SOLUTION: III. Given the linear system 2x1 + 3x2 - x3 = 0 x1 - 4x2 + 5x3 = 0 (a) Verify that x1 = 1, x2 = -1, x3 = - 1 is a solution. (b) Verify that x1 = -2, x2 = 2, x3 = 2 is a so

Algebra ->  Matrices-and-determiminant -> SOLUTION: III. Given the linear system 2x1 + 3x2 - x3 = 0 x1 - 4x2 + 5x3 = 0 (a) Verify that x1 = 1, x2 = -1, x3 = - 1 is a solution. (b) Verify that x1 = -2, x2 = 2, x3 = 2 is a so      Log On


   



Question 1152851: III. Given the linear system
2x1 + 3x2 - x3 = 0
x1 - 4x2 + 5x3 = 0
(a) Verify that x1 = 1, x2 = -1, x3 = - 1 is a solution.
(b) Verify that x1 = -2, x2 = 2, x3 = 2 is a solution.
(c) Adding the corresponding x-values of the solution in parts (a) and (b) gives x1 = -1, x2 = 1, x3 = 1. Is this a solution to the linear system?
(d) Multiply each of the x-values in part (a) by 3. Are the resulting values a solution to the linear system?

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!


2x%5B1%5D+%2B+3x%5B2%5D+-+x%5B3%5D+=+0
x%5B1%5D+-+4x%5B2%5D+%2B+5x%5B3%5D+=+0
(a) Verify that x%5B1%5D+=+1, x%5B2+%5D=+-1, x%5B3%5D+=+-+1 is a solution.
2x%5B1%5D+%2B+3x%5B2%5D+-+x%5B3%5D+=+0......substitute given values
2%2A1+%2B+3%28-1%29+-+%28-1%29=+0
2-3+%2B1=+0
3-3+=+0
0=0->true
and
x%5B1%5D+-+4x%5B2%5D+%2B+5x%5B3%5D+=+0
1+-+4%28-1%29++%2B+5%28-1%29+=+0
1+%2B+4++-5+=+0
5-5=0
0=0 ->true
so, x%5B1%5D+=+1, x%5B2+%5D=+-1, x%5B3%5D+=+-+1 is a solution

(b) Verify that x%5B1%5D+=+-2, x%5B2%5D+=+2,+x%5B3%5D+=+2 is a solution.
2x%5B1%5D+%2B+3x%5B2%5D+-+x%5B3%5D+=+0......substitute given values
2%2A%28-2%29+%2B+3%282%29+-+%282%29=+0
-4+%2B6-2=+0
-6%2B6=0+
0=0->true
and
x%5B1%5D+-+4x%5B2%5D+%2B+5x%5B3%5D+=+0
-2+-+4%282%29++%2B+5%282%29+=+0
-2+-8++%2B10+=+0
-10%2B10=0
0=0 ->true
=>x%5B1%5D+=+-2, x%5B2%5D+=+2, x%5B3%5D+=+2 is a solution
(c) Adding the corresponding x-values of the solution in parts (a) and (b) gives
x%5B1%5D+=+-1, x%5B2%5D+=+1, x%5B3%5D+=+1
Is this a solution to the linear system?
2x%5B1%5D+%2B+3x%5B2%5D+-+x%5B3%5D+=+0......substitute given values
2%2A%28-1%29+%2B+3%281%29+-+%281%29=+0
-2+%2B3-1=+0
-3%2B3=0+
0=0->true
and
x%5B1%5D+-+4x%5B2%5D+%2B+5x%5B3%5D+=+0
-1+-+4%281%29++%2B+5%281%29+=+0
-1+-4++%2B5+=+0
-5%2B5=0
0=0+->true
=>x%5B1%5D+=+-1, x%5B2%5D+=+1, x%5B3%5D+=+1 is a solution to the linear system

(d) Multiply each of the x-values in part (a) by 3. Are the resulting values a solution to the linear system?
x%5B1%5D%2A3+=+1%2A3=3, x%5B2+%5D%2A3=+-1%2A3=-3, x%5B3%5D+%2A3=+-+1%2A3=-3
2x%5B1%5D+%2B+3x%5B2%5D+-+x%5B3%5D+=+0......substitute given values
2%2A%283%29+%2B+3%28-3%29+-+%28-3%29=+0
6-9%2B3=+0
9-9=0+
0=0->true
and
x%5B1%5D+-+4x%5B2%5D+%2B+5x%5B3%5D+=+0
3+-+4%28-3%29++%2B+5%28-3%29+=+0
3+%2B12+-15+=+0
15-15=0
0=0+->true
=>the resulting values are solution to the linear system