Find the values of the parameters a and p for which the syst

Find the values of the parameters a and p for which the system {x + 5y = 6 6y + az = a + 1 - p, x - ay + z = p + 2, has: (i) a unique solution (ii) infinitely many solutions, (iii) no solutions.

Solution

The coefficient matrix of the given linear system is A =

1

5

0

0

6

a

1

-a

1

Now, det(A) = a²+5a+6 = a²+2a+3a+6= a(a+2)+3(a+2) = (a+2)(a+3). We know that a linear system of 3 equations in 3 variables has a unique solution if the coefficient matrix of the system is invertible. Here, A is invertible if det(A) 0, i.e. if a -2 or -3. Thus, the given system has a unique solution if a -2 or -3. If the coefficient matrix of the system is not invertible, then the system has either no solution or infinite solutions. Thus, if a =-2 or, a = -3, the the given linear system has either no solution or infinite solutions.

The augmented matrix of the given linear system is B =

1

5

0

6

0

6

a

a+1-p

1

-a

1

p+2

To solve the given linear system of equations, we will reduce B to its RREF as under:

3.Add (a+5) time 2^nd row to the3rd row;

4.Multiply the 3^rd row by 6/(a²+5a+6)

Then the RREF of B is

1

0

0

(6a²-5ap-5a+5p+31)/(a²+5a+6)

0

1

0

(ap+7a-p+1)/( a²+5a+6)

0

0

1

(a²-ap+6a-11p-31)/( a²+5a+6)

In the 4^th row operation above, we have multiplied the 3^rd row by 6/(a²+5a+6). This is possible only if (a²+5a+6)0 i.e. if a -2 or -3. If, a -2 or -3, then B has rank 3 so that Nullity(B) =0 and there are no free variables. This means that there is a unique solution in such cases.

1	5	0
0	6	a
1	-a	1