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!
$\displaystyle \displaystyle \begin{bmatrix}
a& 3\\
2 & a-1
\end{bmatrix} $
so to that matrix to be singular it's determinant must be equal to zero .....
$\displaystyle D = \begin{vmatrix}
a&3 \\
2 & a-1
\end{vmatrix} = 0$
so you have
$\displaystyle D = \begin{vmatrix}
a&3 \\
2 & a-1
\end{vmatrix} = a\cdot (a-1) - 3\cdot 2 = a^2-a-6 =(a-3)(a+2) $
now just solve system of equations
$\displaystyle ax+3y=2a $
$\displaystyle 2x +(a-1)y = a^2 $
in any way that you know to do .... (do you know how to do that ?)
$\displaystyle ax+3y=2a $
$\displaystyle 2x +(a-1)y = a^2 $
for a= -2
$\displaystyle -2x+3y=-4 $
$\displaystyle 2x -3y = 4 $
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 :
$\displaystyle 2x-3y= 4 $
$\displaystyle 2x -3y = 4 $
so you have
$\displaystyle 4x-6y = 8 $
so you have one equation with 2 unknowns....
for a = 3
$\displaystyle 3x+3y=6 $
$\displaystyle 2x +2y = 9 $
Edit : here you have :
$\displaystyle x+y = 2 $
$\displaystyle x+y = 4.5 $
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