![(a)= \left[\begin{array}{cc}-1&2\\-1/3&-4/3\\14/3&4/3\end{array}\right]](/tpl/images/0524/3141/bf04e.png)
,![(b) = \left[\begin{array}{cc}-1/3&2/3\\-1&-4/3\\-3&-2\end{array}\right]](/tpl/images/0524/3141/ac1c6.png)
, ![(c) - = \left[\begin{array}{cc}4&-8\\1&5\\-23&-8\end{array}\right]](/tpl/images/0524/3141/039e1.png)
![and (d) = \left[\begin{array}{cc}14/6&-28/6\\10/6&4\\4/6&4\end{array}\right]](/tpl/images/0524/3141/98145.png)
![\left[\begin{array}{cc}2&-4\\1&3\\-5&0\end{array}\right]](/tpl/images/0524/3141/7c8df.png)
and B= ![\left[\begin{array}{cc}1&-2\\1&2\\4&4\end{array}\right]](/tpl/images/0524/3141/29eee.png)
![X= \frac{1}{3}(B-2A) = \frac{1}{3}(\left[\begin{array}{cc}1&-2\\1&2\\4&4\end{array}\right]-2\left[\begin{array}{cc}2&-4\\1&3\\-5&0\end{array}\right])](/tpl/images/0524/3141/2214c.png)
![= \frac{1}{3}(B-2A) = \frac{1}{3}(\left\left[\begin{array}{cc}1&-2\\1&2\\4&4\end{array}\right]-\left[\begin{array}{cc}4&-8\\2&6\\-10&0\end{array}\right])](/tpl/images/0524/3141/94127.png)
![= \frac{1}{3}(B-2A) = \frac{1}{3}\left[\begin{array}{cc}-3&6\\-1&-4\\14&4\end{array}\right] = \left[\begin{array}{cc}-1&2\\-1/3&-4/3\\14/3&4/3\end{array}\right]](/tpl/images/0524/3141/9c89f.png)
![X= \frac{1}{3}(2A-5B) = \frac{1}{3}(2\left[\begin{array}{cc}2&-4\\1&3\\-5&0\end{array}\right]-5\left[\begin{array}{cc}1&-2\\1&2\\4&4\end{array}\right])](/tpl/images/0524/3141/93179.png)
![\frac{1}{3}(\left[\begin{array}{cc}4&-8\\2&6\\-10&0\end{array}\right]-\left[\begin{array}{cc}5&-10\\5&10\\20&20\end{array}\right]) = \frac{1}{3}\left[\begin{array}{cc}-1&2\\-3&-4\\-30&-20\end{array}\right] = \left[\begin{array}{cc}-1/3&2/3\\-1&-4/3\\-3&-2\end{array}\right]](/tpl/images/0524/3141/64098.png)
![X= 3A-2) = 3\left[\begin{array}{cc}2&-4\\1&3\\-5&0\end{array}\right]-2\left[\begin{array}{cc}1&-2\\1&2\\4&4\end{array}\right] = \left[\begin{array}{cc}6&-12\\3&9\\-15&0\end{array}\right]-\left[\begin{array}{cc}2&-4\\2&4\\8&8\end{array}\right]](/tpl/images/0524/3141/b12b1.png)
![= \left[\begin{array}{cc}4&-8\\1&5\\-23&-8\end{array}\right]](/tpl/images/0524/3141/8dab6.png)
![X= \frac{1}{6}(4A+3B) = \frac{1}{6}(4\left[\begin{array}{cc}2&-4\\1&3\\-5&0\end{array}\right]+6\left[\begin{array}{cc}1&-2\\1&2\\4&4\end{array}\right])](/tpl/images/0524/3141/7f8e9.png)
![=\frac{1}{6} (\left[\begin{array}{cc}8&-16\\4&12\\-20&0\end{array}\right]+\left[\begin{array}{cc}6&-12\\6&12\\24&24\end{array}\right]) =](/tpl/images/0524/3141/3419e.png)
![=\frac{1}{6}\left[\begin{array}{cc}14&-28\\10&24\\4&24\end{array}\right] = \left[\begin{array}{cc}14/6&-28/6\\10/6&4\\4/6&4\end{array}\right]](/tpl/images/0524/3141/4a259.png)
Solve for X in the equation, given A = 2 1 −5 −4 −3 0 and B = 1 1 4 −2 2 4 .
(a) 3X + 2A...
Mathematics, 26.02.2020 02:25, mike2910
Solve for X in the equation, given A = 2 1 −5 −4 −3 0 and B = 1 1 4 −2 2 4 .
(a) 3X + 2A = B X = 5 5 5 5 5 5 Incorrect: Your answer is incorrect. seenKey [-1, -1/3; 14/3, 2; 8/3, 4/3]
(b) 2A − 5B = 3X X = 5 5 5 5 5 5 Incorrect: Your answer is incorrect. seenKey [-1/3, -1; -10, 2/3; -16/3, -20/3]
(c) X − 3A + 2B = O X = 5 5 5 5 5 5 Incorrect: Your answer is incorrect. seenKey [4, 1; -23, -8; -13, -8]
(d) 6X − 4A − 3B = O X = 5 5 5 5 5 5 Incorrect: Your answer is incorrect. seenKey [11/6, 7/6; -4/3, -11/3; -1, 2]
Answers: 2
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