Linear Algebra: linear equations

Amy-Lee
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Show that if u and v are solutions of the linear system Ax = b, then w = 1/4 u + 3/4 v is also a solution of Ax = b.
 
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Just plug it in and use the fact that matrix multiplication is linear.
 
Plug what? where?

Ax=b, then b=uv ? and uv=w? uv = 1/4u + 3/4v?
 
You want to show that w "is also a solution of Ax= b".

So "plug" w in for x: Aw= what? Use the fact that w= (1/4)u+ (3/4)v and that A is a linear transformation.
 
Amy-Lee said:
Show that if u and v are solutions of the linear system Ax = b, then w = 1/4 u + 3/4 v is also a solution of Ax = b.
Amy-Lee said:
Plug what? where?

Ax=b, then b=uv ? and uv=w? uv = 1/4u + 3/4v?

Amy-Lee, forget x …

you know that Au = b and Av = b,

so just carry on from there. :wink:
 
tiny-tim said:
Amy-Lee, forget x …

you know that Au = b and Av = b,

so just carry on from there. :wink:

thanks TT, I will try that
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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...
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