How do you solve the system $3a-3b+4c=-23$ , $a+2b-3c=25$ , and $4a-b+c=25$ ?

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Answer 1 · 2017-08-06T08:18:13

The solution is $S = O/$

We perform the Gauss Jordan elimination with the augmented matrix

$((1,2,-3,:,25),(3,-3,4,:,-23),(4,-1,1,:,25))$

$R3larrR3-4R1$ , $=>$ , $((1,2,-3,:,25),(3,-3,4,:,-23),(0,-9,13,:,-75))$

$R2larrR2-3R1$ , $=>$ , $((1,2,-3,:,25),(0,-9,13,:,-98),(0,-9,13,:,-75))$

$R3larrR3-R2$ , $=>$ , $((1,2,-3,:,25),(0,-9,13,:,-98),(0,0,0,:,23))$

$R2larr(R2)/(-9)$ , $=>$ , $((1,2,-3,:,25),(0,1,-13/9,:,98/9),(0,0,0,:,23))$

$R1larrR1-2R2$ , $=>$ , $((1,0,-1/9,:,29/9),(0,1,-13/9,:,98/9),(0,0,0,:,23))$

The system of equations is inconsistent.

How do you solve the system 3a-3b+4c=-233a-3b+4c=-23, a+2b-3c=25a+2b-3c=25, and 4a-b+c=254a-b+c=25?