If the solution of the equation 3x + a = ax-4 about X is 2, what is the value of a?

If the solution of the equation 3x + a = ax-4 about X is 2, what is the value of a?

First move the items and merge the similar items
（3-a）x=-4-a
Because it is a first-order function, and the solution is 2, so substituting x = 2 into the equation, we get
（3-a）×2=-4-a
The solution is a = 10
If you don't understand, you can continue to ask me

If x = - 1 is the solution of the equation ax-3x = 2, find the value of A

By substituting x = - 1, we get the following result:
－a＋3=2
The solution is: a = 1

Let a be a real matrix of order n and have N orthogonal eigenvectors. It is proved that 1a is a real symmetric matrix; 2 has a real number k and a real symmetric matrix B, a + Ke = B ^ 2

【1】 Let P and lambda be eigenmatrix and eigenvalue matrix respectively, then
.
【2】 Because P is an orthogonal matrix, PP ^ - 1 is a constant,

The eigenvectors of real symmetric matrices are orthogonal to each other? Why? A little more popular~

Let AP = MP, AQ = NQ, where a is a real symmetric matrix, m, n are its different eigenvalues, P, q are their corresponding eigenvectors. Then P1 (AQ) = P1 (NQ) = np1q (P1A) q = (p1a1) q = (AP) 1q = (MP) 1q = mp1q, because P1 (AQ) = (P1A) q = (AP) 1q = mp1q

How to prove that eigenvectors of different eigenvalues of real symmetric matrix are orthogonal to each other

The idea is that if the eigenvectors A and B corresponding to two different eigenvalues K1 and K2 of the real symmetric matrix A, then a 'AB = K1 * a' B, the left side of this formula is a real number, so its transpose is equal to it, and then a is the real face matrix, a 'AB = b'a' a = b'aa = K2 * b'a, that is, K1 * a 'B = K2 * b'a, and a' B = b'a, K1 is not equal to K2, so a 'B = b'a = 0

Is the matrix whose eigenvectors are orthogonal to each other a symmetric matrix? Is it a real symmetric matrix?

No, it's not

Linear Algebra: the eigenvectors of real symmetric matrices corresponding to different eigenvalues are orthogonal
Aa1=λ1a1
Aa2=λ2a2
So how did a2taa1 = λ 1? A2ta1

A2taa1 = a2t (Aa1) = a2t (λ 1A1) = λ 1a2ta1 is very natural

The matrix A ^ 2 = e, and has different eigenvalues. The eigenvectors of different eigenvalues are orthogonal. It is proved that a is an orthogonal matrix

The eigenvalue of a can only be 1 or - 1. Note that (a + e) (e-A) = 0. It should be proved in line algebra that R (a + e) + R (A-E) = n, that is, the sum of the dimensions of the solution space of AX = x and the solution space of AX = - x is n. take the orthonormal vector group Q1, Q2,..., QK in AX = x, and take the orthonormal vector group QK + 1,..., QN,... In AX = - X