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Solve each one of the following systems by Cramer's method:

$2 x+2 y+z=2$
$x+2 y-2 z=1$
$3 x-6 y+6 z=3$

$\Rightarrow A x=b \Rightarrow A=\left(\begin{array}{ccc}{2} & {2} & {1} \\ {1} & {2} & {-2} \\ {3} & {-6} & {6}\end{array}\right), x=\left(\begin{array}{c}{x} \\ {y} \\ {z}\end{array}\right), b=\left(\begin{array}{c}{2} \\ {1} \\ {3}\end{array}\right)$

$det(A)=\left|\begin{array}{ccc}{2} & {2} & {1} \\ {1} & {2} & {-2} \\ {3} & {-6} & {6}\end{array}\right|=-36 \neq 0$

$det\left(A_{1}\right)=\left|\begin{array}{ccc}{2} & {2} & {1} \\ {1} & {2} & {-2} \\ {3} & {-6} & {6}\end{array}\right|=-36$

$\Rightarrow x=\frac{det(A_{1})}{det(A)}=\frac{-36}{-36}=1$

$det\left(A_{2}\right)=\left|\begin{array}{lll}{2} & {2} & {1} \\ {1} & {1} & {-2} \\ {3} & {3} & {6}\end{array}\right|=0 \quad \Rightarrow \quad y=\frac{det(A_{2})}{det(A)}=\frac{0}{-36}=0$

$det\left(A_{3}\right)=\left|\begin{array}{lll}{2} & {2} & {2} \\ {1} & {2} & {1} \\ {3} & {6} & {3}\end{array}\right|=0 \Rightarrow z=\frac{det\left(A_{3}\right)}{det(A)}=\frac{0}{-36}=0$

$\left(\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right)=\left(\begin{array}{l}{1} \\ {0} \\ {0}\end{array}\right)$

Solve each one of the following systems by Cramer's method:

$x+y+z=3$
$2 x+3 y+3 z=7$
$4 x-3 y-2 z=3$

$A x=b, A=\left(\begin{array}{ccc}{1} & {1} & {1} \\ {2} & {3} & {3} \\ {4} & {-3} & {-2}\end{array}\right), x=\left(\begin{array}{l}{x} \\ {y} \\ {2}\end{array}\right), b=\left(\begin{array}{c}{3} \\ {7} \\ {3}\end{array}\right)$

$det(A)=\left|\begin{array}{ccc}{1} & {1} & {1} \\ {2} & {3} & {3} \\ {4} & {-3} & {-2}\end{array}\right|=1 \neq 0$

$det\left(A_{1}\right)=\left|\begin{array}{ccc}{3} & {1} & {1} \\ {7} & {3} & {3} \\ {3} & {-3} & {-2}\end{array}\right|=2 \Rightarrow x=\frac{det\left(A_{1}\right)}{det(A)}=\frac{2}{1}=2$

$det\left(A_{2}\right)=\left|\begin{array}{lll}{1} & {3} & {1} \\ {2} & {7} & {3} \\ {4} & {3} & {-2}\end{array}\right|=3 \quad \Rightarrow y=\frac{det\left(A_{2}\right)}{det(A)}=\frac{3}{1}=3$

$det\left(A_{3}\right)=\left|\begin{array}{ccc}{1} & {1} & {3} \\ {2} & {3} & {7} \\ {4} & {-3} & {3}\end{array}\right|=-2 \Rightarrow z=\frac{det(A_{3})}{det(A)}=\frac{-2}{1}=-2$

$\left(\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right)=\left(\begin{array}{r}{2} \\ {3} \\ {-2}\end{array}\right)$