Your n times n, color(red)(A) color(blue)( mathbf x) = color(green)(mathbf b), system in matrix form is:
color(red)( ((1,2,-3), (3,-1,5), (4, 1, a^2 - 14)) ) color(blue)( ((x), (y),(z)) )= color(green)( ((4), (2),(a+2)) )
When there is no obvious linear pattern - in terms of the row and column vectors dependence - the easiest thing to do is to start row reducing. I cannot get Gaussian elimination working on Socratic so will just stack the numbers up in brackets:
((1,2,-3), (3,-1,5), (4, 1, a^2 - 14)) ((4), (2),(a+2))
R2 to R2 - 3R1, R3 to R3 - 4 R1
((1,2,-3), (0,-7,14), (0, -7, a^2 - 2)) ((4), (-10),(a-14))
R3 to R3 - R2
((1,2,-3), (0,-7,14), (0, 0, a^2 - 16)) ((4), (-10),(a-4))
So we have this:
((1,2,-3), (0,-7,14), (0, 0, a^2 - 16)) ((x), (y),(z))= ((4), (-10),(a-4))
If we start at the bottom:
(a^2 - 16)z = a-4
implies (a-4)(a + 4)z = a-4
So we go through the scenarios:
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If a = 4, then the solution is everything in \mathcal R^2. There are an infinite number of solutions for z.
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If a ne 4, that last equation becomes: (a + 4)z = 1 implies z = 1/(a + 4). In that event, x and y are just linear combinations of z.
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However, if a = -4, there is no solution to the matrix equation.
We can explore this a bit more.
If a = 4, we have this:
((1,2,-3), (0,-7,14), (0, 0, 0)) ((x), (y),(z))= ((4), (-10),(0)). Look at the bottom row !
If a = -4, we have this:
((1,2,-3), (0,-7,14), (0, 0,0)) ((x), (y),(z))= ((4), (-10),(-8)). Again, look at the bottom row. This version of a that insists that z = -8.
