Solve the matrix equation x-xa = B, where a = (101; 210; - 32 - 3), B = (1 - 21; - 341)

Solve the matrix equation x-xa = B, where a = (101; 210; - 32 - 3), B = (1 - 21; - 341)

Solution: from x-xa = B, X (e-A) = B ((e-A) ^ t, B ^ t) = 0 - 231 - 300 - 2 - 24 - 10411r3 * (- 1), R2 * (- 1 / 2), r1-3r20 - 20 - 230011 - 2100 - 4 - 1 - 1r1 * (- 1 / 2), R3 + 4r201 - 3 / 20011 - 21003

Let a and X satisfy the relation XA + e = a ^ 2-x, where a = (120,340,567) matrix X

XA+E=A^2-X =>
XA+X=A^2-E =>
X(A+E)=A^2-E^2 =>
X(A+E)=(A+E)(A-E) =>X=A-E
So a = (0 2 0, 3 3 0, 5 6 6)

Under what conditions do matrices A and B have ab = Ba
Under what conditions do matrices A and B have ab = Ba
Under what condition (a + b) square = a square + b square + 2Ab

When matrices A, B and ab are symmetric matrices of order n, a and B are commutative, that is, ab = ba
prove:
A. B and ab are symmetric matrices, that is, at = a, BT = B, (AB) t = ab
So AB = (AB) t = (BT) (at) = ba
When a and B are commutative, satisfy (a + b) & sup2; = A & sup2; + B & sup2; + 2Ab
prove:
A. B is commutative, that is ab = ba
(A+B)²
=A²+AB+BA+B²
=A²+AB+AB+B²
=A²+B²+2AB