Use Cramer’s Rule to solve (if possible) the system of linear equations. 4x-y-z=1 2x+2y+3z=10 5x-2y-2z=-1

Step 1
Given system of equations are
4x-y-z=1
2x+2y+3z=10
5x-2y-2z=-1
Firstly, we will find determinant of coefficient matrix.
${\displaystyle D=\left[\begin{array}{ccc}4& -1& -1\\ 2& 2& 3\\ 5& -2& -2\end{array}\right]=4\left|\begin{array}{cc}2& 3\\ -2& -2\end{array}\right|-(-1)\left|\begin{array}{cc}2& 3\\ 5& -2\end{array}\right|-1\left|\begin{array}{cc}2& 2\\ 5& -2\end{array}\right|}$
D=4(-4+6)+(-4-15)-(-4-10)=8-19+14=3
Step 2
Now, we will find the following determinants.
${\displaystyle D_1=\left[\begin{array}{ccc}1& -1& -1\\ 10& 2& 3\\ -1& -2& -2\end{array}\right]=1\left|\begin{array}{cc}2& 3\\ -2& -2\end{array}\right|-(-1)\left|\begin{array}{cc}10& 3\\ -1& -2\end{array}\right|-1\left|\begin{array}{cc}10& 2\\ -1& -2\end{array}\right|}$
${\displaystyle D_1=(-4+6)+(-20+3)-(-20+2)=2-17+18=3}$
${\displaystyle D_2=\left[\begin{array}{ccc}4& 1& -1\\ 2& 10& 3\\ 5& -1& -2\end{array}\right]=4\left|\begin{array}{cc}10& 3\\ -1& -2\end{array}\right|-1\left|\begin{array}{cc}2& 3\\ 5& -2\end{array}\right|-1\left|\begin{array}{cc}2& 10\\ 5& -1\end{array}\right|}$
${\displaystyle D_2=4(-20+3)-(-4-15)-(-2-50)=-68+19+52=3}$
${\displaystyle D_3=\left[\begin{array}{ccc}4& -1& 1\\ 2& 2& 10\\ 5& -2& -1\end{array}\right]=4\left|\begin{array}{cc}2& 10\\ -2& -1\end{array}\right|-(-1)\left|\begin{array}{cc}2& 10\\ 5& -1\end{array}\right|+1\left|\begin{array}{cc}2& 2\\ 5& -2\end{array}\right|}$
${\displaystyle D_3=4(-2+20)+(-2-50)+(-4-10)=6}$
Now, solution of given system of equations is
${\displaystyle x=\frac{D_1}{D},y=\frac{D_2}{D},z=\frac{D_3}{D}}$
x=1,y=1,z=2
Step 3
Result: Solution of given system of equations is
x=1, y=1, z=2