Understanding Why (A+B)(A+B) is Not Valid for Matrices: Linear Algebra Homework

ephemeral1
Messages
28
Reaction score
0

Homework Statement


Explain why the formula is not valid for matrices.
(A+B)(A+B)=A^2 + 2AB + B^2


Homework Equations


none.


The Attempt at a Solution



I don't know really know how to start this. I don't really know why that is not valid. Please help me understand. Thank you.
 
Physics news on Phys.org
(A+B)(A+B) = A^2 + AB + BA + B^2. That much is true for matrices as well as real numbers. What goes wrong in between this line and (A+B)(A+B) = A^2 + 2 AB + B^2?
 
remember matrix multiplication is not like normal multiplication, in general it is noncommutative which means it matters which side you multiply on. another example of noncommutative algebra is the curl or cross product of vectors. If you are discussing the composition of linear transformations remember that when you multiply the matrix representation of a linear transf. it is analagous to composition of functions. obviously T(F(x)) not equal to F(T(x)) for all F,T.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top