Before we solve the system, let's verify if it has a solution, let's do the determinant test:
![\det \begin{bmatrix}{6} & {-4} & {-5} \\ {-1} & {-3} & {6} \\ {7} & {-6} & {-4}\end{bmatrix}=1](https://img.qammunity.org/2023/formulas/mathematics/college/rcczwjeybqhqz8owweqmmiomwzvozqrpto.png)
The determinant is 1, so it's different from zero, then the system has a unique solution.
Now let's determine the unique solution, I'll solve it using Cramer's Rule, because we already have the determinant of the coefficient matrix
Then
![\Delta=\mleft|\begin{matrix}6 & -4 & -5 \\ -1 & -3 & 6 \\ 7 & -6 & -4\end{matrix}\mright|=1](https://img.qammunity.org/2023/formulas/mathematics/college/vbth21ktmu44w01e4x8ty8bi9mvoxztcin.png)
Now let's evaluate the determinants with the coefficients
![\Delta_x=\mleft|\begin{matrix}2 & -4 & -5 \\ -2 & -3 & 6 \\ 4 & -6 & -4\end{matrix}\mright|=-88](https://img.qammunity.org/2023/formulas/mathematics/college/cdfqa5j1gv5m112t3rr68hli4h9s5vhg4e.png)
Now the variable y
![\Delta_y=\mleft|\begin{matrix}6 & 2 & -5 \\ -1 & -2 & 6 \\ 7 & 4 & -4\end{matrix}\mright|=-70](https://img.qammunity.org/2023/formulas/mathematics/college/qvqhqohn9uvl720set2afta9efp5696nz2.png)
And the last one
![\Delta_z=\mleft|\begin{matrix}6 & -4 & 2 \\ -1 & -3 & -2 \\ 7 & -6 & 4\end{matrix}\mright|=-50](https://img.qammunity.org/2023/formulas/mathematics/college/6g9rjx995v2fb8xih4rj1gck71vmx8to58.png)
The solution will be
![\begin{gathered} x=(\Delta_x)/(\Delta)=(-88)/(1)=-88 \\ \\ y=(\Delta_y)/(\Delta)=(-70)/(1)=-70 \\ \\ z=(\Delta_z)/(\Delta)=(-50)/(1)=-50 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/kwtqhaym0w4b8acq6ao5c8us2aa4pz6qt9.png)
Final answer:
Only one solution
x = -88
y = -70
z = -50