# Given the system, what is the determinant D(a)? x+y+z=1 x+2y+4z=2 x+ay+a^2*z=3 Calculate D(2^x)=0.

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The determinant of the system of equations is formed by the coefficients of x, y and z which is a function of a.

Therefore the determinant D(a)= | [(1 1 1),( 1 2 4),(1 a a^2 )|

detrminant D(a) = 1 (2a^2 -4a) - 1(a^2 -4) +1(a-2)

D(a) = 2a^2-4a-a^2+4 -a-2

= a^2 -5a +4

Therefore D(a) = a^2-5a+4

D(a) = (a-1)((a-4).

Now to find the solution of D(2^x):

(This x should not be confused up with the x in the system of given equations).

Therefore D(2^x) = 0 gives:

(2^x-1) ((2^x-4) = 0.

So 2^x = 1 = 0 .

or 2^x =2^0 .

So x = 0.

2^x -4 = 0 gives 2^x = 2^2.

So x = 2.

Therefore x = 0 and x = 2.

The determinant of thesystem is formed from the coefficients of the variables x, y, z.

|1 1 1|

D(a) = |1 2 4|

|1 a a^2|

D(a) = 1*2*a^2 + 1*a*a + 1*4*1 - 1*2*1 - 4*a*1 - 1*1*a^2

D(a) = 2a^2 + a^2 + 4 - 2 - 4a - a^2

We'll combine like terms:

D(a) = 2a^2 - 4a + 2

Now, we'll calculate D(2^x), substituting a by 2^x in the expression of D(a):

D(2^x) = 2(2^x)^2 - 4(2^x) + 2

We'll note 2^x = a

2a^2 - 4a + 2 = 0

We'll divide by 2:

a^2 - 2a + 1 = 0

(a-1)^2 = 0

a = 1

2^x = 1

We'll write 1 as a power of 2:

2^x = 2^0

Since the bases are matching, we'll apply one to one property:

**x = 0**