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Tipe:

StandarSNBTHOTS


No.

Jika matriks \(A\cdot B=\begin{pmatrix}4&6\\-7&-14\end{pmatrix}\) dan det A = 7, maka det(7BA−1) adalah....
  1. −16
  2. −15
  3. −14
  1. −13
  2. −12
Matematika Zone Indonesia
ALTERNATIF PENYELESAIAN
\(\begin{aligned} A\cdot B&=\begin{pmatrix}4&6\\-7&-14\end{pmatrix}\\[4pt] |AB|&=(4)(-14)-(6)(-7)\\ |A||B|&=-56+42\\ 7|B|&=-14\\ |B|&=-2 \end{aligned}\)

\(\begin{aligned} \left|7BA^{-1}\right|&=7^2|B|\left|A^{-1}\right|\\ &=49(-2)\left(\dfrac1{|A|}\right)\\ &=-98\left(\dfrac17\right)\\ &=\boxed{\boxed{-14}} \end{aligned}\)
Jadi, det(7BA−1) = −14.
JAWAB: C

No.

Jika \({A=\begin{pmatrix}-1&-1&0\\-1&1&2\end{pmatrix}}\), \({B=\begin{pmatrix}-4&x\\-8&y\\-6&z\end{pmatrix}}\) dan \({AB=\begin{pmatrix}12&18\\-16&4\end{pmatrix}}\), maka nilai z − x adalah ....
Matematika Zone Indonesia
ALTERNATIF PENYELESAIAN
\(\begin{aligned} AB&=\begin{pmatrix}12&18\\-16&4\end{pmatrix}\\ \begin{pmatrix}-1&-1&0\\-1&1&2\end{pmatrix}\begin{pmatrix}-4&x\\-8&y\\-6&z\end{pmatrix}&=\begin{pmatrix}12&18\\-16&4\end{pmatrix}\\ \begin{pmatrix}12&-x-y\\-16&-x+y+2z\end{pmatrix}&=\begin{pmatrix}12&18\\-16&4\end{pmatrix} \end{aligned}\)

\(\begin{aligned} -x-y&=18\\ -x+y+2z&=4&\qquad+\\\hline -2x+2z&=22\\ 2z-2x&=22\\ z-x&=\boxed{\boxed{11}} \end{aligned}\)
Jadi, z − x = 11.

No.

Diketahui matriks \({A=\begin{pmatrix}-6&-9\\9&-8\end{pmatrix}}\) dan \({B=\begin{pmatrix}1&y\\x&3\end{pmatrix}}\). Jika determinan AB adalah 903 maka xy = ....
Matematika Zone Indonesia
ALTERNATIF PENYELESAIAN
\(\begin{aligned} |AB|&=903\\ |A||B|&=903\\ (129)(3-xy)&=903\\ 3-xy&=7\\ xy&=\boxed{\boxed{-4}} \end{aligned}\)
Jadi, xy = −4.

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 ≠ 0 dan p ≠ 0. Matriks PQ tidak mempunyai invers apabila nilai p = ....
Matematika Zone Indonesia
ALTERNATIF PENYELESAIAN
\(\begin{aligned} |PQ|&=0\\ |P||Q|&=0\\ ((2)(3)-(-1)(4))((2r)(p+1)-(1)(r))&=0\\ (10)(2pr+2r-r)&=0\\ 2pr+r&=0\\ r(2p+1)&=0\\ 2p+1&=0\\ 2p&=-1\\ p&=-\dfrac12 \end{aligned}\)
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 ....
Matematika Zone Indonesia
ALTERNATIF PENYELESAIAN
\(\begin{aligned} \begin{pmatrix}2&k\\1&0\end{pmatrix}A&=\begin{pmatrix}2&4k\\1&0\end{pmatrix}\\ A&=\begin{pmatrix}2&k\\1&0\end{pmatrix}^{-1}\begin{pmatrix}2&4k\\1&0\end{pmatrix}\\ &=\dfrac1{2\cdot0-k\cdot1}\begin{pmatrix}0&-k\\-1&2\end{pmatrix}\begin{pmatrix}2&4k\\1&0\end{pmatrix}\\ &=\dfrac1{-k}\begin{pmatrix}-k&0\\0&-4k\end{pmatrix}\\ &=\begin{pmatrix}1&0\\0&4\end{pmatrix} \end{aligned}\)
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. \(\begin{pmatrix}2&7\\9&-3\end{pmatrix}\)
  1. \(\begin{pmatrix}2&7\\9&3\end{pmatrix}\)
  2. \(\begin{pmatrix}2&9\\7&-3\end{pmatrix}\)
Matematika Zone Indonesia
ALTERNATIF PENYELESAIAN
\(\begin{aligned} 2K-L+M&=2\begin{pmatrix}-1&4\\3&-2\end{pmatrix}-\begin{pmatrix}5&1\\3&-2\end{pmatrix}+\begin{pmatrix}9&0\\6&-1\end{pmatrix}\\ &=\begin{pmatrix}-2&8\\6&-4\end{pmatrix}-\begin{pmatrix}5&1\\3&-2\end{pmatrix}+\begin{pmatrix}9&0\\6&-1\end{pmatrix}\\ &=\boxed{\boxed{\color{blue}\begin{pmatrix}2&7\\9&-3\end{pmatrix}}} \end{aligned}\)
Jadi, \(2K-L+M=\begin{pmatrix}2&7\\9&-3\end{pmatrix}\).
JAWAB: E

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
  1. −30
  2. −36
ALTERNATIF PENYELESAIAN
\(\begin{aligned}C&=AB\\ |C|&=|A||B|\\ &=\left[(-1)(-2)-(2)(3)\right]\cdot[(1)(0)-(-2)(3)]\\ &=[2-6][0-(-6)]\\ &=(-4)(6)\\ &=-24\end{aligned}\)
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
  1. 1
  2. −1
ALTERNATIF PENYELESAIAN
\(\begin{aligned} 2\begin{pmatrix}x&2\\6&6\end{pmatrix}+\begin{pmatrix}2&4\\-2&0\end{pmatrix}&=\begin{pmatrix}1&3\\2&4\end{pmatrix}\begin{pmatrix}-1&2\\3&y\end{pmatrix}\\ \begin{pmatrix}2x&4\\12&12\end{pmatrix}+\begin{pmatrix}2&4\\-2&0\end{pmatrix}&=\begin{pmatrix}8&3y+2\\10&4y+4\end{pmatrix}\\ \begin{pmatrix}2x+2&8\\10&12\end{pmatrix}&=\begin{pmatrix}8&3y+2\\10&4y+4\end{pmatrix} \end{aligned}\)
Jadi, x − y = 1.
JAWAB: D

No.

Misalkan AT 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 = 3AT + 16I, maka nilai 10a + 4b adalah
  1. 32
  2. 34
  3. 35
  1. 36
  2. 40
ALTERNATIF PENYELESAIAN
\(A^T=\begin{pmatrix}8&a\\0&b\end{pmatrix}\)

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

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

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

\(\begin{aligned} 10a+4b&=10(0)+32\\ &=\boxed{\boxed{32}} \end{aligned}\)
Jadi, 10a + 4b = 32.
JAWAB: A

No.

Diketahui matriks A berordo 2×2 dan \(B=\begin{pmatrix}-3&5\\-1&2\end{pmatrix}\) dan \(C=\begin{pmatrix}4&5\\2&3\end{pmatrix}\). Jika A memenuhi B ⋅ A = C, maka det (2A−1) adalah
  1. −2
  2. −1
  3. \(-\dfrac12\)
  1. $\dfrac12$
  2. 2
ALTERNATIF PENYELESAIAN
\(\begin{aligned} \det B&=(-3)(2)-(5)(-1)\\ &=-6-(-5)\\ &=-1 \end{aligned}\)

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

\(\begin{aligned} B\cdot A&=C\\ A&=B^{-1}\cdot C\\ \det A&=\det\left(B^{-1}\cdot C\right)\\ &=\dfrac1{\det B}\cdot\det C\\ &=\dfrac1{-1}\cdot2\\ &=-2 \end{aligned}\)
\(\begin{aligned} \det\left(2A^{-1}\right)&=2^2\det A^{-1}\\ &=4\cdot\dfrac1{\det A}\\ &=4\cdot\dfrac1{-2}\\ &=\boxed{\boxed{-2}} \end{aligned}\)
Jadi, det (2A−1) = −2.
JAWAB: A


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