How would I solve for this variable so that the equation has no unique solution?

Find the value of a for which the following system of equations does not have a unique solution.

4x-y+2z=1
2x+3y=-6
x-2y+ az=3.5

1 Answer
Somebody N.
Apr 11, 2018

#color(blue)(a=1)#

Explanation:

Let #bbA# be the coefficient matrix of the system:

#bbA=[(4,-1,2),(2,3,0),(1,-2,a)]#

For the system:

#bb(AX)=bb(b)#

This has a unique solution if #bb(A^-1)# exists. If #bb(A^-1)# dosen't exist, and the system is consistent then there will be an infinite number of solutions.

#bb(A^-1)# dosen't exist if:

#|bbA|=0#

We can use this idea to find #a#.

We need to find the determinant of #bbA#:

#|bbA|=|(4,-1,2),(2,3,0),(1,-2,a)|#

Expanding about row 1:

#4(3a-(-2(0)))+(2a-(1(0)))+2(2(-2)-(1(3)))#

#12a+2a-14#

#14a-14#

This needs to equal zero:

#14a-14=0=>a=1#

So:

#bbA=[(4,-1,2),(2,3,0),(1,-2,1)]#

And the system:

#4x-y+2z=1#

#2x+3y \ \ \ \ \ \ \ \=-6#

#x \ \ -2y+z=3.5#

Has infinitely many solutions:

Related questions
Impact of this question
430 views around the world
You can reuse this answer
Creative Commons License