# Solve the system of equations x_1−3x_2+4x_3=-4 3x_1−7x_2+7x3=-8 −4x_1+6x_2−x_3=7

Solve the system of equations ${x}_{1}-3{x}_{2}+4{x}_{3}=-4$
$3{x}_{1}-7{x}_{2}+7x3=-8$
$-4{x}_{1}+6{x}_{2}-{x}_{3}=7$
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cyhuddwyr9

Given:
${x}_{1}-3{x}_{2}+4{x}_{3}=-4$
$3{x}_{1}-7{x}_{2}+7x3=-8$
$-4{x}_{1}+6{x}_{2}-{x}_{3}=7$
Step 1
Find the value of ${\delta }_{1}$ & ${\delta }_{2}$
${\delta }_{1}=\left[\begin{array}{ccc}-4& -3& 4\\ 3& -7& 7\\ -4& 6& -1\end{array}\right]$
=1(7-42)+3(-3+28)+4(18-28)
=-35+75-40
=0
Step 3
${\delta }_{1}=\left[\begin{array}{ccc}-4& -3& 4\\ -8& -7& 7\\ 7& 6& -1\end{array}\right]$
=-4(7-42)+3(8-49)+(-48+49)
=140-123
$=21\ne 0$
according to cramers rule if value of is zero and if any one of is non zero then the system of linear equations has no solution.
therefore the given system of linear equations has no solution