If
\[
\begin{bmatrix}
x + y & 2 \\
5 & xy
\end{bmatrix}
=
\begin{bmatrix}
6 & 2 \\
5 & 8
\end{bmatrix},
\]
then the value of
\[
\left( \frac{24}{x} + \frac{24}{y} \right)
\]
is:
Show Hint
- 7
- 6
- 8
- 18
The Correct Option is D
Solution and Explanation
From the equality of the matrices: \[ \begin{bmatrix} x + y & 2 \\ 5 & xy \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}, \]
Step 1: Use the relationship between \( x \) and \( y \)
The given equations are: \[ x + y = 6 \quad \text{and} \quad xy = 8. \] These are the sum and product of the roots of a quadratic equation. Let the quadratic equation be: \[ t^2 - (x + y)t + xy = 0. \]
Substitute \( x + y = 6 \) and \( xy = 8 \): \[ t^2 - 6t + 8 = 0. \]
Factorize the equation: \[ t^2 - 6t + 8 = (t - 2)(t - 4) = 0. \] Thus, \( x = 2 \) and \( y = 4 \) (or vice versa).
Step 2: Compute \( \frac{24}{x} + \frac{24}{y} \)
Substitute \( x = 2 \) and \( y = 4 \): \[ \frac{24}{x} + \frac{24}{y} = \frac{24}{2} + \frac{24}{4}. \] Simplify: \[ \frac{24}{2} + \frac{24}{4} = 12 + 6 = 18. \]
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\[ \begin{bmatrix} x + y & 2 \\ 5 & xy \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}, \]
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