Solutions of a system of linear equations

[diogomgf]

If the matrices [MATH]D_{1}[/MATH] and [MATH]Y_{1}[/MATH] are solutions the a system of linear equations of type [MATH]AX = B[/MATH], and [MATH]aD_{1} + bY_{1}[/MATH] is also a solution then [MATH]B = 0[/MATH] (null matrix)?

From what I see: [MATH]D_{1} = A^{-1}B[/MATH] and [MATH]Y_{1} = A^{-1}B[/MATH] also.
Then [MATH]a(AA^{-1}B) + b(AA^{-1}B) = B \Leftrightarrow aB + bB = B[/MATH]. [MATH]B = 0[/MATH] is true only if [MATH]a,b \neq \frac{1}{2}[/MATH].

But the book has it for any [MATH]a,b \in {R}[/MATH].

Sorry for the real numbers LaTeX representation, I don't know any better.

 

[Romsek]

What's missing in the problem statement is that [MATH]a D_1 + b Y_1 [/MATH] is a solution [MATH]\forall a,b \in \mathbb{R}[/MATH]
Add that in and it's clear that [MATH](a+b)B=B \Rightarrow B=0[/MATH]

 

[diogomgf]

What's missing in the problem statement is that [MATH]a D_1 + b Y_1 [/MATH] is a solution [MATH]\forall a,b \in \mathbb{R}[/MATH]
Add that in and it's clear that [MATH](a+b)B=B \Rightarrow B=0[/MATH]

Not sure I understand

 

[Romsek]

The problem fails to state that \(\displaystyle a D_1 + bY_1\) is a solution given ANY \(\displaystyle a,b\)

[MATH]A(a D_1 + b Y_1) = \\ a A D_1 + b A Y_1 = \\ a B + bB = \\ (a+b)B = B[/MATH]
If this is true for ANY \(\displaystyle a,b\) then it's clear that \(\displaystyle B=0\)

 

[diogomgf]

The problem fails to state that \(\displaystyle a D_1 + bY_1\) is a solution given ANY \(\displaystyle a,b\)

[MATH]A(a D_1 + b Y_1) = \\ a A D_1 + b A Y_1 = \\ a B + bB = \\ (a+b)B = B[/MATH]
If this is true for ANY \(\displaystyle a,b\) then it's clear that \(\displaystyle B=0\)

Got it !

 

[Steven G]

For example, if 7B = B then B=0.

 

[diogomgf]

For example, if 7B = B then B=0.

Yep, but in the case [MATH]aB + bB = B[/MATH] it's different... if [MATH]a = \frac{1}{2}, b = \frac{1}{2}[/MATH] for example, the case is valid... that's why the "for any" was necessary...