The following system of equations 2x-y+3z = 5,3x+2y-z = 7,4x+5y-λz=μ…

The correct answer is Option (3) → λ = 5 and μ = 9

Given system:

2x - y + 3z = 5

3x + 2y - z = 7

$4x + 5y - \lambda z = \mu$

For consistency in the dependent case, the third equation must be a linear combination of the first two. Use the ratio method:

Form the determinant of the coefficient matrix:

$\Delta = \begin{vmatrix} 2 & -1 & 3 \\ 3 & 2 & -1 \\ 4 & 5 & -\lambda \end{vmatrix} = 0$

Expand along row 1:

$\Delta = 2 \begin{vmatrix}2 & -1 \\ 5 & -\lambda \end{vmatrix} - (-1) \begin{vmatrix}3 & -1 \\ 4 & -\lambda \end{vmatrix} + 3 \begin{vmatrix}3 & 2 \\ 4 & 5 \end{vmatrix}$

$ = 2(2(-\lambda) - (-1)(5)) + (3(-\lambda) - (-1)(4)) + 3(3(5) - 2(4))$

$ = 2(-2\lambda + 5) + (-3\lambda + 4) + 3(15 - 8)$

$ = -4\lambda + 10 - 3\lambda + 4 + 21$

$ = -7\lambda + 35$

Set $\Delta = 0 \Rightarrow -7\lambda + 35 = 0 \Rightarrow \lambda = 5$

Now use ratio method for constants:

Check if $\frac{4}{2} = \frac{5}{-1} = \frac{-\lambda}{3} = \frac{\mu}{5}$

$\frac{4}{2} = 2,\quad \frac{5}{-1} = -5$ (not equal)

Use linear combination: Let $a(2,-1,3,5) + b(3,2,-1,7) = (4,5,-\lambda,\mu)$

$2a + 3b = 4$    ...(i)
$-a + 2b = 5$    ...(ii)

From (ii): $a = 2b - 5$

Substitute into (i):

$2(2b - 5) + 3b = 4 \Rightarrow 4b - 10 + 3b = 4 \Rightarrow 7b = 14 \Rightarrow b = 2$

$a = 2(2) - 5 = -1$

Now compute:

$-\lambda = a(3) + b(-1) = -1(3) + 2(-1) = -3 - 2 = -5 \Rightarrow \lambda = 5$

$\mu = a(5) + b(7) = -1(5) + 2(7) = -5 + 14 = 9$