Question Solved1 Answer The reduced row-echelon form of the augmented matrix for a system of linear equations with variables x₁, x6 is given below. Determine the solutions for the system and enter them below. 10003 2 1 0 1 0 0 -2 3 0 0 0 1 0 4 -5 3 0 0 0 1 1 0 -2 If the system has infinitely many solutions, select "The system has at least one solution". Your answer may use The reduced row-echelon form of the augmented matrix for a system of linear equations with variables x₁, x6 is given below. Determine the solutions for the system and enter them below. 10003 2 1 0 1 0 0 -2 3 0 0 0 1 0 4 -5 3 0 0 0 1 1 0 -2 If the system has infinitely many solutions, select "The system has at least one solution". Your answer may use expressions involving the parameters r, s, and t. The system has at least one solution x1 = 0 x2 = 0 x3 = 0 x4 = 0 X5 = 0 x6 = 0

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Transcribed Image Text: The reduced row-echelon form of the augmented matrix for a system of linear equations with variables x₁, x6 is given below. Determine the solutions for the system and enter them below. 10003 2 1 0 1 0 0 -2 3 0 0 0 1 0 4 -5 3 0 0 0 1 1 0 -2 If the system has infinitely many solutions, select "The system has at least one solution". Your answer may use expressions involving the parameters r, s, and t. The system has at least one solution x1 = 0 x2 = 0 x3 = 0 x4 = 0 X5 = 0 x6 = 0
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Transcribed Image Text: The reduced row-echelon form of the augmented matrix for a system of linear equations with variables x₁, x6 is given below. Determine the solutions for the system and enter them below. 10003 2 1 0 1 0 0 -2 3 0 0 0 1 0 4 -5 3 0 0 0 1 1 0 -2 If the system has infinitely many solutions, select "The system has at least one solution". Your answer may use expressions involving the parameters r, s, and t. The system has at least one solution x1 = 0 x2 = 0 x3 = 0 x4 = 0 X5 = 0 x6 = 0
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The Rove reduad materix,{:[[[1,0,0,0,3,2,1],[0,1,0,0,-2,3,0],[0,0,1,0,4,-5,3],[0,0,0,1,1,0,-2]]],[" Variables "x_(1)","x_(2)","x_(3)","x_(4)","x_(5)","x_(6)]:}The system of Equation becomes:{:[x_(1)+3x_(5)+2x_(6)=1],[x_(2)-2x_(5)+3x_(6)=0],[x_(3)+4x_(5)-5x_(6)=3],[x_(4)+4x_(5)=-2],[[[x_(1)],[x_(2)],[x_(3)],[x_(4)],[x_(5)],[x_(6)]]=[[x_(2)=2x_(5)-3x_(6)],[x_(3)=3-4x_(5)+5x_ ... See the full answer