Exercise Zone : Matriks [3]

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Tipe:

Standar SBMPTN HOTS

No.

Diketahui matriks \(A=\pmatrix{1&2\\3&7}\), \(B=\pmatrix{a&b\\2&3}\), dan \(A^T\cdot B^T=\pmatrix{13&11\\30&25}\) maka nilai 12a+9b=

1. 44
2. 46
3. 48

4. 50
5. 52

SKMP Juli 2020

\(\eqalign{ A^T\cdot B^T&=\pmatrix{13&11\\30&25}\\ \pmatrix{1&3\\2&7}\pmatrix{a&2\\b&3}&=\pmatrix{13&11\\30&25}\\ \pmatrix{a+3b&11\\2a+7b&25}&=\pmatrix{13&11\\30&25} }\)

\(\eqalign{ 2a+7b&=30\\ a+3b&=13\qquad&{\color{red}\times2} }\)

\(\eqalign{ 2a+7b&=30\\ 2a+2b&=26\qquad&{\color{red}-}\\\hline b&=4 }\)

\(\eqalign{ a+3b&=13\\ a+3(4)&=13\\ a+12&=13\\ a&=1 }\)

\(\eqalign{ 12a+9b&=12(1)+9(4)\\ &=12+36\\ &=\boxed{\boxed{48}} }\)

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. \(\boxed{\boxed{\pmatrix{-2&3\\8&4}\pmatrix{x\\y}=\pmatrix{5\\7}}}\)
2. \(\boxed{\boxed{\pmatrix{-5\\-6}=\pmatrix{7&8\\3&-4}\pmatrix{x\\y}}}\)
3. 3x+4y-10=0
   3x+4y=10
   
   7x-8y+12=0
   7x-8y=-12
   
   \(\boxed{\boxed{\pmatrix{3&4\\7&-8}\pmatrix{x\\y}=\pmatrix{10\\-12}}}\)

4. \(\boxed{\boxed{\pmatrix{2&5&-3\\3&-7&-1\\5&-9&6}\pmatrix{x\\y\\z}=\pmatrix{6\\10\\12}}}\)
5. 5y-10=0
   5y=10
   
   \(\boxed{\boxed{\pmatrix{4&0\\0&5\\2&3}\pmatrix{x\\y}=\pmatrix{20\\10\\16}}}\)

Jadi, \(\pmatrix{-2&3\\8&4}\pmatrix{x\\y}=\pmatrix{5\\7}\) \(\pmatrix{-5\\-6}=\pmatrix{7&8\\3&-4}\pmatrix{x\\y}\) \(\pmatrix{3&4\\7&-8}\pmatrix{x\\y}=\pmatrix{10\\-12}\) \(\pmatrix{2&5&-3\\3&-7&-1\\5&-9&6}\pmatrix{x\\y\\z}=\pmatrix{6\\10\\12}\) \(\pmatrix{4&0\\0&5\\2&3}\pmatrix{x\\y}=\pmatrix{20\\10\\16}\)

No.

Nyatakan matriks \(\pmatrix{6&3\\0&8}\) sebagai kombinasi linear dari matriks berikut:
\(\pmatrix{1&2\\-1&3}\), \(\pmatrix{0&1\\2&4}\), dan \(\pmatrix{4&-2\\0&-2}\).

Grup Telegram

\(\eqalign{ \pmatrix{6&3\\0&8}&=k_1\pmatrix{1&2\\-1&3}+k_2\pmatrix{0&1\\2&4}+k_3\pmatrix{4&-2\\0&-2}\\ &=\pmatrix{k_1+4k_3&2k_1+k_2-2k_3\\-k_1+2k_2&3k_1+4k_2-2k_3} }\)

\(\eqalign{ k_1&&&+&4k_3&=6\\ 2k_1&+&k_2&-&2k_3&=3\\ -k_1&+&2k_2&&&=0\\ 3k_1&+&4k_2&-&2k_3&=8 }\)
\(\eqalign{ \pmatrix{1&0&4&6\\2&1&-2&3\\-1&2&0&0\\3&4&-2&8}&\eqalign{-2b_1+b_2\\b_1+b_3\\-3b_1+b_3}\\ \pmatrix{1&0&4&6\\0&1&-10&-9\\0&2&4&6\\0&4&-14&-10}&\eqalign{\dfrac12b_3\\\dfrac12b_4}\\ \pmatrix{1&0&4&6\\0&1&-10&-9\\0&1&2&3\\0&2&-7&-5}&\eqalign{-b_2+b_3\\-2b_2+b_4}\\ \pmatrix{1&0&4&6\\0&1&-10&-9\\0&0&12&12\\0&0&13&13}&\eqalign{\dfrac1{12}b_3\\\dfrac1{13}b_4}\\ \pmatrix{1&0&0&2\\0&1&0&1\\0&0&1&1\\0&0&0&0} }\)
k_1=2, k_2=1, k_3=1

Jadi, \(\pmatrix{6&3\\0&8}=2\pmatrix{1&2\\-1&3}+\pmatrix{0&1\\2&4}+\pmatrix{4&-2\\0&-2}\).

No.

Diketahui \(A=\pmatrix{-3&2\\1&0}\), \(B=\pmatrix{4&-1\\2&5}\) dan \(C=\pmatrix{-2&-2\\3&3}\), maka A-(B-C)=...

1. \(\pmatrix{1&5\\0&1}\)
2. \(\pmatrix{4&3\\-3&2}\)
3. \(\pmatrix{2&1\\3&-2}\)

4. \(\pmatrix{-9&1\\2&-2}\)
5. \(\pmatrix{-3&1\\-3&2}\)

Grup Telegram

\(\eqalign{ A-(B-C)&=\pmatrix{-3&2\\1&0}-\left(\pmatrix{4&-1\\2&5}-\pmatrix{-2&-2\\3&3}\right)\\ &=\pmatrix{-3&2\\1&0}-\pmatrix{6&1\\-1&2}\\ &=\boxed{\boxed{\pmatrix{-9&1\\2&-2}}} }\)

Jadi, \(A-(B-C)=\pmatrix{-9&1\\2&-2}\). JAWAB: D

No.

Diketahui \(P=\pmatrix{-3&1\\4&-2}\) dan \(Q=\pmatrix{3&4\\1&2}\). Hitunglah \left(PQ\right)^{-1}

Grup Telegram

\(\eqalign{ PQ&=\pmatrix{-3&1\\4&-2}\pmatrix{3&4\\1&2}\\ &=\pmatrix{-8&-10\\10&12}\\ \left(PQ\right)^{-1}&=\dfrac1{(-8)(12)-(-10)(10)}\pmatrix{12&10\\-10&-8}\\ &=\dfrac1{-96+100}\pmatrix{12&10\\-10&-8}\\ &=\dfrac14\pmatrix{12&10\\-10&-8}\\ &=\boxed{\boxed{\pmatrix{3&\dfrac52\\-\dfrac52&-2}}} }\)

Jadi, \(\left(PQ\right)^{-1}=\pmatrix{3&\dfrac52\\-\dfrac52&-2}\).

No.

X adalah matriks persegi berordo 2\times2 yang memenuhi \(X\pmatrix{1&2\\3&4}=\pmatrix{4&8\\5&8}\). Matriks X adalah

1. \(\pmatrix{3&2\\-2&1}\)
2. \(\pmatrix{3&2\\2&1}\)
3. \(\pmatrix{-4&0\\-1&-2}\)

4. \(\pmatrix{4&0\\2&1}\)
5. \(\pmatrix{4&0\\-1&2}\)

Grup Telegram

\(\eqalign{ X\pmatrix{1&2\\3&4}&=\pmatrix{4&8\\5&8}\\ X&=\pmatrix{4&8\\5&8}\pmatrix{1&2\\3&4}^{-1}\\ &=\pmatrix{4&8\\5&8}\cdot\dfrac1{1\cdot4-2\cdot3}\pmatrix{4&-2\\-3&1}\\ &=\pmatrix{4&8\\5&8}\cdot\dfrac1{-2}\pmatrix{4&-2\\-3&1}\\ &=\dfrac1{-2}\pmatrix{4&8\\5&8}\pmatrix{4&-2\\-3&1}\\ &=\dfrac1{-2}\pmatrix{-8&0\\-4&-2}\\ &=\boxed{\boxed{\pmatrix{4&0\\2&1}}} }\)

Jadi, \(X=\pmatrix{4&0\\2&1}\). 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

Grup Telegram

\(\pmatrix{3&4&-3\\-2&7&-6\\5&8&-1}\)

Jadi, matriks koefisien sistem persamaan linear tersebut adalah \(\pmatrix{3&4&-3\\-2&7&-6\\5&8&-1}\)

No.

Diketahui matriks \(A=\pmatrix{5&2\\0&3}\) dan \(B=\pmatrix{-3&1\\17&0}\), 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

\(A^T=\pmatrix{5&0\\2&3}\)

\(B+A^T=\pmatrix{-3&1\\17&0}+\pmatrix{5&0\\2&3}=\pmatrix{2&1\\19&3}\)

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

\(\eqalign{ AX&=B+A^T\\ |A||X|&=\left|B+A^T\right|\\ 15|X|&=-13\\ |X|&=\boxed{\boxed{-\dfrac{13}{15}}} }\)

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

No.

Untuk persamaan {2\begin{pmatrix}x&3y\\3&y\end{pmatrix}+\begin{pmatrix}3&x-6\\1&x\end{pmatrix}=\begin{pmatrix}11&10\\7&8\end{pmatrix}}, harga {x+y} adalah

1. -2
2. 2
3. 4

4. 6
5. 7

Grup Telegram

\begin{aligned} 2\begin{pmatrix}x&3y\\3&y\end{pmatrix}+\begin{pmatrix}3&x-6\\1&x\end{pmatrix}&=\begin{pmatrix}11&10\\7&8\end{pmatrix}\\ \begin{pmatrix}2x&6y\\6&2y\end{pmatrix}+\begin{pmatrix}3&x-6\\1&x\end{pmatrix}&=\begin{pmatrix}11&10\\7&8\end{pmatrix}\\ \begin{pmatrix}2x+3&6y+x-6\\7&2y+x\end{pmatrix}&=\begin{pmatrix}11&10\\7&8\end{pmatrix} \end{aligned}

\begin{aligned} 2x+3&=11\\ 2x&=8\\ x&=4 \end{aligned}

\begin{aligned} 2y+x&=8\\ 2y+4&=8\\ 2y&=4\\ y&=2 \end{aligned}

\begin{aligned} x+y&=2+4\\ &=\boxed{\boxed{6}} \end{aligned}

Jadi, {x+y=6}.

No.

Jika \begin{pmatrix}5&4\\1&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}6\\1\end{pmatrix}, nilai {3x+y} adalah ....

1. 5
2. 6
3. 7

4. 8
5. 9

\begin{aligned} \begin{pmatrix}5&4\\1&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}&=\begin{pmatrix}6\\1\end{pmatrix}\\ \begin{pmatrix}5x+4y\\x+y\end{pmatrix}&=\begin{pmatrix}6\\1\end{pmatrix} \end{aligned}
\begin{aligned} 5x+4y&=6\\ x+y&=1\qquad&{\color{red}\times4} \end{aligned}

\begin{aligned} 5x+4y&=6\\ 4x+4y&=4\qquad&{\color{red}-}\\\hline\\[-10pt] x&=2 \end{aligned}

\begin{aligned} x+y&=1\\ 2+y&=1\\ y&=1-2\\ &=-1\end{aligned}

\begin{aligned} 3x+y&=3(2)+(-1)\\ &=6-1\\ &=\boxed{\boxed{5}}\end{aligned}

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