Jika A= \left[\begin{array}{ccc}-5&2\\2&-1\\\end{array}\right] , [/tex]1\\\end{array}\right] [/tex] .... Question from @danielcarlos897

October 2018 2 62 Report

Jika A= \left[\begin{array}{ccc}-5&2\\2&-1\\\end{array}\right] , [/tex]1\\\end{array}\right] [/tex] ,
$\left[\begin{array}{ccc}-5&2\\2&-1\\\end{array}\right] , A^{t}$
menyatakan transpose dari A, dan A^(-1) menyatakan invers dari A.
maka A^(t) + A^(-1) adalah...

jackdaniel47 $A^{t} = \left[\begin{array}{ccc}-5&2\\2&-1\\\end{array}\right] \\A^{-1}= \left[\begin{array}{ccc}-1&-2\\-2&-5\\\end{array}\right] \\\\A^{t} + A^{-1} =....?
\\\\= \left[\begin{array}{ccc}-5-1&&2-2\\2-2&&-1-5\\\end{array}\right] \\\\= \left[\begin{array}{ccc}-6&0\\0&-6\\\end{array}\right]$

smoga bisa membantu... :)

semangat matematika... (y)

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izharulhaq78 $\displaystyle A=\left[\begin{array}{ccc}-5&2\\2&-1\\\end{array}\right]\\\\a_{11}=-5\\a_{12}=2\\a_{21}=2\\a_{22}=-1\\\\\text{cari matriks tranpose-nya}\\A^t=\left[\begin{array}{ccc}a_{11}&a_{21}\\a_{12}&a_{22}\\\end{array}\right]\\A^t=\left[\begin{array}{ccc}-5&2\\2&-1\\\end{array}\right]\\\\\text{cari inversnya}\\A^{-1}=\frac{1}{|A|}\left[\begin{array}{ccc}a_{22}&-a_{12}\\-a_{21}&a_{11}\\\end{array}\right]$
$\displaystyle A^{-1}=\frac{1}{a_{11}a_{22}-a_{12}a_{21}}\left[\begin{array}{ccc}a_{22}&-a_{12}\\-a_{21}&a_{11}\\\end{array}\right]\\A^{-1}=\frac{1}{-5\times(-1)-2\times2}\left[\begin{array}{ccc}-1&-2\\-2&-5\\\end{array}\right]\\A^{-1}=\frac{1}{5-4}\left[\begin{array}{ccc}-1&-2\\-2&-5\\\end{array}\right]\\A^{-1}=\frac{1}{1}\left[\begin{array}{ccc}-1&-2\\-2&-5\\\end{array}\right]\\A^{-1}=\left[\begin{array}{ccc}-1&-2\\-2&-5\\\end{array}\right]$

$\displaystyle \text{masuk ke soal}\\A^t+A^{-1}=\left[\begin{array}{ccc}-5&2\\2&-1\\\end{array}\right]+\left[\begin{array}{ccc}-1&-2\\-2&-5\\\end{array}\right]\\A^t+A^{-1}=\left[\begin{array}{ccc}-5-1&2-2\\2-2&-1-5\\\end{array}\right]\\\boxed{\boxed{A^t+A^{-1}=\left[\begin{array}{ccc}-6&0\\0&-6\\\end{array}\right]\text{ atau }-6I}}$

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