1. jikaA=(12) (34)B =(2 3) (0 1)C =(5 2) (-1 0)sederhana dari(A+C)-(A+B) adl...2. jika diketahuimatr.... Question from @AnnaNurJannah

5dregensleyer jawaban soal nomer 1
$$\begin{align} matriks\ A = \left[\begin{array}{ccc}1&2\\3&4\end{array}\right]\ ,\ b=\left[\begin{array}{ccc}2&3\\0&1\end{array}\right]\ ,\ c=\left[\begin{array}{ccc}5&2\\-1&0\end{array}\right] \end$

maka nilai dari A + C
$$\begin{align} A+C&= \left[\begin{array}{ccc}1&2\\3&4\end{array}\right]\ +\ \left[\begin{array}{ccc}5&2\\-1&0\end{array}\right]\\&= \left[\begin{array}{ccc}1+5&2+2\\3-1&4+0\end{array}\right]\ \\&= \left[\begin{array}{ccc}6&4\\2&4\end{array}\right]\ \end$

maka nilai dari A + B
$$\begin{align} A+B &= \left[\begin{array}{ccc}1&2\\3&4\end{array}\right]\ + \ \left[\begin{array}{ccc}2&3\\0&1\end{array}\right]\ \\&= \left[\begin{array}{ccc}1+2&2+3\\3+0&4+1\end{array}\right]\ \\&= \left[\begin{array}{ccc}3&5\\3&5\end{array}\right]\ \end$

sederhana dari (A+C)-(A+B)
$$\begin{align} jawab &= (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]\ \\&= \left[\begin{array}{ccc}6-3&4-5\\2-3&4-5\end{array}\right]\ \\&= \left[\begin{array}{ccc}3&-1\\-1&-1\end{array}\right]\ \end$

jawaban soal nomer 2
$$\begin{align} matriks\ A = \left[\begin{array}{ccc}1&-1\\2&-2\end{array}\right]\ ,\ b=\left[\begin{array}{ccc}1&1\\4&-2\end{array}\right]\ \end$

maka nilai dari A+B
$$\begin{align} A+B &= \left[\begin{array}{ccc}1&-1\\2&-2\end{array}\right]\ + \ \left[\begin{array}{ccc}1&1\\4&-2\end{array}\right]\ \\&= \left[\begin{array}{ccc}1+1&-1+1\\2+4&-2-2\end{array}\right]\ \\&= \left[\begin{array}{ccc}2&0\\6&-4\end{array}\right]\ \end$

jawaban soal nomer 3
$$\begin{align} A&= \left[\begin{array}{ccc}4&-8\\1&-3\end{array}\right]\ \\A^{-1}&= \frac{1}{determinan\ A}\ \ Adjoin\ A \\&= \frac{1}{(4\times -3)-(1\times -8)}\ \ \left[\begin{array}{ccc}-3&8\\-1&-4\end{array}\right]\ \\&= \frac{1}{-12-(-8)}\ \ \left[\begin{array}{ccc}-3&8\\-1&-4\end{array}\right]\ \\&= \frac{1}{-4}\ \ \left[\begin{array}{ccc}-3&8\\-1&-4\end{array}\right]\ \\&= \left[\begin{array}{ccc}^3/_4&-2\\^1/_4&1\end{array}\right]\ \end$

2 votes Thanks 2

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