Suppose that the augmented matrix of

Irum Tariq

Irum Tariq

Answered question

2022-05-08

        Suppose that the augmented matrix of a linear equations has been partially reduced using elementary row operations to

 

1135
011r
0s12

Find all the value(s) (if any) of r and s for which the given system has:

(a)   a unique solution,       (b)  an infinitely many solutions,    (c)    No solutions.

Answer & Explanation

Jazz Frenia

Jazz Frenia

Skilled2023-05-05Added 106 answers

We are given the following partially reduced augmented matrix:
[1135011r0s12]
We want to determine the values of r and s for which the system of linear equations corresponding to this augmented matrix has a unique solution, infinitely many solutions, or no solutions.
(a) To have a unique solution, we need to have all variables as leading variables, which means that there can be no free variables. From the given matrix, we can see that the first two columns have leading variables, but the third column does not. Therefore, we need to ensure that the coefficient in the third column for the third row is non-zero. This means that we must have s0. If s=0, then the third row reduces to 0=2, which is a contradiction. Thus, we have a unique solution if and only if s0.
(b) To have infinitely many solutions, we need to have at least one free variable. From the given matrix, we can see that the variable x3 is a free variable. Therefore, we need to ensure that the coefficient in the third column for the second row is equal to the coefficient in the third column for the third row. This means that we must have s=1. If s1, then we would have a leading variable in the third column, which means that x3 would not be a free variable. Thus, we have infinitely many solutions if and only if s=1.
(c) To have no solutions, we need to have a contradiction in the equations. This can happen if the augmented matrix reduces to a row of the form [00a], where a0. From the given matrix, we can see that if s=2 and r3, then the third row reduces to 0=1, which is a contradiction. Thus, we have no solutions if and only if s=2 and r=3.
In summary:
- The system has a unique solution if and only if s0.
- The system has infinitely many solutions if and only if s=1.
- The system has no solutions if and only if s=2 and r=3.

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