Question

Find a solution to the system of equations by finding the reduced row-echelon form of the augmented matrix for the system of equations

Answer 1

Answer:

The following matrix when written in the form of Augmented Matrix

$\begin{bmatrix}4&-2&-5&-51\\4&-1&-2&-36\\4&-2&4&-24\end{bmatrix}\\\\R_{2}\to R_{2}- R_{1}\\\\ R_{3}\to R_{3}- R_{1}\\\\\begin{bmatrix}4&-2&-5&-51\\0&1&3&15\\0&0&9&27\end{bmatrix}\\\\R_{3}\to \frac{R_{3}}{9}\\\\R_{2}\to R_{2}- 3R_{3}\\\\\begin{bmatrix}4&-2&-5&-51\\0&1&0&6\\0&0& 1&3\end{bmatrix}\\\\R_{1}\to R_{1}+2 R_{2}\\\\\begin{bmatrix}4& 0&-5&-39\\0&1&0&6\\0&0&1&3\end{bmatrix}\\\\\\\\R_{1}\to R_{1}+5 R_{3}$

$\begin{bmatrix}4&0&0&-24\\0& 1& 0& 6\\0& 0& 1& 3\end{bmatrix}\\\\R_{1}\to \frac{R_{1}}{4}\\\\\begin{bmatrix} 1&0&0&-6\\0& 1& 0& 6\\0&0&1&3\end{bmatrix}\\\\x= -6, y=6,z=3$

Option A: x= -6,y=6 and z=3

Answer 2

Answer:

Option A. (-6,6,3)

Step-by-step explanation:

Please, see the attached file.

Thanks.