Can you please check my answer for me ?(:

Mathematics

2 answers:

Alexandra [31]4 years ago

8 0

Dont doubt yourself you did good!

Vinvika [58]4 years ago

7 0

I don't believe you would shade anywhere, so your work is correct :D

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Question 5 (5 points)

Tpy6a [65]

Answer:

The solution of the system of equations is, (1,-1,2)

Step-by-step explanation:

Given system equation;

x + 5y - 3z = -10

-5x + 6y – 5z = -21

-x + 8y - 8z = -25

Matrix form is written as;

$\left[\begin{array}{ccc}1&5&-3\\-5&6&-5\\-1&8&-8\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}-10\\-21\\-25\end{array}\right] \\\\\\det. = 1\left[\begin{array}{cc}\\6&-5\\8&-8\end{array}\right] -5\left[\begin{array}{cc}\\-5&-5\\-1&-8\end{array}\right] -3\left[\begin{array}{cc}\\-5&6\\-1&8\end{array}\right] \\\\\\det. = 1(-8) -5(35)-3(-34)= -8 - 175+ 102 = -81$

Cofactor;

$First \ row \left[\begin{array}{cc}+\\ 6&-5\\\ 8&-8\end{array}\right \left\begin{array}{cc}-\\ -5&-5\\-1&-8\end{array}\right \left\begin{array}{cc}+\\-5&6\\-1&8\end{array}\right] = [-8 \ \ -35 \ \ -34]\\\\\\\ Second \ row \left[\begin{array}{cc}-\\ 5&-3\\\ 8&-8\end{array}\right \left\begin{array}{cc}+\\ 1&-3\\-1&-8\end{array}\right \left\begin{array}{cc}-\\1&5\\-1&8\end{array}\right] = [16\ \ -11 \ \ -13]\\\\\\$

$Third \ row \left[\begin{array}{cc}+\\ 5&-3\\\ 6&-5\end{array}\right \left\begin{array}{cc}-\\ 1&-3\\-5&-5\end{array}\right \left\begin{array}{cc}+\\1&5\\-5&6\end{array}\right]= [-7 \ \ 20\ \ 31]$

$Cofactor =$ $\left[\begin{array}{ccc}-8&-35&-34\\16&-11&-13\\-7&20&31\end{array}\right]$

$inverse \ matrix =-\frac{1}{81} \left[\begin{array}{ccc}-8&16&-7\\-35&-11&20\\-34&-13&31\end{array}\right] \\\\\\$

Solution of the matrix:

$\left[\begin{array}{c}x\\y\\z\end{array}\right] = -\frac{1}{81} \left[\begin{array}{ccc}-8&16&-7\\-35&-11&20\\-34&-13&31\end{array}\right] X \left[\begin{array}{c}-10\\-21\\-25\end{array}\right] = \left[\begin{array}{c}\frac{-8*-10 }{-81 } +\frac{16*-21 }{-81 } + \frac{-7*-25 }{-81 }\\\\\frac{-35*-10 }{-81 } +\frac{-11*-21 }{-81 }+ \frac{20*-25 }{-81 }\\\\\frac{-34*-10 }{-81 }+ \frac{-13*-21 }{-81 }+ \frac{31*-25 }{-81 }\end{array}\right] \\\\\$

$\left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}\frac{-81}{-81} \\\\\frac{81}{-81} \\\\\frac{-162}{-81} \end{array}\right] = \left[\begin{array}{c}1\\-1\\2\end{array}\right]$