The system of equations
$x-3y-8z=-10$
$2x+5y + λz = 13$
$3x + y -4z = 0$
has infinite number of solutions if the value of A is equal to:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 6 **
Given system:
$x - 3y - 8z = -10$
$2x + 5y + \lambda z = 13$
$3x + y - 4z = 0$
For infinite solutions: determinant of coefficient matrix = 0 and consistency must hold.
Coefficient matrix:
$\begin{vmatrix} 1 & -3 & -8\\ 2 & 5 & \lambda\\ 3 & 1 & -4 \end{vmatrix}$
Compute determinant:
$1\begin{vmatrix}5 & \lambda\\ 1 & -4\end{vmatrix} - (-3)\begin{vmatrix}2 & \lambda\\ 3 & -4\end{vmatrix} + (-8)\begin{vmatrix}2 & 5\\ 3 & 1\end{vmatrix}$
= $1(5(-4) - \lambda\cdot 1)$ + $3(2(-4) - \lambda\cdot 3)$ + $-8(2\cdot 1 - 5\cdot 3)$
= $1(-20 - \lambda)$ + $3(-8 - 3\lambda)$ + $-8(2 - 15)$
= $-20 - \lambda + 3(-8 - 3\lambda) -8(-13)$
= $-20 - \lambda - 24 - 9\lambda + 104$
= $60 - 10\lambda$
Set determinant = 0:
$60 - 10\lambda = 0$
$\lambda = 6$
Final Answer: $\lambda = 6$