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

Question
Equations
asked 2021-02-05
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

Answers (1)

2021-02-06
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.
\(\displaystyle{D}={\left[\begin{array}{ccc} {4}&-{1}&-{1}\\{2}&{2}&{3}\\{5}&-{2}&-{2}\end{array}\right]}={4}{\left|\begin{array}{cc} {2}&{3}\\-{2}&-{2}\end{array}\right|}-{\left(-{1}\right)}{\left|\begin{array}{cc} {2}&{3}\\{5}&-{2}\end{array}\right|}-{1}{\left|\begin{array}{cc} {2}&{2}\\{5}&-{2}\end{array}\right|}\)
D=4(-4+6)+(-4-15)-(-4-10)=8-19+14=3
Step 2
Now, we will find the following determinants.
\(\displaystyle{D}_{{1}}={\left[\begin{array}{ccc} {1}&-{1}&-{1}\\{10}&{2}&{3}\\-{1}&-{2}&-{2}\end{array}\right]}={1}{\left|\begin{array}{cc} {2}&{3}\\-{2}&-{2}\end{array}\right|}-{\left(-{1}\right)}{\left|\begin{array}{cc} {10}&{3}\\-{1}&-{2}\end{array}\right|}-{1}{\left|\begin{array}{cc} {10}&{2}\\-{1}&-{2}\end{array}\right|}\)
\(\displaystyle{D}_{{1}}={\left(-{4}+{6}\right)}+{\left(-{20}+{3}\right)}-{\left(-{20}+{2}\right)}={2}-{17}+{18}={3}\)
\(\displaystyle{D}_{{2}}={\left[\begin{array}{ccc} {4}&{1}&-{1}\\{2}&{10}&{3}\\{5}&-{1}&-{2}\end{array}\right]}={4}{\left|\begin{array}{cc} {10}&{3}\\-{1}&-{2}\end{array}\right|}-{1}{\left|\begin{array}{cc} {2}&{3}\\{5}&-{2}\end{array}\right|}-{1}{\left|\begin{array}{cc} {2}&{10}\\{5}&-{1}\end{array}\right|}\)
\(\displaystyle{D}_{{2}}={4}{\left(-{20}+{3}\right)}-{\left(-{4}-{15}\right)}-{\left(-{2}-{50}\right)}=-{68}+{19}+{52}={3}\)
\(\displaystyle{D}_{{3}}={\left[\begin{array}{ccc} {4}&-{1}&{1}\\{2}&{2}&{10}\\{5}&-{2}&-{1}\end{array}\right]}={4}{\left|\begin{array}{cc} {2}&{10}\\-{2}&-{1}\end{array}\right|}-{\left(-{1}\right)}{\left|\begin{array}{cc} {2}&{10}\\{5}&-{1}\end{array}\right|}+{1}{\left|\begin{array}{cc} {2}&{2}\\{5}&-{2}\end{array}\right|}\)
\(\displaystyle{D}_{{3}}={4}{\left(-{2}+{20}\right)}+{\left(-{2}-{50}\right)}+{\left(-{4}-{10}\right)}={6}\)
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
0

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