Question: 10 6 3. (3 points) Consider the symmetric matrix A = 6 10 (i) Prove that A is positive definite, that is x Ax >

10 6 3. (3 points) Consider the symmetric matrix A = 6 10 (i) Prove that A is positive definite, that is x Ax > 0 for every vector x * 0. (ii) Find the eigenvalues A1, 12, and check that they are strictly positive. (iii) Find eigenvectors v1, v2, so that Av = Arv and Av2 = 12v2. Check that v, and v2 are perpendicular to each other

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