For which values of a and b does the system have a unique solution?

Question

Given the system of equations

x+z=3

ax+y=2

3x + y + bz = 7

Answer 1

We will proceed by using Gaussian elimination on the given system of three linear equations. Adding -a times the first equation to the second equation and adding -3 times the first equation to the third equation gives us

x + z = 3

y - az = 2 - 3a

y + (b - 3)z = 7 - 9 = 2.

Adding -1 times the second equation of the new system of linear equations to the third equation of the new system of linear equations gives us

x + z = 3

y - az = 2 - 3a

(b - 3 + a)z = 2 - (2 - 3a) = 2 - 2 + 3a = 3a.

Since we want our new system of linear equations to have a unique solution, we must have that b - 3 + a ≠ 0 since having b - 3 + a = 0 would make the new system of linear equations either have no solution or an infinite number of solutions. When b - 3 + a ≠ 0, then choosing any values of a and b such that b - 3 + a ≠ 0 will lead to a unique solution of the original system of linear equations. Specifically, we will have that z = 3a/(b - 3 + a), y = 2 - 3a + a[3a/(b - 3 + a)] = 2 - 3a + 3a²/(b - 3 + a), and x = 3 - 3a/(b - 3 + a) will be the unique solution to the original system of linear equations whenever values for a and b are chosen such that b - 3 + a ≠ 0.