Solve system of equations using Cramer’s Rule.

$$\begin{gathered} 6x-5y+2z=3 \\ 2x+y-4z=5 \\ 3x-3y+z=-1 \end{gathered}$$

The system of linear equations is

$$\begin{gathered} 6x-5y+2z=3 \\ 2x+y-4z=5 \\ 3x-3y+z=-1 \end{gathered}$$

We will take the coefficients of the variables to form the determinant $D$

$D=\left|\begin{array}{ccc}6& -5& 2\\ 2& 1& -4\\ 3& -3& 1\end{array}\right|$

Evaluate it,

$$\begin{gathered} D=6\left[\right(1\left)\right(1)-(-4\left)\right(-3\left)\right]-(-5)\left[\right(2\left)\right(1)-(-4\left)\right(3\left)\right]+2\left[\right(2\left)\right(-3)-(1\left)\right(3\left)\right] \\ D=6[-11]+5\left[14\right]+2[-9] \\ D=-66+70-18 \\ D=-14 \end{gathered}$$

We will use the constants in place of the $x$coefficients to find the determinant $D_x$

$D_x=\left|\begin{array}{ccc}3& -5& 2\\ 5& 1& -4\\ -1& -3& 1\end{array}\right|$

Evaluate it,

$$\begin{gathered} D_x=3[1-12]-(-5)[5-4]+2[-15+1] \\ {D}_x=3[-11]+5\left[1\right]+2[-14] \\ {D}_x=-33+5-28 \\ {D}_x=-56 \end{gathered}$$

We will use the constants in place of the $y$coefficients to find the determinant $D_y$

$D_y=\left|\begin{array}{ccc}6& 3& 2\\ 2& 5& -4\\ 3& -1& 1\end{array}\right|$

Evaluate it,

$$\begin{gathered} D_y=6[5-4]-3[2+12]+2[-2-15] \\ {D}_y=6-42-34 \\ {D}_y=-70 \end{gathered}$$

We will use the constants in place of the $z$coefficients to find the determinant$D_z$

role="math" localid="1644492077015" $D_z=\left|\begin{array}{ccc}6& -5& 3\\ 2& 1& 5\\ 3& -3& -1\end{array}\right|$

Evaluate it,

$$\begin{gathered} D_z=6[-1+15]-(-5)[-2-15]+3[-6-3] \\ {D}_z=6\left[14\right]+5[-17]+3[-9] \\ {D}_z=84-85-27 \\ {D}_z=-28 \end{gathered}$$

To find $x$,

$$\begin{gathered} x=\frac{D_x}{D} \\ x=\frac{-56}{-14} \\ x=4 \end{gathered}$$

To find $y$,

$$\begin{gathered} y=\frac{D_y}{D} \\ y=\frac{-70}{-14} \\ y=5 \end{gathered}$$

To find $z$,

$$\begin{gathered} z=\frac{D_z}{D} \\ z=\frac{-28}{-14} \\ z=2 \end{gathered}$$

The solution for the system of equation is$(4,5,2)$

Substitute $x=4,y=5,z=2$in all the equations,

$$\begin{gathered} 6x-5y+2z=3 \\ 6\left(4\right)-5\left(5\right)+2\left(2\right)=3 \\ 24-25+4=3 \\ 3=3 \end{gathered}$$

This is true,

$$\begin{gathered} 2x+y-4z=5 \\ 2\left(4\right)+5-4\left(2\right)=5 \\ 8+5-8=5 \\ 5=5 \end{gathered}$$

This is also true,

$$\begin{gathered} 3x-3y+z=-1 \\ 3\left(4\right)-3\left(5\right)+2=-1 \\ 12-15+2=-1 \\ -1=-1 \end{gathered}$$

This is also true.

Therefore, the solution for the system of equation is $(4,5,2)$