# 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?

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### 1 Answer

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]]`