# Math Help - simultaneous equations/matrices

1. ## simultaneous equations/matrices

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

2. Originally Posted by Oasis1993
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

$\displaystyle \begin{bmatrix}
a& 3\\
2 & a-1
\end{bmatrix}$

so to that matrix to be singular it's determinant must be equal to zero .....

$D = \begin{vmatrix}
a&3 \\
2 & a-1
\end{vmatrix} = 0$

so you have

$D = \begin{vmatrix}
a&3 \\
2 & a-1
\end{vmatrix} = a\cdot (a-1) - 3\cdot 2 = a^2-a-6 =(a-3)(a+2)$

now just solve system of equations

$ax+3y=2a$
$2x +(a-1)y = a^2$

in any way that you know to do .... (do you know how to do that ?)

3. thanks!
I do know how to solve the equation however i can't get the answer correct..

My book says that if we use 3 there's no solution but i can't seem to continue with -2 either...

thanks

4. Originally Posted by Oasis1993
thanks!
I do know how to solve the equation however i can't get the answer correct..

My book says that if we use 3 there's no solution but i can't seem to continue with -2 either...

thanks
$ax+3y=2a$
$2x +(a-1)y = a^2$

for a= -2
$-2x+3y=-4$
$2x -3y = 4$

Edit : here you have :

two same equations :

$2x-3y= 4$
$2x -3y = 4$

so you have

$4x-6y = 8$

so you have one equation with 2 unknowns....
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)

for a = 3
$3x+3y=6$
$2x +2y = 9$

Edit : here you have :

$x+y = 2$
$x+y = 4.5$

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