If [ [a+b 2] [5 ab] ] = [ [6 2] [5 8] ], find values of a and b

Shankar

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Oct 25, 2015
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Hello Everyone! Please help Me!

\(\displaystyle \mbox{If }\, \left[\,\begin{array}{cc}a\, +\, b&2\\5&ab\end{array}\,\right]\, =\, \left[\,\begin{array}{cc}6&2\\5&8\end{array}\,\right],\, \mbox{ find the value of }\, a\, \mbox{ and }\, b.\)

So far i have done:


So, we have
By equality of matrices:
a+b=6 ...(i)
ab = 8 ...(ii)

I know the answer is a=2, b=4 or a=4, b=2. But i don't know the process of finding it.
 
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ksdhart

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Well, you know that ab = 8. That means that you need two numbers that multiply to 8. And you also know that a + b = 6, so the two numbers have to add up to 6. So, just list of all the factors of 8:

1 * 8 = 8 ; 1 + 8 = 9. So that's not the answer.
2 * ? = 8 ; 2 + ? = ?. And so on...
 

Shankar

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Oct 25, 2015
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Thanks a lot for helping.
I found the clue by to solve the above problem using (x + a)(x + b) = x^2 + (a + b)x + ab = x^2 + 6x + 8. Factorizing x^2 + 6x + 8 gives x^2 + 6x + 9 - 1 = (x + 3)^2 - 1^2 = (x + 4)(x + 2).

so (a, b) = (2, 4) or (4, 2).
 

stapel

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\(\displaystyle \mbox{If }\, \left[\,\begin{array}{cc}a\, +\, b&2\\5&ab\end{array}\,\right]\, =\, \left[\,\begin{array}{cc}6&2\\5&8\end{array}\,\right],\, \mbox{ find the value of }\, a\, \mbox{ and }\, b.\)

By equality of matrices:
a+b=6 ...(i)
ab = 8 ...(ii)
I found the clue by to solve the above problem using (x + a)(x + b) = x^2 + (a + b)x + ab = x^2 + 6x + 8.
What was the logic by which you generated this equation?

Instead, using what you'd been given (which is more likely to be useful in the future):

. . . . .\(\displaystyle ab\, =\, 8\)

. . . . .\(\displaystyle a\, =\, \dfrac{8}{b}\)

. . . . .\(\displaystyle a\, +\, b\, =\, 6\)

. . . . .\(\displaystyle \dfrac{8}{b}\, +\, b\, =\, 6\)

. . . . .\(\displaystyle 8\, +\, b^2\, =\, 6b\)

. . . . .\(\displaystyle b^2\, -\, 6b\, +\, 8\, =\, 0\)

. . . . .\(\displaystyle (b\, -\, 2)(b\, -\, 4)\, =\, 0\)

Solve the quadratic for the values of b. Then back-solve for the corresponding values of a. ;)
 

Shankar

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Oct 25, 2015
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I got it.
 
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