## Linear Algebra and Its Applications, exercise 1.2.3

Exercise 1.2.3. For the following three equations (in 4-space):

$\setlength\arraycolsep{0.2em}\begin{array}{rrrrrrrrl} u&+&v&+&w&+&z&=&6 \\ u&+&&&w&+&z&=&4 \\ u&+&&&w&&&=&2 \end{array}$

what is the intersection of the three planes described by the equations? If we add a fourth equation u = -1, describe the intersection of the plane associated with this equation with the previous three planes.

Answer: First, subtract (u + w) from both sides of the first two equations to obtain the following

$\setlength\arraycolsep{0.2em}\begin{array}{rcl} v + z&=&6 - (u + w) \\ z&=&4 - (u + w) \\ u + w&=&2 \end{array} \Rightarrow \begin{array}{rcl} v&=&6 - z - (u + w) \\ z&=&4 - (u + w) \\ u + w&=&2 \end{array}$

Then substitute the value of (u + w) (from the third equation) into the first two equations:

$\setlength\arraycolsep{0.2em}\begin{array}{rcl} v&=&4 - z \\ z&=&2 \\ u + w&=&2 \end{array}$

Solve for w in the third equation, and substitute the value of z into the first equation:

$\setlength\arraycolsep{0.2em}\begin{array}{rcl} v&=&2 \\ z&=&2 \\ w&=&-u + 2 \end{array}$

Since v and z have constant values, this corresponds to a line in the u-w plane in 4-space.

If we also assume the equation u = -1 then we have w = 3 by virtue of the third equation. This means that the four planes (in 4-space) corresponding to the four equations intersect in the point (-1, 2, 3, 2).

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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