If the system of equations x + ay = 0, az + y | vedclass

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(A) The given system of equations is homogeneous: $x + ay + 0z = 0$ $0x + y + az = 0$ $ax + 0y + z = 0$ For a homogeneous system to have infinite solu

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$-1$

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(A) The given system of equations is homogeneous: $x + ay + 0z = 0$ $0x + y + az = 0$ $ax + 0y + z = 0$ For a homogeneous system to have infinite solutions (non-trivial solutions),the determinant of the coefficient matrix must be zero: $\begin{vmatrix} 1 & a & 0 \ 0 & 1 & a \ a & 0 & 1 \end{vmatrix} = 0$ Expanding the determinant along the first row: $1(1 	imes 1 - a 	imes 0) - a(0 	imes 1 - a 	imes a) + 0(0 	imes 0 - 1 	imes a) = 0$ $1(1) - a(-a^2) + 0 = 0$ $1 + a^3 = 0$ $a^3 = -1$ Taking the cube root on both sides,we get $a = -1$.

List-$I$	List-$II$
$(A)$ $\lambda=8, \mu \neq 15$	$1$. Infinitely many solutions
$(B)$ $\lambda \neq 8, \mu \in R$	$2$. No solution
$(C)$ $\lambda=8, \mu=15$	$3$. Unique solution

If the system of equations $x + ay = 0,$ $az + y = 0$ and $ax + z = 0$ has infinite solutions,then the value of $a$ is

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