Need help with matrix algebra, should be simple in theory?

Question

Find the line of intersection of the following planes:

3x +2y +z = -1
2x -y + 4z = 5

To my understanding, in theory this should be a simple augmented matrix that just needs to be row reduced. But no matter how I try to do it, I get the following answer:

x = -9t/7 + 9/7
y = -10t/7 - 17/7
z = t

And the correct answer is supposedly:

x = 9t
y = -10t -1
z = -7t +1

I'm extremely frustrated here, can anyone lend a hand?

PaulR2 · Accepted Answer

(The ··· is just to space stuff out.)

The t's will have to cancel out so what if we had:

[ 3 ··· 2 ··· 1 ··· ¦ 0 ]
[ 2 ·· -1 ··· 4 ··· ¦ 0 ]

-R2 + R1

[ 1 ··· 3 ··· -3 ··· ¦ 0 ]
[ 2 ·· -1 ···· 4 ··· ¦ 0 ]

-2R1 + R2

[ 1 ··· 3 ··· -3 ··· ¦ 0 ]
[ 0 ·· -7 ·· 10 ··· ¦ 0 ]

(3/7)R2 + R1, then do (-1/7)R2

[ 1 ··· 0 ····· (9/7) ··· ¦ 0 ]
[ 0 ··· 1 ·· (-10/7) ··· ¦ 0 ]

So now we have for the ratio of coefficients:

x = (-9/7)z
y = (10/7)z

Multiply to get integers

7x = -9z
7y = 10z

Therefore:

x = -9t + A, or x = 9t + A
y = 10t + B, or y = -10t + B
z = 7t + C, or z = -7t + C

Where A, B, C, are constants that may or may not be equal. This was just a thought on how to get the coefficients for t. Maybe from here you can find a way to get the constants but I'm not seeing one.

Josh Swanson · Answer

The two answers are the same (up to a probable typo), they're just different parameterizations of the same line. Using your first parameterization, replace t with 1-7s, which gives

x = -9(1-7s)/7 + 9/7
= -9/7 + 9s + 9/7 = 9s

y = 10(1-7s)/7 - 17/7
= 10/7 - 10s - 17/7 = -10s - 1

z = 1-7s
= -7s + 1

(I've changed y=-10t/7 - 17/7 to y=10t/7 - 17/7.)

You can check these points lie on both planes, so regardless of the derivation, the answer is correct.

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Need help with matrix algebra, should be simple in theory?

2 Answers