## 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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