Question

For which values of a do the pair of linear equations ax+y=a^2 and x+ay=1 have no solution, infinitely many solution,and unique solution​

Answer 1

concepts : if $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ are two linear equations.

equations have no solution when

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

equations have infinitely many solutions when $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

equations have an unique solution when $\frac{a_1}{a_2}\neq\frac{b_1}{b_2}$

here equations are ; ax + y - a² = 0 and x + ay -1 = 0

for no solution

a/1 = 1/a ≠ -a²/-1

or, if a = -1 then, -1/1 = 1/-1 ≠ -(-1)²/-1

hence, a = -1

for infinitely solutions

a/1 = 1/a = -a²/-1

or, a/1 = 1/a = a²/1

a/1 = a²/1 => a(a - 1) = 0, a = 0, 1

1/a = a²/1 => a³ -1 = 0, a = 1

common value of a is 1

so, value of a will be 1

for an unique solution

a/1 ≠ 1/a

or, a² ≠ 1

or, (a² - 1) ≠ 0

or, a ≠ -1 and 1

hence, a belongs to all real numbers except 1 and -1

we can write it , $a\in\mathbb{R}-\{-1,1\}$

Answer 2

Answer:

For no solution,

a is equal to -1.

For infinitely many solution,

a is equal to 1.

For unique solution,

a is equal to-1,1.