Solution :

We write the the given sets of equations in AX=B form

$ \begin{bmatrix} 1&2&-1\\ 1&8&-3 \\ 2&1&-3\\ \end{bmatrix} \begin{bmatrix} x\\y\\z\\ \end{bmatrix}=\begin{bmatrix} 5\\1\\7\\ \end{bmatrix} $

Here, we have augmented matrix

$ C=[A:B] ~ \begin{bmatrix} 1&2&-1&:5\\ 1&8&-3&:1\\ 2&1&-3&:7\\ \end{bmatrix} $

We reduce the augmented matrix into triangular matrix,

$ C=[A:B] ~ \begin{bmatrix} 1&2&-1&:5\\ 1&8&-3&:1\\ 2&1&-3&:7\\ \end{bmatrix} $

$R_2 \rightarrow R_1-R_2 , \space R_3 \rightarrow 2R_1-R_3 $

$ C=[A:B] ~ \begin{bmatrix} 1&2&-1&:5\\ 0&-6&2&:4\\ 0&0&4&:10\\ \end{bmatrix} $

Number of non-zero rows = Rank of matrix

R(C) = R(A) = 3

Hence, the given system is consistent and possesses a unique solution.

In matrix form the system reduces to

$ \begin{bmatrix} 1&2&-1\\ 0&-6&2 \\ 0&0&4\\ \end{bmatrix} \begin{bmatrix} x\\y\\z\\ \end{bmatrix}=\begin{bmatrix} 5\\4\\10\\ \end{bmatrix} \\ $ $$ \begin {aligned} x+2y-z &=5, \\ -6y+2z &=4, \\ 4z &=10 \end{aligned} \\ $$

Solving above equations, we get

$$ x= \frac{43}{6} ,y=\frac{5}{2} , z=\frac{5}{2} $$