Before we solve the system, let's verify if it has a solution, let's do the determinant test:
[tex]\det \begin{bmatrix}{6} & {-4} & {-5} \\ {-1} & {-3} & {6} \\ {7} & {-6} & {-4}\end{bmatrix}=1[/tex]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
[tex]\Delta=\mleft|\begin{matrix}6 & -4 & -5 \\ -1 & -3 & 6 \\ 7 & -6 & -4\end{matrix}\mright|=1[/tex]Now let's evaluate the determinants with the coefficients
[tex]\Delta_x=\mleft|\begin{matrix}2 & -4 & -5 \\ -2 & -3 & 6 \\ 4 & -6 & -4\end{matrix}\mright|=-88[/tex]Now the variable y
[tex]\Delta_y=\mleft|\begin{matrix}6 & 2 & -5 \\ -1 & -2 & 6 \\ 7 & 4 & -4\end{matrix}\mright|=-70[/tex]And the last one
[tex]\Delta_z=\mleft|\begin{matrix}6 & -4 & 2 \\ -1 & -3 & -2 \\ 7 & -6 & 4\end{matrix}\mright|=-50[/tex]The solution will be
[tex]\begin{gathered} x=\frac{\Delta_x}{\Delta}=\frac{-88}{1}=-88 \\ \\ y=\frac{\Delta_y}{\Delta}=\frac{-70}{1}=-70 \\ \\ z=\frac{\Delta_z}{\Delta}=\frac{-50}{1}=-50 \end{gathered}[/tex]Final answer:
Only one solution
x = -88
y = -70
z = -50
