Answer:
D
Step-by-step explanation:
The augmented matrix for the system of three equaitons is
[tex]\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\5&2&-2&|&11\\5&-4&4&|&-7\end{array}\right)[/tex]
Multiply the first row by 5, the second row by -3 and add these two rows:
[tex]\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\5&-4&4&|&-7\end{array}\right)[/tex]
Subtract the third row from the second:
[tex]\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\0&6&-6&|&18\end{array}\right)[/tex]
Divide the third row by 6:
[tex]\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\0&1&-1&|&3\end{array}\right)[/tex]
Now multiply the third equation by 26 and add it to the second row:
[tex]\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\0&0&-45&|&-90\end{array}\right)[/tex]
You get the system of three equations:
[tex]\left\{\begin{array}{r}3x-4y-5z=-27\\-26y-19z=-168\\-45z=-90\end{array}\right.[/tex]
From the third equation
[tex]z=\dfrac{90}{45}=2.[/tex]
Substitute z=2 into the second equation:
[tex]-26y-19\cdot 2=-168\\ \\-26y-38=-168\\ \\-26y=-168+38=-130\\ \\y=\dfrac{130}{26}=5.[/tex]
Now substitute z=2 and y=5 into the first equation:
[tex]3x-4\cdot 5-5\cdot 2=-27\\ \\3x-20-10=-27\\ \\3x-30=-27\\ \\3x=-27+30=3\\ \\x=1.[/tex]
The solution is (1,5,2)
