For what values of $ \lambda $ and $ \mu $, the following system of equations has a unique solution?
$ 2x + 3y + 5z = 9 $
$ 7x + 3y - 2z = 8 $
$ 2x + 3y + \lambda z = \mu $
Step 1:
The given system of equations is:
\[ 2x + 3y + 5z = 9 \quad \text{(1)} \] \[ 7x + 3y - 2z = 8 \quad \text{(2)} \] \[ 2x + 3y + \lambda z = \mu \quad \text{(3)} \] We need to find the values of \( \lambda \) and \( \mu \) for which this system has a unique solution.
Step 2:
The condition for a unique solution in a system of linear equations is that the determinant of the coefficient matrix should be non-zero. The coefficient matrix is: \[ \begin{pmatrix} 2 & 3 & 5 7 & 3 & -2 2 & 3 & \lambda \end{pmatrix} \] The determinant of this matrix is: \[ \text{Determinant} = 2 \begin{vmatrix} 3 & -2 3 & \lambda \end{vmatrix} - 3 \begin{vmatrix} 7 & -2 \\ 2 & \lambda \end{vmatrix} + 5 \begin{vmatrix} 7 & 3 \\ 2 & 3 \end{vmatrix} \]
Step 3:
Calculate each of the 2x2 determinants: \[ \begin{vmatrix} 3 & -2 3 & \lambda \end{vmatrix} = 3\lambda - (-6) = 3\lambda + 6 \] \[ \begin{vmatrix} 7 & -2 2 & \lambda \end{vmatrix} = 7\lambda - (-4) = 7\lambda + 4 \] \[ \begin{vmatrix} 7 & 3 2 & 3 \end{vmatrix} = 21 - 6 = 15 \]
Step 4:
Substitute these values back into the determinant expression: \[ \text{Determinant} = 2(3\lambda + 6) - 3(7\lambda + 4) + 5(15) \] \[ = 6\lambda + 12 - 21\lambda - 12 + 75 \] \[ = -15\lambda + 75 \]
Step 5:
For the system to have a unique solution, the determinant must not be zero: \[ -15\lambda + 75 \neq 0 \] \[ \lambda \neq 5 \]
Step 6:
Substitute \( \lambda = 5 \) into any equation, say equation (3), and solve for \( \mu \): \[ 2x + 3y + 5z = \mu \] From the other equations, it follows that \( \mu = 9 \).
Step 7:
Therefore, the values for \( \lambda \) and \( \mu \) for which the system has a unique solution are \( \lambda = 5 \) and \( \mu = 9 \).
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If equilibrium constant for the equation $ A_2 + B_2 \rightleftharpoons 2AB \quad \text{is} \, K_p, $ then find the equilibrium constant for the equation $ AB \rightleftharpoons \frac{1}{2} A_2 + \frac{1}{2} B_2. $
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