Question

If the system of linear equations: \[ x + y + 2z = 6, \] \[ 2x + 3y + az = a + 1, \] \[ -x - 3y + bz = 2b, \] where \( a, b \in \mathbb{R} \), has infinitely many solutions, then \( 7a + 3b \) is equal to:

• A. 22
• B. 9
• C. 16
• D. 12

Hint:

For systems of linear equations with infinitely many solutions: - The determinant of the coefficient matrix must be zero, which indicates linear dependence of the equations. - Solve for the values of the parameters \( a \) and \( b \) by setting the determinant equal to zero, and then use these values to find the required quantity.

Answer 1

The Correct Option is C

Given the system of equations:

\(\Delta = \Delta_1 = \Delta_2 = \Delta_3 = 0\) 

For the system to have infinite solutions, the determinant of the coefficient matrix should be zero:

\(\Delta = 0\) 

First, solve for the values of \(a\) and \(b\) by working with the following determinants:

\[ \begin{vmatrix} 1 & 1 & 2 \\ 2 & 3 & a \\ -1 & -3 & b \end{vmatrix} = 0 \] 

Expanding the determinant:

\[ (3b + 3a) - 1(2b + a) + 2(-6 + 3) = 0 \] 

\( 2a + b = 6 \) (Equation 1) 

Now, for the second determinant:

\[ \begin{vmatrix} 1 & 1 & 6 \\ 2 & 3 & a+1 \\ -1 & -3 & 2b \end{vmatrix} = 0 \] 

Expanding the determinant:

\[ (6b + 3a + 3) - 1(4b + a + 1) + 2(6 - 6 + 3) = 0 \] 

\( 2a + 2b = 16 \) (Equation 2) 

Now solve the system of equations (1) and (2):

From Equation (1): \( a + b = 8 \)

From Equation (2): \( 2a + b = 6 \)

By solving this system, we find:

\( a = -2, \quad b = 10 \)

Finally, verify the solution:

\( 7a + 3b = -14 + 30 = 16 \)