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Berikut ini adalah pertanyaan dari kidolpati pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

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Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

$\purple{\huge{1.}}$

$\sf A+C=\left[\begin{array}{ccc}1&2\\3&4\end{array}\right]+\left[\begin{array}{ccc}5&2\\-1&0\end{array}\right]$

$\sf A+C=\left[\begin{array}{ccc}1+5&2+2\\3+(-1)&4+0\end{array}\right]$

$\sf A+C=\left[\begin{array}{ccc}6&4\\2&4\end{array}\right]$

$\sf A+B=\left[\begin{array}{ccc}1&2\\3&4\end{array}\right]+\left[\begin{array}{ccc}2&3\\0&1\end{array}\right]$

$\sf A+B=\left[\begin{array}{ccc}1+2&2+3\\3+0&4+1\end{array}\right]$

$\sf A+B=\left[\begin{array}{ccc}3&5\\3&5\end{array}\right]$

$\sf (A+C)-(A+B)=\left[\begin{array}{ccc}6&4\\2&4\end{array}\right]-\left[\begin{array}{ccc}3&5\\3&5\end{array}\right]$

$\sf (A+C)-(A+B)=\left[\begin{array}{ccc}6-3&4-5\\2-3&4-5\end{array}\right]$

$\boxed{\boxed{\red{\huge{\begin{array}{ccc}\sf (A+C)-(A+B)\\=\left[\begin{array}{ccc}3&-1\\-1&-1\end{array}\right]\end{array}}}}}$

$\\$

$\purple{\huge{2.}}$

$\sf AB=\left[\begin{array}{ccc}4&-3\\3&-4\end{array}\right]\left[\begin{array}{ccc}3&4\\2&-3\end{array}\right]$

$\sf AB=\left[\begin{array}{ccc}(4\times 3)+(-3\times 2)&(4\times 4)+(-3\times -3)\\(3\times 3)+(-4\times 2)&(3\times 4)+(-4\times -3)\end{array}\right]$

$\red{\huge{\sf AB=\left[\begin{array}{ccc}6&25\\1&24\end{array}\right]}}$

$\\$

$\purple{\huge{3.}}$

$det~\sf P=(1\times 3)-(-2\times 5)=3-(-10)$

$\red{\huge{det~\sf P=13}}$

$\\$

$\purple{\huge{4.}}$

$\sf A^{-1}=\frac{1}{(4\times -3)-(-8\times 1)}\left[\begin{array}{ccc}-3&8\\-1&4\end{array}\right]$

$\sf A^{-1}=-\frac{1}{4}\left[\begin{array}{ccc}-3&8\\-1&4\end{array}\right]$

$\red{\huge{\sf A^{-1}=\left[\begin{array}{ccc}\frac{3}{4}&-2\\\frac{1}{4}&-1\end{array}\right]}}$

$\\$

$\purple{\huge{5.}}$

Cara perhitungannya perhatikan gambar terlampir :

$\left|\begin{array}{ccc}2&3&4\\1&2&2\\2&1&3\end{array}\right|$

$=(2\times 2\times 3)+(3\times 2\times 2)+(4\times 1\times 1)-(4\times 2\times 2)-(2\times 2\times 1)-(3\times 1\times 3)$

$=12+12+4-16-4-9$

$\red{\huge{\left|\begin{array}{ccc}2&3&4\\1&2&2\\2&1&3\end{array}\right|=-1}}$

$[tex]\purple{\huge{1.}}[/tex][tex]\sf A+C=\left[\begin{array}{ccc}1&2\\3&4\end{array}\right]+\left[\begin{array}{ccc}5&2\\-1&0\end{array}\right][/tex][tex]\sf A+C=\left[\begin{array}{ccc}1+5&2+2\\3+(-1)&4+0\end{array}\right][/tex][tex]\sf A+C=\left[\begin{array}{ccc}6&4\\2&4\end{array}\right][/tex][tex]\sf A+B=\left[\begin{array}{ccc}1&2\\3&4\end{array}\right]+\left[\begin{array}{ccc}2&3\\0&1\end{array}\right][/tex][tex]\sf A+B=\left[\begin{array}{ccc}1+2&2+3\\3+0&4+1\end{array}\right][/tex][tex]\sf A+B=\left[\begin{array}{ccc}3&5\\3&5\end{array}\right][/tex][tex]\sf (A+C)-(A+B)=\left[\begin{array}{ccc}6&4\\2&4\end{array}\right]-\left[\begin{array}{ccc}3&5\\3&5\end{array}\right][/tex][tex]\sf (A+C)-(A+B)=\left[\begin{array}{ccc}6-3&4-5\\2-3&4-5\end{array}\right][/tex][tex]\boxed{\boxed{\red{\huge{\begin{array}{ccc}\sf (A+C)-(A+B)\\=\left[\begin{array}{ccc}3&-1\\-1&-1\end{array}\right]\end{array}}}}}[/tex][tex]\\[/tex][tex]\purple{\huge{2.}}[/tex][tex]\sf AB=\left[\begin{array}{ccc}4&-3\\3&-4\end{array}\right]\left[\begin{array}{ccc}3&4\\2&-3\end{array}\right][/tex][tex]\sf AB=\left[\begin{array}{ccc}(4\times 3)+(-3\times 2)&(4\times 4)+(-3\times -3)\\(3\times 3)+(-4\times 2)&(3\times 4)+(-4\times -3)\end{array}\right][/tex][tex]\red{\huge{\sf AB=\left[\begin{array}{ccc}6&25\\1&24\end{array}\right]}}[/tex][tex]\\[/tex][tex]\purple{\huge{3.}}[/tex][tex]det~\sf P=(1\times 3)-(-2\times 5)=3-(-10)[/tex][tex]\red{\huge{det~\sf P=13}}[/tex][tex]\\[/tex][tex]\purple{\huge{4.}}[/tex][tex]\sf A^{-1}=\frac{1}{(4\times -3)-(-8\times 1)}\left[\begin{array}{ccc}-3&8\\-1&4\end{array}\right][/tex][tex]\sf A^{-1}=-\frac{1}{4}\left[\begin{array}{ccc}-3&8\\-1&4\end{array}\right][/tex][tex]\red{\huge{\sf A^{-1}=\left[\begin{array}{ccc}\frac{3}{4}&-2\\\frac{1}{4}&-1\end{array}\right]}}[/tex][tex]\\[/tex][tex]\purple{\huge{5.}}[/tex]Cara perhitungannya perhatikan gambar terlampir :[tex]\left|\begin{array}{ccc}2&3&4\\1&2&2\\2&1&3\end{array}\right|[/tex][tex]=(2\times 2\times 3)+(3\times 2\times 2)+(4\times 1\times 1)-(4\times 2\times 2)-(2\times 2\times 1)-(3\times 1\times 3)[/tex][tex]=12+12+4-16-4-9[/tex][tex]\red{\huge{\left|\begin{array}{ccc}2&3&4\\1&2&2\\2&1&3\end{array}\right|=-1}}[/tex]$

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Last Update: Fri, 16 Jul 21

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