Question

If the system of equations $ x+2 y+3 z=3$ $ 4 x+3 y-4 z=4 $ $ 8 x+4 y-\lambda z=9+\mu$ has infinitely many solutions, then the ordered pair $(\lambda, \mu)$ is equal to :

• A. $\left(\frac{72}{5},-\frac{21}{5}\right)$
• B. $\left(\frac{72}{5}, \frac{21}{5}\right)$
• C. $\left(-\frac{72}{5} \cdot-\frac{21}{5}\right)$
• D. $\left(-\frac{72}{5}, \frac{21}{5}\right)$

Answer 1

The Correct Option is A

The correct answer is (A) : $\left(\frac{72}{5},-\frac{21}{5}\right)$
$x+2y+3z=\mathrm{3......}$(i)
$4x+3y-4z=\mathrm{4.......}$(ii)
$8x+4y-\lambda z=9+\mu ......$(iii)
$\text{(i)}\times 4\text{- (ii)}\Rightarrow 5y+16z=\mathrm{8......}$(iv)
(ii) $\times 2-$ (iii) $\Rightarrow 2y+(\lambda -8)z=-1-\mu ......$(v)
(iv) $\times 2-$ (iii) $\times 5\Rightarrow (32-5(\lambda -8))z=16-5(-1-\mu )$
For infinite solutions $\Rightarrow 72-5\lambda =0\Rightarrow \lambda =\frac{72}{5}$
$21+5\mu =0\Rightarrow \mu =\frac{-21}{5}$
$\Rightarrow (\lambda ,\mu )\equiv \left(\frac{72}{5},\frac{-21}{5}\right)$

Answer 2

Step 1: Eliminate variables

We begin by eliminating variables. Start with equation (i) and subtract equation (ii):

\[ 4(x + 2y + 3z) - (4x + 3y - 4z) = 12 - 4. \]

Simplifying this expression:

\[ 5y + 16z = 8 \quad \cdots (iv). \]  

 

Next, subtract equation (iii) from equation 2(ii):

\[ 2(4x + 3y - 4z) - (8x + 4y - z) = 8 - (9 + \mu). \]

Simplifying this expression gives:

\[ 2y + 7z = -1 - \mu \quad \cdots (v). \] 

 

The system of equations is now reduced to two equations:

\[ 5y + 16z = 8 \quad \cdots (iv), \] \[ 2y + 7z = -1 - \mu \quad \cdots (v). \] 

Step 2: Solve for \( y \) and \( z \) in terms of \( \mu \)

Multiply equation (v) by 5 and equation (iv) by 2 to align the coefficients of \( y \):

\[ 10y + 35z = -5 - 5\mu \quad \cdots (vi), \] \[ 10y + 32z = 16 \quad \cdots (vii). \] 

 

Now, subtract equation (vii) from equation (vi):

\[ (10y + 35z) - (10y + 32z) = (-5 - 5\mu) - 16. \] Simplifying this gives: \[ 3z = -21 - 5\mu \quad \Rightarrow \quad z = \frac{-21 - 5\mu}{3} \quad \cdots (viii). \] 

 

Substitute \( z \) from equation (viii) into equation (iv):

\[ 5y + 16 \times \frac{-21 - 5\mu}{3} = 8. \] Simplify this equation: \[ 5y + \frac{-336 - 80\mu}{3} = 8. \] Multiply through by 3 to eliminate the fraction: \[ 15y - 336 - 80\mu = 24. \] Simplifying further: \[ 15y = 360 + 80\mu \quad \cdots (ix). \] Therefore: \[ y = \frac{360 + 80\mu}{15} = 24 + \frac{16\mu}{3} \quad \cdots (x). \] 

Step 3: Condition for infinitely many solutions

For the system to have infinitely many solutions, the determinant of the coefficients matrix must be zero. This leads to two conditions:

From equation (iv):

\[ 72 - 5\lambda = 0 \quad \Rightarrow \quad \lambda = \frac{72}{5}. \] 

 

From equation (v):

\[ 21 + 5\mu = 0 \quad \Rightarrow \quad \mu = \frac{-21}{5}. \] 

Step 4: Verify the solution

Finally, we substitute \( \lambda = \frac{72}{5} \) and \( \mu = \frac{-21}{5} \) back into the equations. Both conditions are satisfied, confirming that the solution is correct.