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

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No.

Jika matriks {A=$$\left(\begin{array}{cc}a& -3\\ 1& 1\end{array}\right)$$} 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

$$\begin{array}{rl}det(A)& =a\cdot 1-(-3)\cdot 1\\ & =a+3\end{array}$$

CARA 1

$$\begin{array}{rl}{A}^{-1}& ={\displaystyle \frac{1}{det(A)}}\left(\begin{array}{cc}1& 3\\ -1& a\end{array}\right)\\ & ={\displaystyle \frac{1}{a+3}}\left(\begin{array}{cc}1& 3\\ -1& a\end{array}\right)\\ & =\left(\begin{array}{cc}{\displaystyle \frac{1}{a+3}}& {\displaystyle \frac{3}{a+3}}\\ -{\displaystyle \frac{1}{a+3}}& {\displaystyle \frac{a}{a+3}}\end{array}\right)\\ det\left(A^{-1}\right)& ={\displaystyle \frac{1}{a+3}}\cdot {\displaystyle \frac{a}{a+3}}-{\displaystyle \frac{3}{a+3}}\cdot (-{\displaystyle \frac{1}{a+3}})\\ & ={\displaystyle \frac{a}{(a+3{)}^2}}+{\displaystyle \frac{3}{(a+3{)}^2}}\\ & ={\displaystyle \frac{a+3}{(a+3{)}^2}}\\ & ={\displaystyle \frac{1}{a+3}}\end{array}$$

$$\begin{array}{rl}3det(A)& =det\left(A^{-1}\right)-2\\ 3(a+3)& ={\displaystyle \frac{1}{a+3}}-2\times (a+3)\\ 3(a+3{)}^2& =1-2(a+3)\\ 3(a^2+6a+9)& =1-2a-6\\ 3a^2+18a+27& =-2a-5\\ 3a^2+18a+27+2a+5& =0\\ 3a^2+20a+32& =0\end{array}$$

a_1a_2=\dfrac{32}3

CARA 2

$$\begin{array}{rl}3det(A)& =det\left(A^{-1}\right)-2\\ 3det(A)& ={\displaystyle \frac{1}{det(A)}}-2& \times det(A)\\ 3{(det(A))}^2& =1-2det(A)\\ 3{(det(A))}^2+2det(A)-1& =0\\ (3det(A)-1)(det(A)+1)& =0\\ (3(a+3)-1)(a+3+1)& =0\\ (3a+8)(a+4)& =0\end{array}$$

$$\begin{array}{rl}{a}_1{a}_2& =(-{\displaystyle \frac{8}{3}})(-4)\\ & ={\displaystyle \frac{32}{3}}\end{array}$$

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

No.

Diketahui matriks {A=$$\left(\begin{array}{cc}3& a\\ b& 2\end{array}\right)$$} dan {B=$$\left(\begin{array}{cc}a& b\\ 3& 2\end{array}\right)$$}. 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

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

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

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

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

No.

Jika {$$\left(\begin{array}{cc}1& 2\\ 1& 3\end{array}\right)$$$$\left(\begin{array}{c}x\\ y\end{array}\right)$$=$$\left(\begin{array}{c}a\\ b\end{array}\right)$$} dan {$$\left(\begin{array}{cc}5& 2\\ 3& 1\end{array}\right)$$$$\left(\begin{array}{c}a\\ b\end{array}\right)$$=$$\left(\begin{array}{c}1\\ 4\end{array}\right)$$}, maka nilai {x+y=}

1. 30
2. 31
3. 32

4. 33
5. 34

$$\begin{array}{rl}\left(\begin{array}{cc}5& 2\\ 3& 1\end{array}\right)\left(\begin{array}{c}a\\ b\end{array}\right)& =\left(\begin{array}{c}1\\ 4\end{array}\right)\\ \left(\begin{array}{cc}5& 2\\ 3& 1\end{array}\right)\left(\begin{array}{cc}1& 2\\ 1& 3\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)& =\left(\begin{array}{c}1\\ 4\end{array}\right)\\ \left(\begin{array}{cc}7& 16\\ 4& 9\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)& =\left(\begin{array}{c}1\\ 4\end{array}\right)\\ \left(\begin{array}{c}x\\ y\end{array}\right)& ={\left(\begin{array}{cc}7& 16\\ 4& 9\end{array}\right)}^{-1}\left(\begin{array}{c}1\\ 4\end{array}\right)\\ & ={\displaystyle \frac{1}{(7)(9)-(16)(4)}}\left(\begin{array}{cc}9& -16\\ -4& 7\end{array}\right)\left(\begin{array}{c}1\\ 4\end{array}\right)\\ & ={\displaystyle \frac{1}{(-1}}\left(\begin{array}{c}-55\\ 24\end{array}\right)\\ & =\left(\begin{array}{c}55\\ -24\end{array}\right)\end{array}$$
x=55
y=-24

$$\begin{array}{rl}x+y& =55+(-24)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 31}}}}\end{array}$$

Jadi, x+y=31. JAWAB: B

No.

Diketahui matriks A berukuran {3\times3} dan memenuhi {A$$\left(\begin{array}{c}3\\ 2\\ 1\end{array}\right)$$=$$\left(\begin{array}{c}2\\ 4\\ 2\end{array}\right)$$} dan {A$$\left(\begin{array}{c}3\\ 1\\ 2\end{array}\right)$$=$$\left(\begin{array}{c}2\\ 2\\ 2\end{array}\right)$$}, matriks {A$$\left(\begin{array}{c}6\\ 4\\ 2\end{array}\right)$$=}

1. $\left(\begin{array}{c}4\\ 8\\ 4\end{array}\right)$
2. $\left(\begin{array}{c}8\\ 4\\ 8\end{array}\right)$
3. $\left(\begin{array}{c}12\\ 8\\ 4\end{array}\right)$

4. $\left(\begin{array}{c}4\\ 8\\ 12\end{array}\right)$
5. $\left(\begin{array}{c}12\\ 8\\ 12\end{array}\right)$

CARA BIASA

Misal $A=\left(\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right)$

$$\begin{array}{rl}A\left(\begin{array}{c}3\\ 2\\ 1\end{array}\right)& =\left(\begin{array}{c}2\\ 4\\ 2\end{array}\right)\\ \left(\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right)\left(\begin{array}{c}3\\ 2\\ 1\end{array}\right)& =\left(\begin{array}{c}2\\ 4\\ 2\end{array}\right)\\ \left(\begin{array}{c}3a+2b+c\\ 3d+2e+f\\ 3g+2h+i\end{array}\right)& =\left(\begin{array}{c}2\\ 4\\ 2\end{array}\right)\end{array}$$

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

$$\begin{array}{rl}A\left(\begin{array}{c}6\\ 4\\ 2\end{array}\right)& =\left(\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right)\left(\begin{array}{c}6\\ 4\\ 2\end{array}\right)\\ & =\left(\begin{array}{c}6a+4b+2c\\ 6d+4e+2f\\ 6g+4h+2i\end{array}\right)\\ & =\left(\begin{array}{c}2(3a+2b+c)\\ 2(3d+2e+f)\\ 2(3g+2h+i)\end{array}\right)\\ & =\left(\begin{array}{c}2(2)\\ 2(4)\\ 2(2)\end{array}\right)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle \left(\begin{array}{c}4\\ 8\\ 4\end{array}\right)}}}}\end{array}$$

CARA CEPAT

$$\begin{array}{rl}A\left(\begin{array}{c}6\\ 4\\ 2\end{array}\right)& =A\cdot 2\left(\begin{array}{c}3\\ 2\\ 1\end{array}\right)\\ & =2A\left(\begin{array}{c}3\\ 2\\ 1\end{array}\right)\\ & =2\left(\begin{array}{c}2\\ 4\\ 2\end{array}\right)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle \left(\begin{array}{c}4\\ 8\\ 4\end{array}\right)}}}}\end{array}$$

Jadi, A$$\left(\begin{array}{c}6\\ 4\\ 2\end{array}\right)$$=$$\left(\begin{array}{c}4\\ 8\\ 4\end{array}\right)$$. JAWAB: A

No.

Diketahui matriks A berordo 2 \times 2 dan {A=$$\left(\begin{array}{cc}-2& -5\\ 1& 3\end{array}\right)$$,} {C=$$\left(\begin{array}{cc}4& 6\\ 3& 5\end{array}\right)$$.} 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

$$\begin{array}{rl}A\cdot B& =C\\ B& =A^{-1}\cdot C\end{array}$$

$$\begin{array}{rl}det\left(2B^{-1}\right)& =2^{2}det\left(B^{-1}\right)\\ & ={\displaystyle \frac{4}{detB}}\\ & ={\displaystyle \frac{4}{det(A^{-1}\cdot C)}}\\ & ={\displaystyle \frac{4}{det{A}^{-1}\cdot detC}}\\ & ={\displaystyle \frac{4detA}{detC}}\\ & ={\displaystyle \frac{4(-1)}{2}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -2}}}}\end{array}$$

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