Solve Fraction Expansion: Find A & B

[Phymath]

[tex]\frac{1}{(c^2a_{11}a_{22}-c^2a_{12}a_{21})} = \frac{1}{A} + \frac{1}{B}[/tex]

anyone know how to solve for A and B?

 

[TenaliRaman]

two variables
1 equation
hence many solutions are possible just by setting A = k
unless u are restricting the domains of A and B ...

-- AI

 

[tacman]

To solve for A and B, we can use the fact that when adding fractions with the same denominator, we can simply add the numerators. In this case, we have:

\frac{1}{(c^2a_{11}a_{22}-c^2a_{12}a_{21})} = \frac{1}{A} + \frac{1}{B}

We can rewrite the left side as a single fraction by finding the common denominator. In this case, the common denominator is A*B. So we have:

\frac{1}{(c^2a_{11}a_{22}-c^2a_{12}a_{21})} = \frac{A+B}{AB}

Now, we can equate the numerators and denominators of the two fractions and solve for A and B:

A + B = 1
AB = c^2a_{11}a_{22}-c^2a_{12}a_{21}

We can solve for B in the first equation by subtracting A from both sides:

B = 1 - A

We can substitute this into the second equation to get:

A(1-A) = c^2a_{11}a_{22}-c^2a_{12}a_{21}

Expanding the left side, we get:

A - A^2 = c^2a_{11}a_{22}-c^2a_{12}a_{21}

Bringing all terms to one side, we have:

A^2 - A + (c^2a_{11}a_{22}-c^2a_{12}a_{21}) = 0

This is a quadratic equation in terms of A. We can use the quadratic formula to solve for A:

A = \frac{-b \pm \sqrt{b^2-4ac}}{2a}

In this case, a = 1, b = -1, and c = (c^2a_{11}a_{22}-c^2a_{12}a_{21}). Substituting these values, we get:

A = \frac{1 \pm \sqrt{1-4(c^2a_{11}a_{22}-c^2a_{12}a_{21})}}{2}

Simplifying, we have:

A = \frac{1 \pm \sqrt{1