Exercise Zone : Matriks

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Tipe:

Standar SBMPTN HOTS

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SUMBER

No.

Diberikan matriks {P=\begin{pmatrix}2&-1\\4&3\end{pmatrix}} dan {Q=\begin{pmatrix}2r&1\\r&p+1\end{pmatrix}} dengan {r\neq0} dan {p\neq0}. Matriks PQ tidak mempunyai invers apabila nilai p= ....

\begin{aligned} |PQ|&=0\\ |P||Q|&=0\\ ((2)(3)-(-1)(4))((2r)(p+1)-(1)(r))&=0\\ (10)(2pr+2r-r)&=0\\ 2pr+r&=0\\ r(2p+1)&=0\\ 2p+1&=0\\ 2p&=-1\\ p&=-\dfrac12 \end{aligned}

Jadi, p=-\dfrac12.

No.

Matriks A yang memenuhi {\begin{pmatrix}2&k\\1&0\end{pmatrix}A=\begin{pmatrix}2&4k\\1&0\end{pmatrix}} adalah ....

\begin{aligned} \begin{pmatrix}2&k\\1&0\end{pmatrix}A&=\begin{pmatrix}2&4k\\1&0\end{pmatrix}\\ A&=\begin{pmatrix}2&k\\1&0\end{pmatrix}^{-1}\begin{pmatrix}2&4k\\1&0\end{pmatrix}\\ &=\dfrac1{2\cdot0-k\cdot1}\begin{pmatrix}0&-k\\-1&2\end{pmatrix}\begin{pmatrix}2&4k\\1&0\end{pmatrix}\\ &=\dfrac1{-k}\begin{pmatrix}-k&0\\0&-4k\end{pmatrix}\\ &=\begin{pmatrix}1&0\\0&4\end{pmatrix} \end{aligned}

Jadi, A=\begin{pmatrix}1&0\\0&4\end{pmatrix}.

No.

Diketahui matriks {K=\begin{pmatrix}-1&4\\3&-2\end{pmatrix}}, {L=\begin{pmatrix}5&1\\3&-2\end{pmatrix}} dan {M=\begin{pmatrix}9&0\\6&-1\end{pmatrix}}. Matriks {2K-L+M} adalah....

1. \begin{pmatrix}-2&8\\-8&3\end{pmatrix}
2. \begin{pmatrix}1&7\\-6&3\end{pmatrix}
3. \begin{pmatrix}2&7\\9&-3\end{pmatrix}

4. \begin{pmatrix}2&7\\9&3\end{pmatrix}
5. \begin{pmatrix}2&9\\7&-3\end{pmatrix}

\begin{aligned} 2K-L+M&=2\begin{pmatrix}-1&4\\3&-2\end{pmatrix}-\begin{pmatrix}5&1\\3&-2\end{pmatrix}+\begin{pmatrix}9&0\\6&-1\end{pmatrix}\\ &=\begin{pmatrix}-2&8\\6&-4\end{pmatrix}-\begin{pmatrix}5&1\\3&-2\end{pmatrix}+\begin{pmatrix}9&0\\6&-1\end{pmatrix}\\ &=\boxed{\boxed{\color{blue}\begin{pmatrix}2&7\\9&-3\end{pmatrix}}} \end{aligned}

Jadi, 2K-L+M=\begin{pmatrix}2&7\\9&-3\end{pmatrix}.

No.

Diketahui matriks {A=\begin{pmatrix}-1&2\\3&-2\end{pmatrix}} dan {B=\begin{pmatrix}1&-2\\3&0\end{pmatrix}}. Jika matriks {C=AB}, maka determinan matriks C= ....

1. 18
2. 12
3. -24

4. -30
5. -36

\begin{aligned}C&=AB\\ |C|&=|A||B|\\ &=\left[(-1)(-2)-(2)(3)\right]\cdot[(1)(0)-(-2)(3)]\\ &=[2-6][0-(-6)]\\ &=(-4)(6)\\ &=-24\end{aligned}

Jadi, determinan matriks C adalah -24. JAWAB: C

No.

Diketahui persamaan matriks:
{2\begin{pmatrix}x&2\\6&6\end{pmatrix}+\begin{pmatrix}2&4\\-2&0\end{pmatrix}=\begin{pmatrix}1&3\\2&4\end{pmatrix}\begin{pmatrix}-1&2\\3&y\end{pmatrix}}
Nilai x-y=

1. -2
2. 2
3. 0

4. 1
5. -1

\begin{aligned} 2\begin{pmatrix}x&2\\6&6\end{pmatrix}+\begin{pmatrix}2&4\\-2&0\end{pmatrix}&=\begin{pmatrix}1&3\\2&4\end{pmatrix}\begin{pmatrix}-1&2\\3&y\end{pmatrix}\\ \begin{pmatrix}2x&4\\12&12\end{pmatrix}+\begin{pmatrix}2&4\\-2&0\end{pmatrix}&=\begin{pmatrix}8&3y+2\\10&4y+4\end{pmatrix}\\ \begin{pmatrix}2x+2&8\\10&12\end{pmatrix}&=\begin{pmatrix}8&3y+2\\10&4y+4\end{pmatrix} \end{aligned}

Jadi, x-y=1. JAWAB: D

No.

Misalkan A^T adalah transpose matriks A dan {I=\begin{pmatrix}1&0\\0&1\end{pmatrix}}. Jika {A=\begin{pmatrix}8&0\\a&b\end{pmatrix}} sehingga {5A=3A^T+16I}, maka nilai {10a+4b} adalah

1. 32
2. 34
3. 35

4. 36
5. 40

{A^T=\begin{pmatrix}8&a\\0&b\end{pmatrix}}

\begin{aligned} 5A&=3A^T+16I\\ 5\begin{pmatrix}8&0\\a&b\end{pmatrix}&=3\begin{pmatrix}8&a\\0&b\end{pmatrix}+16\begin{pmatrix}1&0\\0&1\end{pmatrix}\\ \begin{pmatrix}40&0\\5a&5b\end{pmatrix}&=\begin{pmatrix}24&3a\\0&3b\end{pmatrix}+\begin{pmatrix}16&0\\0&16\end{pmatrix}\\ \begin{pmatrix}40&0\\5a&5b\end{pmatrix}&=\begin{pmatrix}40&3a\\0&3b+16\end{pmatrix} \end{aligned}

\begin{aligned} 0&=3a\\ a&=0 \end{aligned}

\begin{aligned} 5b&=3b+16\\ 2b&=16\\ 4b&=32 \end{aligned}

\begin{aligned} 10a+4b&=10(0)+32\\ &=\boxed{\boxed{32}} \end{aligned}

Jadi, 10a+4b=32. JAWAB: A

No.

Diketahui matriks A berordo 2\times2 dan {B=\begin{pmatrix}-3&5\\-1&2\end{pmatrix}} dan {C=\begin{pmatrix}4&5\\2&3\end{pmatrix}}. Jika A memenuhi {B\cdot A=C}, maka \det\left(2A^{-1}\right) adalah

1. -2
2. -1
3. -\dfrac12

4. \dfrac12
5. 2

\begin{aligned} \det B&=(-3)(2)-(5)(-1)\\ &=-6-(-5)\\ &=-1 \end{aligned}

\begin{aligned} \det C&=(4)(3)-(2)(5)\\ &=12-10\\ &=2 \end{aligned}

\begin{aligned} B\cdot A&=C\\ A&=B^{-1}\cdot C\\ \det A&=\det\left(B^{-1}\cdot C\right)\\ &=\dfrac1{\det B}\cdot\det C\\ &=\dfrac1{-1}\cdot2\\ &=-2 \end{aligned}

\begin{aligned} \det\left(2A^{-1}\right)&=2^2\det A^{-1}\\ &=4\cdot\dfrac1{\det A}\\ &=4\cdot\dfrac1{-2}\\ &=\boxed{\boxed{-2}} \end{aligned}

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