Respuesta :
Answer:
[tex](a)= \left[\begin{array}{cc}-1&2\\-1/3&-4/3\\14/3&4/3\end{array}\right][/tex] ,[tex](b) = \left[\begin{array}{cc}-1/3&2/3\\-1&-4/3\\-3&-2\end{array}\right][/tex] , [tex](c) - = \left[\begin{array}{cc}4&-8\\1&5\\-23&-8\end{array}\right][/tex] [tex]and (d) = \left[\begin{array}{cc}14/6&-28/6\\10/6&4\\4/6&4\end{array}\right][/tex]
Step-by-step explanation:
Given that A= [tex]\left[\begin{array}{cc}2&-4\\1&3\\-5&0\end{array}\right][/tex] and B= [tex]\left[\begin{array}{cc}1&-2\\1&2\\4&4\end{array}\right][/tex]
The following are the major concepts applied.
I. Scalar Multiplication- You multiply each element by the number outside
II. Matrix Subtraction and Addition- You subtract/add as may be required elements in the same position on the matrix.
(a) In 3X+2A=B
3X=B-2A
[tex]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])[/tex]
[tex]= \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])[/tex]
[tex]= \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][/tex]
(b) 2A-5B=3X
[tex]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])[/tex]
This equals to:
[tex]\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][/tex]
(c) X-3A+2B=0
Making X the subject f the formula
X=3A-2B
[tex]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][/tex]
On Subtraction:
[tex]= \left[\begin{array}{cc}4&-8\\1&5\\-23&-8\end{array}\right][/tex]
(d) 6X − 4A − 3B = 0
[tex]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])[/tex]
[tex]=\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]) =[/tex][tex]=\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][/tex]
