Question

Solve the system of equations by finding the reduced row-echelon form of the augmented matrix for the system of equations. 2x+y+4z=16 5x-2y+2z=-1 x+2y-3z=-9 A(-1, 2, 22) B(-10, 22, 42) C(-1, 2, 4) D(-10, 2, 22) Please select the best answer from the choices provided A B C D

Answer 1

Answer:

D. (3, 7, -3)

Step-by-step explanation:

Answer 2

Title:

The answer is (7, $\frac{-2}{9}$, $\frac{10}{3}$).

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 $\left[\begin{array}{cccc}2&1&4&16\\5&-2&-2&1\\1&2&-3&-9\end{array}\right]$.

Multiplying the first row with $\frac{1}{2}$ we get, $\left[\begin{array}{cccc}1&\frac{1}{2} &2&8\\5&-2&-2&1\\1&2&-3&-9\end{array}\right]$.

Subtracting the 5 times of the first row from second row we get, $\left[\begin{array}{cccc}1&\frac{1}{2} &2&8\\0&\frac{-9}{2} &-12&-39\\1&2&-3&-9\end{array}\right]$.

Multiplying the second row by $\frac{-2}{9}$, we get $\left[\begin{array}{cccc}1&\frac{1}{2} &2&8\\0&1 &\frac{8}{3} &\frac{26}{3} \\1&2&-3&-9\end{array}\right]$.

Subtracting the first row from the third row we get, $\left[\begin{array}{cccc}1&\frac{1}{2} &2&8\\0&1 &\frac{8}{3} &\frac{26}{3} \\0&\frac{3}{2} &-5&-17\end{array}\right]$.

Subtracting Half of the second row from the first row and one and half of the second row from the third row, we get $\left[\begin{array}{cccc}1&0&\frac{-2}{3} &\frac{11}{3} \\0&1 &\frac{8}{3} &\frac{26}{3} \\0&0&-9&-30\end{array}\right]$.

Dividing the third row by -9 we get, $\left[\begin{array}{cccc}1&0&\frac{-2}{3} &\frac{11}{3} \\0&1 &\frac{8}{3} &\frac{26}{3} \\0&0&1&\frac{10}{3} \end{array}\right]$.

Adding two-third of the third row with the first row and subtracting $\frac{8}{3}$ of the third row from the second row, we get, $\left[\begin{array}{cccc}1&0&0 &7 \\0&1 &0&\frac{-2}{9} \\0&0&1&\frac{10}{3} \end{array}\right]$.