## Linear Algebra and Its Applications, exercise 1.2.4

Exercise 1.2.4. For the following three equations: $\setlength\arraycolsep{0.2em}\begin{array}{rcrcl} x&+&2y&=&2 \\ x&-&y&=&2 \\ &&y&=&1 \end{array}$

sketch the lines associated with the equation and determine whether the equations have a simultaneous solution. What about if the right-hand sides of all three equations were zero? Are there other right-hand sides for which a solution exists?

Answer: I’ll skip the sketch. Instead, let’s solve the first two equations above by subtracting the second equation from the first. This gives us the equation $(x + 2y) - (x - y) = 2 - 2 \Rightarrow 3y = 0 \Rightarrow y = 0$

Substituting y = 0 into both equations gives x = 2. So the point (2, 0) is a solution to both equations and is the point where the lines corresponding to the first two equations meet. If we then add the third equation y = 1 it’s clear that there is no simultaneous solution for all three equations.

Now set the right-hand sides of all equations to zero: $\setlength\arraycolsep{0.2em}\begin{array}{rcrcl} x&+&2y&=&0 \\ x&-&y&=&0 \\ &&y&=&0 \end{array}$

Substituting the value y = 0 into the first two equations gives x = 0. So the point (0, 0) is a solution to all three equations.

Do other solutions exist? Going back to the original equations, start with the third equation y = 1 and then solve for x in the second equation to get x = 3. Substituting x = 3 and y = 1 in the left hand side of the first equation gives $x + 2y = 3 + 2 \cdot 1 = 5$

So the following equations $\setlength\arraycolsep{0.2em}\begin{array}{rcrcl} x&+&2y&=&5 \\ x&-&y&=&2 \\ &&y&=&1 \end{array}$

have the unique solution (3, 1).

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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