The given system of equations is:

1. x + y + z = 1
2. x + (a + 2)y + 2z = 5
3. x + 7y + (a + 1)z = 9

We can solve this system using the method of substitution or elimination, but since we are looking for the values of 'a' that will give us no solution, a unique solution, or infinitely many solutions, it's more efficient to use the determinant method.

The determinant of a system of linear equations will tell us the nature of the solutions. If the determinant is zero, the system has either no solution or infinitely many solutions. If the determinant is non-zero, the system has a unique solution.

The determinant (D) of the coefficients of this system is:

D = | 1 1 1 |
| 1 a+2 2 |
| 1 7 a+1 |

Calculating the determinant, we get:

D = 1(a+2)(a+1) + 1*2*1 + 1*7*1 - 1*7*2 - 1*(a+2)*1 - 1*1*(a+1)
= a^2 + 3a + 2 + 2 + 7 - 14 - a - 2 - a - 1
= a^2 + a - 4

(a) For the system to have no solution, the determinant of the system should be zero and at least one of the determinants obtained by replacing a column with the constants should be non-zero.

Setting D = 0, we get:

a^2 + a - 4 = 0

Solving this quadratic equation, we get:

a = [-1 ± sqrt(1 + 4*4)] / 2
= [-1 ± sqrt(17)] / 2
= -1/2 ± sqrt(17)/2

So, the system has no solution for a = -1/2 ± sqrt(17)/2, provided at least one of Dx, Dy, Dz is non-zero. Dx, Dy, Dz are determinants obtained by replacing x, y, z columns with constants respectively.

(b) For the system to have a unique solution, the determinant of the system should be non-zero. So, the system has a unique solution for all real numbers a except a = -1/2 ± sqrt(17)/2.

(c) For the system to have infinitely many solutions, the determinant of the system should be zero and all the determinants obtained by replacing a column with the constants should also be zero. So, the system has infinitely many solutions for a = -1/2 ± sqrt(17)/2, provided Dx = Dy = Dz = 0.

Question

The given system of equations is:

1. x + y + z = 1
2. x + (a + 2)y + 2z = 5
3. x + 7y + (a + 1)z = 9

We can solve this system using the method of substitution or elimination, but since we are looking for the values of 'a' that will give us no solution, a unique solution, or infinitely many solutions, it's more efficient to use the determinant method.

The determinant of a system of linear equations will tell us the nature of the solutions. If the determinant is zero, the system has either no solution or infinitely many solutions. If the determinant is non-zero, the system has a unique solution.

The determinant (D) of the coefficients of this system is:

D = | 1 1 1 |
    | 1 a+2 2 |
    | 1 7 a+1 |

Calculating the determinant, we get:

D = 1(a+2)(a+1) + 1*2*1 + 1*7*1 - 1*7*2 - 1*(a+2)*1 - 1*1*(a+1)
  = a^2 + 3a + 2 + 2 + 7 - 14 - a - 2 - a - 1
  = a^2 + a - 4

(a) For the system to have no solution, the determinant of the system should be zero and at least one of the determinants obtained by replacing a column with the constants should be non-zero.

Setting D = 0, we get:

a^2 + a - 4 = 0

Solving this quadratic equation, we get:

a = [-1 ± sqrt(1 + 4*4)] / 2
  = [-1 ± sqrt(17)] / 2
  = -1/2 ± sqrt(17)/2

So, the system has no solution for a = -1/2 ± sqrt(17)/2, provided at least one of Dx, Dy, Dz is non-zero. Dx, Dy, Dz are determinants obtained by replacing x, y, z columns with constants respectively.

(b) For the system to have a unique solution, the determinant of the system should be non-zero. So, the system has a unique solution for all real numbers a except a = -1/2 ± sqrt(17)/2.

(c) For the system to have infinitely many solutions, the determinant of the system should be zero and all the determinants obtained by replacing a column with the constants should also be zero. So, the system has infinitely many solutions for a = -1/2 ± sqrt(17)/2, provided Dx = Dy = Dz = 0.

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