Exercise Zone : Matriks [3]

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Tipe:

Standar SBMPTN HOTS

No.

Diketahui matriks $A=\left(\begin{array}{cc}1& 2\\ 3& 7\end{array}\right)$, $B=\left(\begin{array}{cc}a& b\\ 2& 3\end{array}\right)$, dan $A^T\cdot B^T=\left(\begin{array}{cc}13& 11\\ 30& 25\end{array}\right)$ maka nilai 12a+9b=

1. 44
2. 46
3. 48

4. 50
5. 52

SKMP Juli 2020

$\begin{array}{rl}{A}^T\cdot B^T& =\left(\begin{array}{cc}13& 11\\ 30& 25\end{array}\right)\\ \left(\begin{array}{cc}1& 3\\ 2& 7\end{array}\right)\left(\begin{array}{cc}a& 2\\ b& 3\end{array}\right)& =\left(\begin{array}{cc}13& 11\\ 30& 25\end{array}\right)\\ \left(\begin{array}{cc}a+3b& 11\\ 2a+7b& 25\end{array}\right)& =\left(\begin{array}{cc}13& 11\\ 30& 25\end{array}\right)\end{array}$

$\begin{array}{rl}2a+7b& =30\\ a+3b& =13& \times 2\end{array}$

$\begin{array}{rl}2a+7b& =30\\ 2a+2b& =26& -\\ b& =4\end{array}$

$\begin{array}{rl}a+3b& =13\\ a+3(4)& =13\\ a+12& =13\\ a& =1\end{array}$

$\begin{array}{rl}12a+9b& =12(1)+9(4)\\ & =12+36\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 48}}}}\end{array}$

Jadi, 12a+9b=48. JAWAB: C

No.

Tulislah koefisien variabel-variabelnya dari sistem persamaan linear berikut ke dalam bentuk matriks.

1. -2x+3y=5
   8x+4y=7
2. -5=7x+8y
   -6=3x-4y
3. 3x+4y-10=0
   7x-8y+12=0

4. 2x+5y-3z=6
   3x-7y-z=10
   5x-9y+6z=12
5. 4x=20
   5y-10=0
   2x+3y=16

Grup Telegram

1. $\boxed{{\displaystyle \boxed{{\displaystyle \left(\begin{array}{cc}-2& 3\\ 8& 4\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}5\\ 7\end{array}\right)}}}}$
2. $\boxed{{\displaystyle \boxed{{\displaystyle \left(\begin{array}{c}-5\\ -6\end{array}\right)=\left(\begin{array}{cc}7& 8\\ 3& -4\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)}}}}$
3. 3x+4y-10=0
   3x+4y=10
   
   7x-8y+12=0
   7x-8y=-12
   
   $\boxed{{\displaystyle \boxed{{\displaystyle \left(\begin{array}{cc}3& 4\\ 7& -8\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}10\\ -12\end{array}\right)}}}}$

4. $\boxed{{\displaystyle \boxed{{\displaystyle \left(\begin{array}{ccc}2& 5& -3\\ 3& -7& -1\\ 5& -9& 6\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}6\\ 10\\ 12\end{array}\right)}}}}$
5. 5y-10=0
   5y=10
   
   $\boxed{{\displaystyle \boxed{{\displaystyle \left(\begin{array}{cc}4& 0\\ 0& 5\\ 2& 3\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}20\\ 10\\ 16\end{array}\right)}}}}$

Jadi, $\left(\begin{array}{cc}-2& 3\\ 8& 4\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}5\\ 7\end{array}\right)$ $\left(\begin{array}{c}-5\\ -6\end{array}\right)=\left(\begin{array}{cc}7& 8\\ 3& -4\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)$ $\left(\begin{array}{cc}3& 4\\ 7& -8\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}10\\ -12\end{array}\right)$ $\left(\begin{array}{ccc}2& 5& -3\\ 3& -7& -1\\ 5& -9& 6\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}6\\ 10\\ 12\end{array}\right)$ $\left(\begin{array}{cc}4& 0\\ 0& 5\\ 2& 3\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}20\\ 10\\ 16\end{array}\right)$

No.

Nyatakan matriks $\left(\begin{array}{cc}6& 3\\ 0& 8\end{array}\right)$ sebagai kombinasi linear dari matriks berikut:
$\left(\begin{array}{cc}1& 2\\ -1& 3\end{array}\right)$, $\left(\begin{array}{cc}0& 1\\ 2& 4\end{array}\right)$, dan $\left(\begin{array}{cc}4& -2\\ 0& -2\end{array}\right)$.

Grup Telegram

$\begin{array}{rl}\left(\begin{array}{cc}6& 3\\ 0& 8\end{array}\right)& =k_1\left(\begin{array}{cc}1& 2\\ -1& 3\end{array}\right)+k_2\left(\begin{array}{cc}0& 1\\ 2& 4\end{array}\right)+k_3\left(\begin{array}{cc}4& -2\\ 0& -2\end{array}\right)\\ & =\left(\begin{array}{cc}{k}_1+4k_3& 2k_1+k_2-2k_3\\ -k_1+2k_2& 3k_1+4k_2-2k_3\end{array}\right)\end{array}$

$\begin{array}{rlllll}{k}_1& & & +& 4k_3& =6\\ 2k_1& +& k_2& -& 2k_3& =3\\ -k_1& +& 2k_2& & & =0\\ 3k_1& +& 4k_2& -& 2k_3& =8\end{array}$
$\begin{array}{rl}\left(\begin{array}{cccc}1& 0& 4& 6\\ 2& 1& -2& 3\\ -1& 2& 0& 0\\ 3& 4& -2& 8\end{array}\right)& \begin{array}{r}-2b_1+b_2\\ b_1+b_3\\ -3b_1+b_3\end{array}\\ \left(\begin{array}{cccc}1& 0& 4& 6\\ 0& 1& -10& -9\\ 0& 2& 4& 6\\ 0& 4& -14& -10\end{array}\right)& \begin{array}{r}{\displaystyle \frac{1}{2}}{b}_3\\ {\displaystyle \frac{1}{2}}{b}_4\end{array}\\ \left(\begin{array}{cccc}1& 0& 4& 6\\ 0& 1& -10& -9\\ 0& 1& 2& 3\\ 0& 2& -7& -5\end{array}\right)& \begin{array}{r}-b_2+b_3\\ -2b_2+b_4\end{array}\\ \left(\begin{array}{cccc}1& 0& 4& 6\\ 0& 1& -10& -9\\ 0& 0& 12& 12\\ 0& 0& 13& 13\end{array}\right)& \begin{array}{r}{\displaystyle \frac{1}{12}}{b}_3\\ {\displaystyle \frac{1}{13}}{b}_4\end{array}\\ \left(\begin{array}{cccc}1& 0& 0& 2\\ 0& 1& 0& 1\\ 0& 0& 1& 1\\ 0& 0& 0& 0\end{array}\right)\end{array}$
k_1=2, k_2=1, k_3=1

Jadi, $\left(\begin{array}{cc}6& 3\\ 0& 8\end{array}\right)=2\left(\begin{array}{cc}1& 2\\ -1& 3\end{array}\right)+\left(\begin{array}{cc}0& 1\\ 2& 4\end{array}\right)+\left(\begin{array}{cc}4& -2\\ 0& -2\end{array}\right)$.

No.

Diketahui $A=\left(\begin{array}{cc}-3& 2\\ 1& 0\end{array}\right)$, $B=\left(\begin{array}{cc}4& -1\\ 2& 5\end{array}\right)$ dan $C=\left(\begin{array}{cc}-2& -2\\ 3& 3\end{array}\right)$, maka A-(B-C)=...

1. $\left(\begin{array}{cc}1& 5\\ 0& 1\end{array}\right)$
2. $\left(\begin{array}{cc}4& 3\\ -3& 2\end{array}\right)$
3. $\left(\begin{array}{cc}2& 1\\ 3& -2\end{array}\right)$

4. $\left(\begin{array}{cc}-9& 1\\ 2& -2\end{array}\right)$
5. $\left(\begin{array}{cc}-3& 1\\ -3& 2\end{array}\right)$

Grup Telegram

$\begin{array}{rl}A-(B-C)& =\left(\begin{array}{cc}-3& 2\\ 1& 0\end{array}\right)-(\left(\begin{array}{cc}4& -1\\ 2& 5\end{array}\right)-\left(\begin{array}{cc}-2& -2\\ 3& 3\end{array}\right))\\ & =\left(\begin{array}{cc}-3& 2\\ 1& 0\end{array}\right)-\left(\begin{array}{cc}6& 1\\ -1& 2\end{array}\right)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle \left(\begin{array}{cc}-9& 1\\ 2& -2\end{array}\right)}}}}\end{array}$

Jadi, $A-(B-C)=\left(\begin{array}{cc}-9& 1\\ 2& -2\end{array}\right)$. JAWAB: D

No.

Diketahui $P=\left(\begin{array}{cc}-3& 1\\ 4& -2\end{array}\right)$ dan $Q=\left(\begin{array}{cc}3& 4\\ 1& 2\end{array}\right)$. Hitunglah \left(PQ\right)^{-1}

Grup Telegram

$\begin{array}{rl}PQ& =\left(\begin{array}{cc}-3& 1\\ 4& -2\end{array}\right)\left(\begin{array}{cc}3& 4\\ 1& 2\end{array}\right)\\ & =\left(\begin{array}{cc}-8& -10\\ 10& 12\end{array}\right)\\ {\left(PQ\right)}^{-1}& ={\displaystyle \frac{1}{(-8)(12)-(-10)(10)}}\left(\begin{array}{cc}12& 10\\ -10& -8\end{array}\right)\\ & ={\displaystyle \frac{1}{-96+100}}\left(\begin{array}{cc}12& 10\\ -10& -8\end{array}\right)\\ & ={\displaystyle \frac{1}{4}}\left(\begin{array}{cc}12& 10\\ -10& -8\end{array}\right)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle \left(\begin{array}{cc}3& {\displaystyle \frac{5}{2}}\\ -{\displaystyle \frac{5}{2}}& -2\end{array}\right)}}}}\end{array}$

Jadi, ${\left(PQ\right)}^{-1}=\left(\begin{array}{cc}3& {\displaystyle \frac{5}{2}}\\ -{\displaystyle \frac{5}{2}}& -2\end{array}\right)$.

No.

X adalah matriks persegi berordo 2\times2 yang memenuhi $X\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)=\left(\begin{array}{cc}4& 8\\ 5& 8\end{array}\right)$. Matriks X adalah

1. $\left(\begin{array}{cc}3& 2\\ -2& 1\end{array}\right)$
2. $\left(\begin{array}{cc}3& 2\\ 2& 1\end{array}\right)$
3. $\left(\begin{array}{cc}-4& 0\\ -1& -2\end{array}\right)$

4. $\left(\begin{array}{cc}4& 0\\ 2& 1\end{array}\right)$
5. $\left(\begin{array}{cc}4& 0\\ -1& 2\end{array}\right)$

Grup Telegram

$\begin{array}{rl}X\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)& =\left(\begin{array}{cc}4& 8\\ 5& 8\end{array}\right)\\ X& =\left(\begin{array}{cc}4& 8\\ 5& 8\end{array}\right){\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)}^{-1}\\ & =\left(\begin{array}{cc}4& 8\\ 5& 8\end{array}\right)\cdot {\displaystyle \frac{1}{1\cdot 4-2\cdot 3}}\left(\begin{array}{cc}4& -2\\ -3& 1\end{array}\right)\\ & =\left(\begin{array}{cc}4& 8\\ 5& 8\end{array}\right)\cdot {\displaystyle \frac{1}{-2}}\left(\begin{array}{cc}4& -2\\ -3& 1\end{array}\right)\\ & ={\displaystyle \frac{1}{-2}}\left(\begin{array}{cc}4& 8\\ 5& 8\end{array}\right)\left(\begin{array}{cc}4& -2\\ -3& 1\end{array}\right)\\ & ={\displaystyle \frac{1}{-2}}\left(\begin{array}{cc}-8& 0\\ -4& -2\end{array}\right)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle \left(\begin{array}{cc}4& 0\\ 2& 1\end{array}\right)}}}}\end{array}$

Jadi, $X=\left(\begin{array}{cc}4& 0\\ 2& 1\end{array}\right)$. JAWAB: D

No.

Diberikan sistem persamaan linear berikut:
3x-4y-3z=12
-2x+7y-6z=9
5x+8y-z=-10
Nyatakanlah matriks koefisien sistem persamaan linear tersebut

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$\left(\begin{array}{ccc}3& 4& -3\\ -2& 7& -6\\ 5& 8& -1\end{array}\right)$

Jadi, matriks koefisien sistem persamaan linear tersebut adalah $\left(\begin{array}{ccc}3& 4& -3\\ -2& 7& -6\\ 5& 8& -1\end{array}\right)$

No.

Diketahui matriks $A=\left(\begin{array}{cc}5& 2\\ 0& 3\end{array}\right)$ dan $B=\left(\begin{array}{cc}-3& 1\\ 17& 0\end{array}\right)$, Jika A^T transpose matriks A dan {AX=B+A^T}, maka determinan matriks X adalah ....

1. \dfrac{15}{13}
2. \dfrac{13}{15}
3. -\dfrac{13}{15}

4. -1
5. 1

Grup Telegram

|A|=5\cdot3-2\cdot0=15

$A^T=\left(\begin{array}{cc}5& 0\\ 2& 3\end{array}\right)$

$B+A^T=\left(\begin{array}{cc}-3& 1\\ 17& 0\end{array}\right)+\left(\begin{array}{cc}5& 0\\ 2& 3\end{array}\right)=\left(\begin{array}{cc}2& 1\\ 19& 3\end{array}\right)$

\left|B+A^T\right|=2\cdot3-1\cdot19=-13

$\begin{array}{rl}AX& =B+A^T\\ |A||X|& =|B+A^T|\\ 15|X|& =-13\\ |X|& =\boxed{{\displaystyle \boxed{{\displaystyle -{\displaystyle \frac{13}{15}}}}}}\end{array}$

Jadi, determinan matriks X adalah -\dfrac{13}{15}. JAWAB: C

No.

Untuk persamaan {2$$\left(\begin{array}{cc}x& 3y\\ 3& y\end{array}\right)$$+$$\left(\begin{array}{cc}3& x-6\\ 1& x\end{array}\right)$$=$$\left(\begin{array}{cc}11& 10\\ 7& 8\end{array}\right)$$}, harga {x+y} adalah

1. -2
2. 2
3. 4

4. 6
5. 7

Grup Telegram

$$\begin{array}{rl}2\left(\begin{array}{cc}x& 3y\\ 3& y\end{array}\right)+\left(\begin{array}{cc}3& x-6\\ 1& x\end{array}\right)& =\left(\begin{array}{cc}11& 10\\ 7& 8\end{array}\right)\\ \left(\begin{array}{cc}2x& 6y\\ 6& 2y\end{array}\right)+\left(\begin{array}{cc}3& x-6\\ 1& x\end{array}\right)& =\left(\begin{array}{cc}11& 10\\ 7& 8\end{array}\right)\\ \left(\begin{array}{cc}2x+3& 6y+x-6\\ 7& 2y+x\end{array}\right)& =\left(\begin{array}{cc}11& 10\\ 7& 8\end{array}\right)\end{array}$$

$$\begin{array}{rl}2x+3& =11\\ 2x& =8\\ x& =4\end{array}$$

$$\begin{array}{rl}2y+x& =8\\ 2y+4& =8\\ 2y& =4\\ y& =2\end{array}$$

$$\begin{array}{rl}x+y& =2+4\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 6}}}}\end{array}$$

Jadi, {x+y=6}.

No.

Jika $$\left(\begin{array}{cc}5& 4\\ 1& 1\end{array}\right)$$$$\left(\begin{array}{c}x\\ y\end{array}\right)$$=$$\left(\begin{array}{c}6\\ 1\end{array}\right)$$, nilai {3x+y} adalah ....

1. 5
2. 6
3. 7

4. 8
5. 9

$$\begin{array}{rl}\left(\begin{array}{cc}5& 4\\ 1& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)& =\left(\begin{array}{c}6\\ 1\end{array}\right)\\ \left(\begin{array}{c}5x+4y\\ x+y\end{array}\right)& =\left(\begin{array}{c}6\\ 1\end{array}\right)\end{array}$$
$$\begin{array}{rl}5x+4y& =6\\ x+y& =1& \times 4\end{array}$$

$$\begin{array}{rl}5x+4y& =6\\ 4x+4y& =4& -\\ \\ x& =2\end{array}$$

$$\begin{array}{rl}x+y& =1\\ 2+y& =1\\ y& =1-2\\ & =-1\end{array}$$

$$\begin{array}{rl}3x+y& =3(2)+(-1)\\ & =6-1\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 5}}}}\end{array}$$

Jadi, {3x+y=5}. JAWAB: A