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:

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

Answer 2

Step 1: Write coefficient matrix and constants.
The given system is: \[ \begin{aligned} x + y + 2z &= 6, \\ 2x + 3y + a z &= a + 1, \\ - x - 3y + b z &= 2b. \end{aligned} \] The coefficient matrix \(A\) is \[ A = \begin{bmatrix} 1 & 1 & 2 \\ 2 & 3 & a \\ -1 & -3 & b \end{bmatrix}, \] 

Step 2: Condition for infinitely many solutions. 
For a system to have infinitely many solutions, \[ \text{rank}(A) = \text{rank}([A|B]) < 3. \] This means the determinant of \(A\) must be zero.

Step 3: Compute determinant of \(A\).
\[ \det(A) = \begin{vmatrix} 1 & 1 & 2 \\ 2 & 3 & a \\ -1 & -3 & b \end{vmatrix} = 1(3b + 3a) - 1(2b + a) + 2(-6 + 3) \] Wait—let’s compute properly: \[ \det(A) = 1(3b - a(-3)) - 1(2b - a(-1)) + 2(2(-3) - 3(-1)). \] Simplify step-by-step: \[ \det(A) = 1(3b + 3a) - 1(2b + a) + 2(-6 + 3). \] \[ \det(A) = (3b + 3a) - (2b + a) + 2(-3). \] \[ \det(A) = (3b - 2b) + (3a - a) - 6. \] \[ \det(A) = b + 2a - 6. \]

Step 4: Set determinant to zero.
For infinitely many solutions, \[ b + 2a - 6 = 0 \quad \Rightarrow \quad b = 6 - 2a. \]

Step 5: Ensure consistency (augmented matrix).
Substitute \(b = 6 - 2a\) into the system. The third equation becomes: \[ -x - 3y + (6 - 2a)z = 2(6 - 2a), \] i.e. \[ -x - 3y + (6 - 2a)z = 12 - 4a. \]

Now multiply the first equation by 1, second equation by some scalar \(k\), and third equation by another scalar so that they are linearly dependent.

We can express the second row as a linear combination of the first and third rows for consistency. Let’s find the relationship between the equations.

From Equation (1): \(x + y + 2z = 6\). Equation (2): \(2x + 3y + a z = a + 1.\)

Subtract 2×(Eq.1) from Eq.2: \[ (2x - 2x) + (3y - 2y) + (a - 4)z = (a + 1) - 12. \] \[ y + (a - 4)z = a - 11. \quad \text{(Eq.4)} \] Similarly, add Eq.(1) and Eq.(3): \[ (x - x) + (y - 3y) + (2 + b)z = 6 + 2b. \] \[ -2y + (2 + b)z = 6 + 2b. \] \[ y = \frac{(2 + b)z - (6 + 2b)}{2}. \quad \text{(Eq.5)} \] For consistency, Eq.(4) and Eq.(5) must represent the same linear relation between \(y\) and \(z\). Equate coefficients of \(z\) and constants: \[ \text{Coefficient of } z: \quad 1 \text{ (from Eq.4)} = \frac{2 + b}{2}, \] \[ \Rightarrow 2 = 2 + b \ \Rightarrow\ b = 0. \] Then from Eq.(4) and Eq.(5) constant term equality: \[ a - 11 = -\frac{6 + 2b}{2} = -3. \] \[ a = 8. \]

Step 6: Compute \(7a + 3b.\)
\[ 7a + 3b = 7(8) + 3(0) = 56. \] Wait—this doesn’t satisfy the earlier determinant condition \(b = 6 - 2a\). Let’s check consistency:

\[ b = 6 - 2(8) = 6 - 16 = -10. \] Substitute \(b = -10\) into previous step: Eq.(4): \(y + (a - 4)z = a - 11\). Eq.(5): \(y = \frac{(2 + (-10))z - (6 + 2(-10))}{2} = \frac{-8z - (-14)}{2} = -4z + 7.\) Now Eq.(4) → \(y = - (a - 4)z + a - 11.\) Equating coefficients of \(z\) and constants: \[ -(a - 4) = -4 \Rightarrow a - 4 = 4 \Rightarrow a = 8. \] \[ a - 11 = 7 \Rightarrow a = 18. \] Contradiction unless we recheck properly; maybe one equation was sign-misread. Let’s re-evaluate systematically.

 

Better route: Direct consistency via ratio of coefficients.

For infinitely many solutions, ratios of coefficients and constants must be equal:

\[ \frac{1}{2} = \frac{1}{3} = \frac{2}{a} = \frac{6}{a+1}. \] No, that’s between first and second equations only—not all consistent. Instead, we use determinant condition and one more relation.

 

From determinant: \(b = 6 - 2a.\)

For consistency: The three planes must intersect in a line, so any two equations must be consistent.

Subtract Eq.(1) from Eq.(2): \[ (2x - x) + (3y - y) + (a - 2)z = (a + 1 - 6). \] \[ x + 2y + (a - 2)z = a - 5. \quad \text{(Eq.6)} \] Subtract Eq.(1) from Eq.(3): \[ (-x - x) + (-3y - y) + (b - 2)z = (2b - 6). \] \[ -2x - 4y + (b - 2)z = 2b - 6. \] Divide by -2: \[ x + 2y - \frac{(b - 2)}{2}z = 3 - b. \quad \text{(Eq.7)} \] For consistency (infinitely many solutions), Eq.(6) and Eq.(7) must represent the same plane, so their coefficients are proportional: \[ \frac{1}{1} = \frac{2}{2} = \frac{a - 2}{-\frac{b - 2}{2}} = \frac{a - 5}{3 - b}. \] From \(x\) and \(y\) coefficients, ratio = 1. Thus: \[ a - 2 = -\frac{b - 2}{2} \quad \text{and} \quad a - 5 = 3 - b. \] From the first: \[ 2(a - 2) = -(b - 2) \Rightarrow 2a - 4 = -b + 2 \Rightarrow b = 6 - 2a. \] This matches the determinant condition ✔. Now use the second: \[ a - 5 = 3 - b \Rightarrow a + b = 8. \] Substitute \(b = 6 - 2a\): \[ a + (6 - 2a) = 8 \Rightarrow 6 - a = 8 \Rightarrow a = -2. \] Then \(b = 6 - 2(-2) = 10.\)

 

Step 7: Compute \(7a + 3b.\)
\[ 7a + 3b = 7(-2) + 3(10) = -14 + 30 = 16. \]

 

Final Answer:

\[ \boxed{16} \]