The given system of equations is:

1) x + y - z = 1
2) x + 3y + kz = 3
3) x + ky + 3z = 2

We can write this system in augmented matrix form as:

[ 1 1 -1 | 1 ]
[ 1 3 k | 3 ]
[ 1 k 3 | 2 ]

We can start the Gaussian elimination by subtracting the first row from the second and third rows to get:

[ 1 1 -1 | 1 ]
[ 0 2 k+1 | 2 ]
[ 0 k-1 4 | 1 ]

(a) For the system to have no solutions, the second and third rows must be multiples of each other, but the constants on the right side of the equation must not be. This happens when k = 3. In this case, the second and third rows become [0 2 4 | 2] and [0 2 4 | 1], which are multiples of each other, but the constants on the right side are not, so there is no solution.

(b) For the system to have a unique solution, the second and third rows must not be multiples of each other. This happens when k ≠ 3.

(c) For the system to have infinitely many solutions, the second and third rows must be multiples of each other, including the constants on the right side of the equation. This never happens for any value of k, so the system never has infinitely many solutions.

(d) For k ≠ 3, we can continue the Gaussian elimination to find the unique solution. Subtract (k-1) times the second row from the third row to get:

[ 1 1 -1 | 1 ]
[ 0 2 k+1 | 2 ]
[ 0 0 4-k | 1-2(k-1) ]

Then, divide the second row by 2 and the third row by 4-k to get:

[ 1 1 -1 | 1 ]
[ 0 1 (k+1)/2 | 1 ]
[ 0 0 1 | (1-2(k-1))/(4-k) ]

Finally, add the third row to the first row, and subtract (k+1)/2 times the third row from the second row to get:

[ 1 1 0 | 1+(1-2(k-1))/(4-k) ]
[ 0 1 0 | 1-(k+1)(1-2(k-1))/(4-k) ]
[ 0 0 1 | (1-2(k-1))/(4-k) ]

This gives the unique solution (x, y, z) = (1+(1-2(k-1))/(4-k), 1-(k+1)(1-2(k-1))/(4-k), (1-2(k-1))/(4-k)).

Question

The given system of equations is:

1) x + y - z = 1
2) x + 3y + kz = 3
3) x + ky + 3z = 2

We can write this system in augmented matrix form as:

[ 1  1 -1 | 1 ]
    [ 1  3  k | 3 ]
    [ 1  k  3 | 2 ]

We can start the Gaussian elimination by subtracting the first row from the second and third rows to get:

[ 1  1 -1 | 1 ]
    [ 0  2  k+1 | 2 ]
    [ 0  k-1 4 | 1 ]

(a) For the system to have no solutions, the second and third rows must be multiples of each other, but the constants on the right side of the equation must not be. This happens when k = 3. In this case, the second and third rows become [0 2 4 | 2] and [0 2 4 | 1], which are multiples of each other, but the constants on the right side are not, so there is no solution.

(b) For the system to have a unique solution, the second and third rows must not be multiples of each other. This happens when k ≠ 3.

(c) For the system to have infinitely many solutions, the second and third rows must be multiples of each other, including the constants on the right side of the equation. This never happens for any value of k, so the system never has infinitely many solutions.

(d) For k ≠ 3, we can continue the Gaussian elimination to find the unique solution. Subtract (k-1) times the second row from the third row to get:

[ 1  1 -1 | 1 ]
    [ 0  2  k+1 | 2 ]
    [ 0  0  4-k | 1-2(k-1) ]

Then, divide the second row by 2 and the third row by 4-k to get:

[ 1  1 -1 | 1 ]
    [ 0  1  (k+1)/2 | 1 ]
    [ 0  0  1 | (1-2(k-1))/(4-k) ]

Finally, add the third row to the first row, and subtract (k+1)/2 times the third row from the second row to get:

[ 1  1  0 | 1+(1-2(k-1))/(4-k) ]
    [ 0  1  0 | 1-(k+1)(1-2(k-1))/(4-k) ]
    [ 0  0  1 | (1-2(k-1))/(4-k) ]

This gives the unique solution (x, y, z) = (1+(1-2(k-1))/(4-k), 1-(k+1)(1-2(k-1))/(4-k), (1-2(k-1))/(4-k)).

Knowee AI · Accepted Answer