Question

( left[begin{array}{cccc}2 & -1 & 3 & 9 1 & 1 & 1 & 6 1 & -1 & 1 & 2end{array}right] )
( R_{3} rightarrow R_{3}-R_{2} )
( left[begin{array}{cccc}2 & -1 & 3 & 9 1 & 1 & 1 & 6 0 & -2 & 0 & -4end{array}right] )
( R_{2} rightarrow R_{1}-2 R_{2} )
( left[begin{array}{cccc}2 & -1 & 3 & 9 0 & 3 & -1 & 3 0 & -2 & 0 & -4end{array}right] times 2 )
( sumleft[begin{array}{cccc}2 & -1 & 3 & 9 0 & 6 & -2 & 6 0 & -6 & 0 & -12end{array}right] )
( R_{3} rightarrow R_{3}+R_{2} )
( left[begin{array}{cccc}2 & -1 & 3 & 9 0 & 6 & -2 & 6 0 & 0 & -2 & -6end{array}right] )
( therefore quad 27=6 Rightarrow z=3 )
( 6 x y-z z=6 )
( =b+b=12 Rightarrow y=2 )
( x-y+3 z=9 rightarrow 2 x- )

# Ch. In this case, system of equations have unique solution. i.e., X = 3, y = 1, z = 1. The solution of the given system of equations is x = 3, y = 1, z = 1. 3 Solve the equations 2x - y + 3z = 9, x + y + z = 6, x - y + z = 2 by Gauss - Jordan method. Mar. '18 (AP); Mar. '14, '10; May '14 - The given system of linear equations are 2x – y + 32 = 9; x + y + z = 6; x - y + z = 2. To 27 Tot 507

Solution