# 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

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