If the system of equations,a^2 x - ay = 1 - a | vedclass

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(A) Given the system of equations:$a^2 x - ay = 1 - a$  $(1)$$bx + (3 - 2b) y = 3 + a$  $(2)$Since $(x, y) = (1, 1)$ is a solution,substitute these va

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$a = 1, b = -1$

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(A) Given the system of equations:$a^2 x - ay = 1 - a$  $(1)$$bx + (3 - 2b) y = 3 + a$  $(2)$Since $(x, y) = (1, 1)$ is a solution,substitute these values into both equations:For equation $(1)$: $a^2(1) - a(1) = 1 - a \implies a^2 - a = 1 - a \implies a^2 = 1 \implies a = \pm 1$.For equation $(2)$: $b(1) + (3 - 2b)(1) = 3 + a \implies b + 3 - 2b = 3 + a \implies 3 - b = 3 + a \implies b = -a$.Case $I$: If $a = 1$,then $b = -1$. Check for uniqueness: The determinant of the coefficient matrix is $D = \begin{vmatrix} a^2 & -a \ b & 3 - 2b \end{vmatrix} = \begin{vmatrix} 1 & -1 \ -1 & 5 \end{vmatrix} = 5 - 1 = 4 
eq 0$. Thus,$a = 1, b = -1$ gives a unique solution.Case $II$: If $a = -1$,then $b = 1$. Check for uniqueness: $D = \begin{vmatrix} (-1)^2 & -(-1) \ 1 & 3 - 2(1) \end{vmatrix} = \begin{vmatrix} 1 & 1 \ 1 & 1 \end{vmatrix} = 1 - 1 = 0$. Since $D = 0$,the system does not have a unique solution for $a = -1, b = 1$.Therefore,the only valid solution is $a = 1, b = -1$.

Market	$x, y, z$
$I$	$10,000, 2,000, 18,000$
$II$	$6,000, 20,000, 8,000$

If the system of equations,$a^2 x - ay = 1 - a$ and $bx + (3 - 2b) y = 3 + a$ possess a unique solution $x = 1, y = 1$,then:

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