System of linear equation?

a)
determinant of the linear system
= -a^2-35a+78-(13a^2+15+14a)
=-14a^2-49a+63
=-7(2a+9)(a-1), where a≠ -9/2 , 1 because the linear system has an unique solution.

By using cramer's rule (refer to the photo)
x = -a^2-10a +36 -6a^2 -15 -4a / Δ
= 7a^2 -14a + 21 / Δ
= -7(a+3)(a-1) / -7 (2a+9)(a-1)
= (a+3)(2a+9)

y= -2a+42a-39-26a+18+7a /Δ
=21a-21/(2a+9)(a-1)
= -3/ (2a+9)

z=6a-35-52-13a+10+84 /Δ
= -7a+7 /-7 (2a+9)(a-1)
= 1/(2a+9)

b)
from (a), when a=1, it has no unique solution

1 -2 1 1
7 1 -3 2
13 -5 -1 6

1 -2 1 1
0 3 -2 -1
0 3 -2 -1

1 -2 1 1
0 3 -2 -1
0 0 0 0

0=0, it has infinite many solutions
Let z = t
3y - 2z = -1
y = 2t-1 /3

x-2((2t-1)/3) +t =1
x=(1+t)/3

p/3 (1+t)(2t-1)/3 + (2t-1)/3t + t/3(1+t)>0
p(1+t)(2t-1)+(6t-3)t+3t+3t^2>0
(9+2p)t^2 + pt -p >0

so t has no roots and (9+2p) > 0 , p> -9/2
therefore,
Δ<0
p^2-4(-p)(9+2p)<0
p^2+8p^2+36p<0
9p(p+4)<0
When p<-4 and p>0 , p (p+4)>0
When -4<p<0. p(p+4)<0
It also agree with p > -9/2
so
-4<p<0

Try to do question 2 yourself. First see if it has an unique sol. If no, see if it has infinite many or no solution as above. If it has infinite sol. Express x,y,z in t. Then substitute them into x^2 +3y^2 +4z^2. After that you can get an equation with at^2 + bt +c , by using -b/2a in compulsory maths to find t, then find x y z.

3. Consistent linear equation means that there is at least one solution.
1 -1 -1 4
-2 3 7 -9
3 -2 2 11

1 -1 -1 4
0 1 5 -1
0 0 0 0

0=0
there is infinitely many solutions
z=t
y=-1 - 5f
x=3-4t
If ax+by+cz=d,
a(3-4t)+b(-1-5t)+ct = d
t(4a+5b-c)=(3a-b-d)
So
1. 3a-b-d =0
or
2. 3a-b-d ≠ 0 and 4a+5b-c≠ 0
The system will be consistent.



[[[[[[[[(no need to write the following in exam)
Remind that when 3a-b-d ≠ 0 and 4a+5b-c = 0 , t = (3a-b-d)/0 which is undefined and thus no solution.
So the first case, when 3a-b-d ≠ 0 , 4a+5b-c ≠ 0, which forms an unique solution.
The second case, when 3a-b-d =0 and 4a+5b-c ≠ 0, t =0 which forms an unique solution.
The third case ,when 3a-b-d =0,4a+5b-c = 0 ,t can be any real number which is also consistent.]]]]]]]]
so, 1. 3a-b-d =0
or
2. 3a-b-d ≠ 0 and 4a+5b-c≠ 0
The system will be consistent.