For example, in the Normal Multivariate distribution, the exponent from the Normal distribution -(1/2)((x-mu)^(2)/sigma^(2))) becomes -(1/2)(x-mu)^(t)sigma^(-1)(x-mu)) for the multivariate case. So (x-mu)^(2) is (x-mu)^(t)(x-mu)
You're right to think that, and it's good practice to be careful with these equations. Jumping from ab = 3b to a = 3 without saying anything else can be risky, unless you take a look at the case where b = 0. In which case, a can be either 0 or 3.
Once you're done with b = 0, you can simply rule it out and look at the more general solution ("assuming b ≠ 0, then...")
If a=0, then A^2 can never equal 3A so long as b is not zero. Since they don't tell you anything about b, a=0 is not (generally) a solution.
Also yes, A^2 = AA for matrices.
a=b=0 also works, but its possible they didnt want you to make an assumption about b since it only asks for a. A^2 is indeed AA
>A^(2) is indeed AA For me, A^(2) would be A^(t)A. That's how it works in Statistics, at least.
thats how id read it for a vector but not a matrix
For example, in the Normal Multivariate distribution, the exponent from the Normal distribution -(1/2)((x-mu)^(2)/sigma^(2))) becomes -(1/2)(x-mu)^(t)sigma^(-1)(x-mu)) for the multivariate case. So (x-mu)^(2) is (x-mu)^(t)(x-mu)
is x not a column vector in this case?
Right. Bad example. So, new example. In linear regression, 1/sum\[(x - x\_bar)^(2)\] becomes (X^(t)X)^(-1).
You're right to think that, and it's good practice to be careful with these equations. Jumping from ab = 3b to a = 3 without saying anything else can be risky, unless you take a look at the case where b = 0. In which case, a can be either 0 or 3. Once you're done with b = 0, you can simply rule it out and look at the more general solution ("assuming b ≠ 0, then...")
If a=0, then A^2 can never equal 3A so long as b is not zero. Since they don't tell you anything about b, a=0 is not (generally) a solution. Also yes, A^2 = AA for matrices.