The Rouché-Capelli Theorem. This theorem establishes a connection between how a linear system behaves and the ranks of its coefficient matrix (A) and its counterpart the augmented matrix.

$rank(A)=rank\left ( \left [ A|B \right ] \right )\:and\:n=rank(A)$

Then satisfying this theorem the system is consistent and has one single solution.

Explanation:

1) To answer that, you should have to know The Rouché-Capelli Theorem. This theorem establishes a connection between how a linear system behaves and the ranks of its coefficient matrix (A) and its counterpart the augmented matrix.

$rank(A)=rank\left ( \left [ A|B \right ] \right )\:and\:n=rank(A)$

$rank(A)$

Then the system is consistent and has a unique solution.

$\left\{\begin{matrix}x-3y-2z=6 \\ 2x-4y-3z=8 \\ -3x+6y+8z=-5 \end{matrix}\right.$

2) Writing it as Linear system

$A=\begin{pmatrix}1 & -3 &-2 \\ 2& -4 &-3 \\ -3 &6 &8 \end{pmatrix}$ $B=\begin{pmatrix}6\\ 8\\ 5\end{pmatrix}$

$rank(A) =\left(\begin{matrix}7 & 0 & 0 \\0 & 7 & 0 \\0 & 0 & 7\end{matrix}\right)=3$

3) The Rank (A) is 3 found through Gauss elimination

$(A|B)=\begin{pmatrix}1 & -3 &-2 &6 \\ 2& -4 &-3 &8 \\ -3&6 &8 &-5 \end{pmatrix}$

$rank(A|B)=\left(\begin{matrix}1 & -3 & -2 \\0 & 2 & 1 \\0 & 0 & \frac{7}{2}\end{matrix}\right)=3$

4) The rank of (A|B) is also equal to 3, found through Gauss elimination:

So this linear system is consistent and has a unique solution.

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