Question

If \( A \) and \( B \) are the two real values of \( k \) for which the system of equations \( x + 2y + z = 1 \), \( x + 3y + 4z = k \), \( x + 5y + 10z = k^2 \) is consistent, then \( A + B = \): (a) 3

• A. 3
• B. 4
• C. 5
• D. 7

Hint:

For consistency in linear systems, use row reduction or matrix determinant methods. The determinant should be zero for the system to have solutions.

Answer 1

The Correct Option is A

For the system of equations to be consistent, the determinant of the coefficient matrix must be zero. The coefficient determinant \( \Delta \) is given by: 

\[\Delta = \begin{vmatrix} 1 & 2 & 1 \\ 1 & 3 & 4 \\ 1 & 5 & 10 \end{vmatrix}\]

Expanding along the first row: 

\[\Delta = 1 \begin{vmatrix} 3 & 4 \\ 5 & 10 \end{vmatrix} - 2 \begin{vmatrix} 1 & 4 \\ 1 & 10 \end{vmatrix} + 1 \begin{vmatrix} 1 & 3 \\ 1 & 5 \end{vmatrix}\]

\[ = 1(30 - 20) - 2(10 - 4) + 1(5 - 3) \] \[ = 10 - 12 + 2 = 0 \] Since \( \Delta = 0 \), the system is consistent. Now, solving for \( k \), we set up the determinant \( \Delta_1 \): 

\[\Delta_1 = \begin{vmatrix} k & 2 & 1 \\ 1 & 3 & 4 \\ k^2 & 5 & 10 \end{vmatrix} = 0\]

Expanding along the first row:

\[\Delta_1 = k \begin{vmatrix} 3 & 4 \\ 5 & 10 \end{vmatrix} - 2 \begin{vmatrix} 1 & 4 \\ k^2 & 10 \end{vmatrix} + 1 \begin{vmatrix} 1 & 3 \\ k^2 & 5 \end{vmatrix} = 0\]

 Solving, we obtain the quadratic equation: \[ (k - 2)(k - 1) = 0 \] \[ \Rightarrow k = 2, 1 \] Thus, the real values of \( k \) are \( A = 2 \) and \( B = 1 \). \[ A + B = 2 + 1 = 3 \]