HOTS Zone : Matriks
Table of Contents
Tipe:
No.
Express \begin{vmatrix}a_1+b_1&c_1+d_1\\a_2+b_2&c_2+d_2\end{vmatrix} as a sum of four determinants whose entries contain no sums.ALTERNATIF PENYELESAIAN
\begin{aligned}
\begin{vmatrix}a_1+b_1&c_1+d_1\\a_2+b_2&c_2+d_2\end{vmatrix}&=\left(a_1+b_1\right)\left(c_2+d_2\right)-\left(a_2+b_2\right)\left(c_1+d_1\right)\\
&=a_1c_2+a_1d_2+b_1c_2+b_1d_2-a_2c_1-a_2d_1-b_2c_1-b_2d_1\\
&=a_1c_2-a_2c_1+a_1d_2-a_2d_1+b_1c_2-b_2c_1+b_1d_2-b_2d_1\\
&=\begin{vmatrix}a_1&c_1\\a_2&c_2\end{vmatrix}+\begin{vmatrix}a_1&d_1\\a_2&d_2\end{vmatrix}+\begin{vmatrix}b_1&c_1\\b_2&c_2\end{vmatrix}+\begin{vmatrix}b_1&d_1\\b_2&d_2\end{vmatrix}
\end{aligned}
So, \(\begin{vmatrix}a_1+b_1&c_1+d_1\\a_2+b_2&c_2+d_2\end{vmatrix}=\begin{vmatrix}a_1&c_1\\a_2&c_2\end{vmatrix}+\begin{vmatrix}a_1&d_1\\a_2&d_2\end{vmatrix}+\begin{vmatrix}b_1&c_1\\b_2&c_2\end{vmatrix}+\begin{vmatrix}b_1&d_1\\b_2&d_2\end{vmatrix}\).
No.
Jika $A=\begin{pmatrix}2&5\\-1&-2\end{pmatrix}$ maka tentukan matriksALTERNATIF PENYELESAIAN
\(\begin{aligned}
A^2&=\begin{pmatrix}2&5\\-1&-2\end{pmatrix}\cdot\begin{pmatrix}2&5\\-1&-2\end{pmatrix}\\
&=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}\\
&=-\begin{pmatrix}1&0\\0&1\end{pmatrix}\\
&=-I\\
A&=-A^{-1}
\end{aligned}\)
\(\begin{aligned} A^3&=A^2\cdot A\\ &=-I\cdot A\\ &=-A\\ A^3+A&=0 \end{aligned}\)
\(\begin{aligned} A^4&=A^3\cdot A\\ &=\left(-A\right)\cdot\left(-A^{-1}\right)\\ &=I \end{aligned}\)
\(\begin{aligned} A^{2017}+A^{2020}+A^{2023}&=A^{(1\mod4)}+A^{(0\mod4)}+A^{(3\mod4)}\\ &=A+I+A^3\\ &=I\\ &=\begin{pmatrix}1&0\\0&1\end{pmatrix} \end{aligned}\)
\(\begin{aligned} A^3&=A^2\cdot A\\ &=-I\cdot A\\ &=-A\\ A^3+A&=0 \end{aligned}\)
\(\begin{aligned} A^4&=A^3\cdot A\\ &=\left(-A\right)\cdot\left(-A^{-1}\right)\\ &=I \end{aligned}\)
\(\begin{aligned} A^{2017}+A^{2020}+A^{2023}&=A^{(1\mod4)}+A^{(0\mod4)}+A^{(3\mod4)}\\ &=A+I+A^3\\ &=I\\ &=\begin{pmatrix}1&0\\0&1\end{pmatrix} \end{aligned}\)
Jadi, $A^{2017}+A^{2020}+A^{2023}=\begin{pmatrix}1&0\\0&1\end{pmatrix}$.
Post a Comment