Hello Everyone,
a) Find the values of a for which the matrix ( a 3 ) is singular.
2 a-1
b) for these values of a solve the equation ax+3y=2a
2x+(a-1)y= a square
Thanks for your time!
for a= -2
you have infinity of the solutions and equations are dependent (meaning for each new x or y you have another solution for that second unknown)Edit : here you have :
two same equations :
so you have
so you have one equation with 2 unknowns....
for a = 3
Edit : here you have :
now here it's easy to see there are no solutions because you have ones that x+y = 2 and another saying that x+y = 4.5 which cant be true for the both of the equations
Edit: hm... this is moved to the university sub forum ( I think it was in pre-algebra at the start .... perhaps i'm wrong but ... ) , well if here you must show that based on determinants of that system you don't have / have solutions and if you have why and which , and if don't why ... and so on ... there are few cases by which you conclude this