Question

solve the system of equations by finding the reduced row-echelon form of the augmented matrix for the system of equations. x-2y+3z=-2. 6x+2y+2z=-48. x+4y+3z=-38​

Answer 1

Answer:

x = -5, y = -6, z = -3

Step-by-step explanation:

Given the system of three equations:

$\left\{\begin{array}{l}x-2y+3z=-2\\6x+2y+2z=-48\\x+4y+3z=-38\end{array}\right.$

Write the augmented matrix for the system of equations

$\left(\begin{array}{ccccc}1&-2&3&|&-2\\6&2&2&|&-48\\1&4&3&|&-38\end{array}\right)$

Find the reduced row-echelon form of the augmented matrix for the system of equations:

$\left(\begin{array}{ccccc}1&-2&3&|&-2\\6&2&2&|&-48\\1&4&3&|&-38\end{array}\right)\sim \left(\begin{array}{ccccc}1&-2&3&|&-2\\0&-14&16&|&36\\0&-6&0&|&36\end{array}\right)\sim \left(\begin{array}{ccccc}1&3&-2&|&-2\\0&16&-14&|&36\\0&0&-6&|&36\end{array}\right)$

Thus, the system of three equations is

$\left\{\begin{array}{r}x+3z-2y=-2\\16z-14y=36\\-6y=36\end{array}\right.$

From the last equation:

$y=-6$

Substitute it into the second equation:

$16z-14\cdot (-6)=36\\ \\16z=36-84\\ \\16z=-48\\ \\z=-3$

Substitute y = -6 and z = -3 into the first equation:

$x+3\cdot (-3)-2\cdot (-6)=-2\\ \\x=-2+9-12\\ \\x=-5$