Exercise Zone : Invers Matriks

Berikut ini adalah kumpulan soal mengenai Invers Matriks. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.

Tipe:

Standar SBMPTN HOTS

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No.

Jika {P=\begin{pmatrix}1&2\\1&3\end{pmatrix}} dan {\begin{pmatrix}x&y\\-z&z\end{pmatrix}=2P^{-1}} dengan P^{-1} menyatakan invers matriks P, maka {x+y=} ....

\begin{aligned} |P|&=(1)(3)-(2)(1)\\ &=1 \end{aligned}

\begin{aligned} P^{-1}&=\dfrac1{|P|}\begin{pmatrix}3&-2\\-1&1\end{pmatrix}\\ &=\dfrac11\begin{pmatrix}3&-2\\-1&1\end{pmatrix}\\ &=\begin{pmatrix}3&-2\\-1&1\end{pmatrix} \end{aligned}

\begin{aligned} \begin{pmatrix}x&y\\-z&z\end{pmatrix}&=2P^{-1}\\ \begin{pmatrix}x&y\\-z&z\end{pmatrix}&=2\begin{pmatrix}3&-2\\-1&1\end{pmatrix}\\ \begin{pmatrix}x&y\\-z&z\end{pmatrix}&=\begin{pmatrix}6&-4\\-2&2\end{pmatrix} \end{aligned}
x=6 dan y=-4

\begin{aligned} x+y&=6+(-4)\\ &=2 \end{aligned}

Jadi, x+y=2.

No.

Jika matriks A=\begin{pmatrix}3&-2\\7&-5\end{pmatrix}, invers matriks A adalah A^{-1}= ....

\begin{aligned} \det A&=\begin{vmatrix}3&-2\\7&-5\end{vmatrix}\\ &=3\cdot(-5)-(-2)\cdot7\\ &=-15+14\\ &=-1 \end{aligned}

\begin{aligned} A^{-1}&=\dfrac1{\det A}\begin{pmatrix}-5&2\\-7&3\end{pmatrix}\\ &=\dfrac1{-1}\begin{pmatrix}-5&2\\-7&3\end{pmatrix}\\ &=\boxed{\boxed{\begin{pmatrix}5&-2\\7&-3\end{pmatrix}}}\end{aligned}

Jadi, A^{-1}=\begin{pmatrix}5&-2\\7&-3\end{pmatrix}.