det\left(A^{-1}\right)-2} adalah

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

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

\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=\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.

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.

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

No.

Diketahui matriks A berordo 2 \times 2 dan {A=\begin{pmatrix}-2&-5\\1&3\end{pmatrix},} {C=\begin{pmatrix}4&6\\3&5\end{pmatrix}.} Jika B memenuhi {A\cdot B=C,} maka \det\left(2B^{-1}\right) adalah

1. 1
2. 2
3. -2

4. -1
5. -4

SKMP Februari 2021

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