6.solve the system of equation, using determinants
x + 3y - 2 = 4
3x - 2y + 5z = - 4
5x - y - 4z = -9
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Sandra A. asked • 07/27/18
I want to solve this, i need help. Justify your answer please.
Consider the linear system:
x - 2y + 3z = 4
2x - 3y + az = 5
3x - 4y + 5z = b
For which values of a and b does the linear system have
a) no solution,
b) a unique solution,
c) infinitely many solutions.
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Mark M. answered • 07/27/18
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x-2y+3z = 4
2x-3y+az = 5
3x-4y+5z = b
Multiply the first equation by -2 and add to the second equation. Multiply the first equation by -3 and add to the third equation.
x - 2y + 3z = 4
y + (a-6)z = -3
2y - 4z = (b-12)
Multiply the second equation by -2 and add to the third equation:
x - 2y + 3z = 4
y + (a-6)z = -3
(-2a+8)z = (b-6)
Infinitely many solutions when -2a+8 = 0 and b-6 = 0
So, a = 4 and b = 6
No solution if -2a+8 = 0 and b-6 ≠ 0
So, a = 4 and b ≠ 6
Unique solution if a ≠ 4 and b = any real number