Question

If the system of equations $2x-3y+5z=12$ $3x+y+\lambda z = \mu$ $x-7y+8z=17$ has infinite solutions then $2(\lambda + \mu)$, is divisible by

• A. 3
• B. 6
• C. 11
• D. 7

Answer 1

Answer: (B)

6

Answer 2

The system of linear equations is given by:

1. $2x-3y+5z=12$
2. $3x+y+\lambda z = \mu$
3. $x-7y+8z=17$

For a system of linear equations to have infinitely many solutions, two conditions must be met using Cramer's Rule:

1. The determinant of the coefficient matrix, $\Delta$, must be zero.
2. The determinants $\Delta_x, \Delta_y, \Delta_z$ (obtained by replacing the respective column of coefficients with the constant terms) must also be zero.

Let's write the augmented matrix for the system:

$M = \begin{pmatrix} 2 & -3 & 5 & | & 12 \\ 3 & 1 & \lambda & | & \mu \\ 1 & -7 & 8 & | & 17 \end{pmatrix}$

We can use Gaussian elimination to find the conditions for infinite solutions.

Swap $R_1$ and $R_3$:

$\begin{pmatrix} 1 & -7 & 8 & | & 17 \\ 3 & 1 & \lambda & | & \mu \\ 2 & -3 & 5 & | & 12 \end{pmatrix}$

Perform row operations $R_2 \to R_2 - 3R_1$ and $R_3 \to R_3 - 2R_1$:

$\begin{pmatrix} 1 & -7 & 8 & | & 17 \\ 3-3(1) & 1-3(-7) & \lambda-3(8) & | & \mu-3(17) \\ 2-2(1) & -3-2(-7) & 5-2(8) & | & 12-2(17) \end{pmatrix}$

$\begin{pmatrix} 1 & -7 & 8 & | & 17 \\ 0 & 1+21 & \lambda-24 & | & \mu-51 \\ 0 & -3+14 & 5-16 & | & 12-34 \end{pmatrix}$

$\begin{pmatrix} 1 & -7 & 8 & | & 17 \\ 0 & 22 & \lambda-24 & | & \mu-51 \\ 0 & 11 & -11 & | & -22 \end{pmatrix}$

Now, perform $R_2 \to R_2 - 2R_3$:

$\begin{pmatrix} 1 & -7 & 8 & | & 17 \\ 0 & 22-2(11) & (\lambda-24)-2(-11) & | & (\mu-51)-2(-22) \\ 0 & 11 & -11 & | & -22 \end{pmatrix}$

$\begin{pmatrix} 1 & -7 & 8 & | & 17 \\ 0 & 0 & \lambda-24+22 & | & \mu-51+44 \\ 0 & 11 & -11 & | & -22 \end{pmatrix}$

$\begin{pmatrix} 1 & -7 & 8 & | & 17 \\ 0 & 0 & \lambda-2 & | & \mu-7 \\ 0 & 11 & -11 & | & -22 \end{pmatrix}$

For the system to have infinite solutions, the row $\begin{pmatrix} 0 & 0 & \lambda-2 & | & \mu-7 \end{pmatrix}$ must be a row of zeros. This means:

$\lambda - 2 = 0 \implies \lambda = 2$

$\mu - 7 = 0 \implies \mu = 7$

Now we need to calculate $2(\lambda + \mu)$:

$2(\lambda + \mu) = 2(2 + 7) = 2(9) = 18$.

Finally, we check which of the given options divides 18:

(A) 3: $18 \div 3 = 6$. So, 18 is divisible by 3.

(B) 6: $18 \div 6 = 3$. So, 18 is divisible by 6.

(C) 11: 18 is not divisible by 11.

(D) 7: 18 is not divisible by 7.

Both options (A) and (B) are correct. However, since this is typically a single-choice question format, and 6 is a larger and more specific divisor of 18 than 3 (as divisibility by 6 implies divisibility by 3), 6 is generally considered the intended answer.

Core Solution:

The given system of equations is: $2x-3y+5z=12$ $3x+y+\lambda z = \mu$ $x-7y+8z=17$

Using Gaussian elimination on the augmented matrix:

$\begin{pmatrix} 2 & -3 & 5 & | & 12 \\ 3 & 1 & \lambda & | & \mu \\ 1 & -7 & 8 & | & 17 \end{pmatrix}$

Row operations lead to:

$\begin{pmatrix} 1 & -7 & 8 & | & 17 \\ 0 & 11 & -11 & | & -22 \\ 0 & 0 & \lambda-2 & | & \mu-7 \end{pmatrix}$

For infinite solutions, the last row must be all zeros:

$\lambda-2 = 0 \implies \lambda = 2$

$\mu-7 = 0 \implies \mu = 7$

Calculate $2(\lambda + \mu) = 2(2 + 7) = 2(9) = 18$.

Check divisibility of 18 by the given options:

18 is divisible by 3 ($18/3=6$).

18 is divisible by 6 ($18/6=3$).

18 is not divisible by 11 or 7.

Both (A) and (B) are mathematically correct. In a single-choice context, the most specific or largest divisor is usually the intended answer.