x1+ (a-1)x2+x4=0 (а-2)x1-ах2-x4 a1+(a-1)a2+ax3+a4=-1 axi+(a–1)a2+(a+4)x3+x4=0 prove may be solved by Cramer's Method.

Question

x1+ (a-1)x2+x4=0 

(а—2)x1 - ax2-x4=1

x1+(a-1)x2+ax3+x4=-1 

ax1+(a–1)x2+(a+4)x3+x4=0 

may be solved by Cramer's Method.

Justify

Answer 1

Cramer's Method uses a determinant of coefficients which must not be zero if there's to be a solution. So let's check:

⎡  1   a-1   0   1 ⎤

⎢ a-2  -1    0  -1⎥

⎢  1   a-1   a   1⎥

⎣  a   a-1 a+4 1⎦

To evaluate this we need to evaluate 3 3×3 determinants:

⎡ -1     0  -1⎥

⎢a-1    a   1⎥= -1(-4)-1[(a-1)(a+4)-a²+a]=4-(4a-4)=8-4a, let A=8-4a,

⎣ a-1 a+4 1⎦

⎡ a-2   0  -1⎤

⎢  1     a   1⎥=(a-2)(-4)-(-a²+a+4)=-4a+8+a²-a-4=a²-5a+4, let B=a²-5a+4,

⎣  a   a+4 1⎦

⎡ a-2  -1    0  ⎤

⎢  1   a-1   a  ⎥=(a-2)(4a-4)+a+4-a²=3a²-11a+12, let C=3a²-11a+12.

⎣  a   a-1 a+4⎦

Now we can evaluate Cramer's determinant: A-(a-1)B-C=8-4a-(a-1)(a²-5a+4)-3a²+11a-12=

8-4a-a³+5a²-4a+a²-5a+4-3a²+11a-12=-a³+3a²-2a=-a(a²-3a+2)=-a(a-2)(a-1).

The determinant is zero when a=0, 1 or 2, and under these conditions Cramer's Method would find inconsistency in the system of equations (or many solutions).

It follows that if a is none of these values, Cramer's Method would find a unique solution.