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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$a = 1, b = -1$

Vedclass · Answer

(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 $a = 1, b = -1$.

$List-I$	$List-II$
$(I)$ If $\frac{q}{r}=10$,then the system of linear equations has	$(P)$ $x=0, y=\frac{10}{9}, z=-\frac{1}{9}$ as a solution
$(II)$ If $\frac{p}{r} \neq 100$,then the system of linear equations has	$(Q)$ $x=\frac{10}{9}, y=-\frac{1}{9}, z=0$ as a solution
$(III)$ If $\frac{p}{q} \neq 10$,then the system of linear equations has	$(R)$ infinitely many solutions
$(IV)$ If $\frac{p}{q}=10$,then the system of linear equations has	$(S)$ no solution
	$(T)$ at least one solution

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:

Similar Questions

The number of real values $\lambda$,such that the system of linear equations $2x - 3y + 5z = 9$,$x + 3y - z = -18$,and $3x - y + (\lambda^2 - |\lambda|)z = 16$ has no solution,is :-

If the system of simultaneous linear equations $3x - 4y + kz + 13 = 0$,$x + 2y - z - 9 = 0$,and $kx - y + 3z + 7 = 0$ has a unique solution $x = \alpha, y = \beta, z = \gamma$ for $k \neq m$ and $2\beta - \gamma = 8$,then $\alpha + m =$

If the system of equations $(\lambda-1) x+(\lambda-4) y+\lambda z=5$,$\lambda x+(\lambda-1) y+(\lambda-4) z=7$,and $(\lambda+1) x+(\lambda+2) y-(\lambda+2) z=9$ has infinitely many solutions,then $\lambda^2+\lambda$ is equal to

Let $\alpha, \beta \in R$ be such that the system of linear equations $x+2y+z=5, 2x+y+\alpha z=5, 8x+4y+\beta z=18$ has no solution. Then $\frac{\beta}{\alpha}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module