Prove that if A is an n x n matrix, then B = A+A' is symmetric?

Question

a, Prove that if A is an n x n matrix, then B = A+A' is symmetric

b, Prove that if A and B are n x n symmetric matrices, then AB is symmetric if and only if AB = BA

2, Solve the system of equations

2x1 + 12x2 = 40
8x1 + 4 x2 = 28

by applying repeated elementary row operations to the augmented matrix until the coefficient matrix is reduced to an identity matrix.

Answer 1

A: I assume that A' is meant to be the transpose of A? If so, then note that for any i, j≤n, we have that B_(ij) = A_(ij) + A'_(ij) = A_(ij) + A_(ji), and B(ji) = A_(ji) + A'_(ji) = A_(ji) + A_(ij) = A_(ij) + A_(ji), so by transitivity, B_(ij) = B(ji). Since this is true for all i, j, B is symmetric.

B: Note that (AB)' = B'A', and since A and B are symmetric, B'A' = BA. Thus (AB)' = BA, and so AB = (AB)' if and only if AB = BA.

Problem 2:

Really basic, but since I have time --

2x₁ + 12x₂ = 40
8x₁ + 4x₂ = 28

Write the coefficient matrix:

[2, 12 | 40]
[8, 4 | 28]

Divide row 1 by 2:

[1, 6 | 20]
[8, 4 | 28]

Subtract 8*row 1 from row 2:

[1, 6 | 20]
[0, -44 | -132]

Divide row 2 by -44:

[1, 6 | 20]
[0, 1 | 3]

Subtract 6*row 2 from row 1:

[1, 0 | 2]
[0, 1 | 3]

Thus x₁ = 2 and x₂ = 3

Answer 2

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