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anonymous

  • 5 years ago

finding the value of a in the given equations through gaussian elimination x+y+z=2 x+3y+2z=5 x+y+(a^2-1)z=a

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  1. anonymous
    • 5 years ago
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    (a^2)z + 2z - a - 2 = 0.

  2. anonymous
    • 5 years ago
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    You can't find a specific numerical value for a, since there will be infinitely many solutions, a different solution parametrized by a different value of a. Let’s write this system in matrix form: 1 1 1 2 1 3 2 5 1 1 (a^2 – 1) a. Then reduce. 1 1 1 2 0 2 1 3 0 0 (a^2 – 2) (a – 2). So we get: 1 1 1 2 0 2 1 3 0 0 1 (a – 2)/(a^2 – 2). We can let “a” be anything such that the denominator isn’t equal to 0. For example, let a = 2. Then that forces z to be 0. Then back substitute into 2y + z = 3 to see that y = 3/2. Then back substitute into x + y + z = 2 to get x = 1/2. We can check that (1/2 , 3/2, 0) solves our system. But we can let a = 1 as well. If we do, we get z = 1. Back substituting into 2y + z = 3 gives y = 1. Back subbing into x + y + z = 3 gives x = 0. So (0, 1, 1) is another solution corresponding to a = 1.

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