Answer :
Title:
The answer is (7, [tex]\frac{-2}{9}[/tex], [tex]\frac{10}{3}[/tex]).
Step-by-step explanation:
The system of equations is given by, 2x+y+4z=16; 5x-2y+2z=-1 and x+2y-3z=-9.
The augmented matrix is [tex]\left[\begin{array}{cccc}2&1&4&16\\5&-2&-2&1\\1&2&-3&-9\end{array}\right][/tex].
Multiplying the first row with [tex]\frac{1}{2}[/tex] we get, [tex]\left[\begin{array}{cccc}1&\frac{1}{2} &2&8\\5&-2&-2&1\\1&2&-3&-9\end{array}\right][/tex].
Subtracting the 5 times of the first row from second row we get, [tex]\left[\begin{array}{cccc}1&\frac{1}{2} &2&8\\0&\frac{-9}{2} &-12&-39\\1&2&-3&-9\end{array}\right][/tex].
Multiplying the second row by [tex]\frac{-2}{9}[/tex], we get [tex]\left[\begin{array}{cccc}1&\frac{1}{2} &2&8\\0&1 &\frac{8}{3} &\frac{26}{3} \\1&2&-3&-9\end{array}\right][/tex].
Subtracting the first row from the third row we get, [tex]\left[\begin{array}{cccc}1&\frac{1}{2} &2&8\\0&1 &\frac{8}{3} &\frac{26}{3} \\0&\frac{3}{2} &-5&-17\end{array}\right][/tex].
Subtracting Half of the second row from the first row and one and half of the second row from the third row, we get [tex]\left[\begin{array}{cccc}1&0&\frac{-2}{3} &\frac{11}{3} \\0&1 &\frac{8}{3} &\frac{26}{3} \\0&0&-9&-30\end{array}\right][/tex].
Dividing the third row by -9 we get, [tex]\left[\begin{array}{cccc}1&0&\frac{-2}{3} &\frac{11}{3} \\0&1 &\frac{8}{3} &\frac{26}{3} \\0&0&1&\frac{10}{3} \end{array}\right][/tex].
Adding two-third of the third row with the first row and subtracting [tex]\frac{8}{3}[/tex] of the third row from the second row, we get, [tex]\left[\begin{array}{cccc}1&0&0 &7 \\0&1 &0&\frac{-2}{9} \\0&0&1&\frac{10}{3} \end{array}\right][/tex].
The solution of this system of linear equations is [tex](x,y,z) = (1,2,4)[/tex]. (Correct choice: C)
How to solve a system of equations by augmented matrix
The system of equations is represented by the following augmented matrix:
[tex]\left[\begin{array}{cccc}2&1&4&16\\5&-2&2&-1\\1&2&-3& -9\end{array}\right][/tex] (1)
Where the fourth column represents the vector column of independent coefficients. After making some operations, we find that the reduced row-echelon form of the augmented matrix is:
[tex]\left[\begin{array}{cccc}1&0&0&1\\0&1&0&2\\0&0&1& 4\end{array}\right][/tex]
The solution of this system of linear equations is [tex](x,y,z) = (1,2,4)[/tex]. (Correct choice: C) [tex]\blacksquare[/tex]
To learn more on augmented matrices, we kindly invite to check this verified question: https://brainly.com/question/4945755