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Advanced Topics:
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Example
Solve a System of Equations
Eigenvalues Problem
π Solution:
From second equation:
x = y + 1
Substitute into first:
2(y + 1) + y = 5
2y + 2 + y = 5
3y = 3 β y = 1
Then:
x = 2
β Final Answer: (x, y) = (2, 1)
π Characteristic equation:
|A β Ξ»I| = 0
(2 β Ξ»)Β² β 1 = 0
λ² β 4Ξ» + 3 = 0
π Ξ» = 1, 3
β Eigenvalues: 1 and 3
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