## Intermediate Algebra for College Students (7th Edition)

No solution or $\varnothing$.
The given system of equations is $3x-y+2z=4$ $-6x+2y-4z=1$ $5x-3y+8z=0$ The augmented matrix is $\Rightarrow \left[\begin{array}{ccc|c} 3 & -1 & 2& 4\\ -6 & 2 & -4& 1 \\ 5&-3&8&0 \end{array}\right]$ Perform $R_1\rightarrow R_1/3$. $\Rightarrow \left[\begin{array}{ccc|c} 3/3 & -1/3 & 2/3& 4/3\\ -6 & 2 & -4& 1 \\ 5&-3&8&0 \end{array}\right]$ Simplify. $\Rightarrow \left[\begin{array}{ccc|c} 1 & -1/3 & 2/3& 4/3\\ -6 & 2 & -4& 1 \\ 5&-3&8&0 \end{array}\right]$ Perform $R_2\rightarrow R_2+6R_1$ and $R_3\rightarrow R_3-5 R_1$. $\Rightarrow \left[\begin{array}{ccc|c} 1 & -1/3 & 2/3& 4/3\\ -6+6(1) & 2+6(-1/3) & -4+6(2/3)& 1+6(4/3) \\ 5-5(1)&-3-5(-1/3)&8-5(2/3)&0-5(4/3) \end{array}\right]$ Simplify. $\Rightarrow \left[\begin{array}{ccc|c} 1 & -1/3 & 2/3& 4/3\\ 0 & 0 & 0& 9 \\ 0&-4/3&14/3&-20/3 \end{array}\right]$ Use back substitution for the second row. $x(0)+y(0)+z(0)=9$ There are no values of $x,y$ and $z$ for which the above equation satisfy. Hence, the system is inconsistent and has no solution.