## 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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