What is the matrix in the reduced echelon form for the equations of the planes: x+4y+z=5 ; x-4y+3z=-1 ; 2x+8y+2z= 6. How many solution(s) are there?

1 Answer | Add Yours

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The equations of the planes are x+4y+z=5, x-4y+3z=-1 and 2x+8y+2z= 6.

In matrix form the equation of the three planes would be:

`[[1, 4, 1],[1, -4, 3],[2,8, 2]]*[[x],[y],[z]]=[[5],[-1],[6]]`

The three planes do not intersect at a point as two of them are parallel to each other. There is no solution.

The matrix `[[1, 4, 1],[1, -4, 3],[2, 8, 2]] ` can be converted to the reduced echelon form by the following steps.

Multiply the 1st row by 1 and add to row 2. This gives:

`[[1, 4, 1],[0, -8, 2],[2, 8, 2]]`

Multiply the 1st row by -2 and add to row 3. This gives:

`[[1, 4, 1],[0, -8, 2],[0, 0, 0]]`

Multiply the 2nd row by -1/8. This gives:

`[[1, 4, 1],[0, 1, -1/4],[0, 0, 0]]`

Multiply row 3 my -4 and add to row 2. This gives:

`[[1, 0, 2],[0, 1, -1/4],[0, 0, 0]]`

The matrix in the reduced echelon form is:

`[[1, 0, 2],[0, 1, -1/4],[0, 0, 0]]`

We’ve answered 318,926 questions. We can answer yours, too.

Ask a question