## Linear Algebra and Its Applications, Review Exercise 1.16

Review exercise 1.16. Given the following system of equations: $\setlength\arraycolsep{0.2em}\begin{array}{rcrcr} kx&+&y&=&1 \\ x&+&ky&=&1 \end{array}$

what must $k$ be for the system to have no solution? One solution? An infinite number of solutions?

Answer: If $k = 0$ then the system reduces to $\setlength\arraycolsep{0.2em}\begin{array}{rcrcr} &&y&=&1 \\ x&&&=&1 \end{array}$

for which $x = 1$ and $y = 1$ is (obviously) a solution.

If $k \ne 0$ then we can multiply the first equation by $1/k$ and subtract it from the second equation to obtain the following system: $\setlength\arraycolsep{0.2em}\begin{array}{rcrcl} kx&+&y&=&1 \\ &&(k - \frac{1}{k})y&=&1 - \frac{1}{k} \end{array}$

If $k = 1$ then this reduces to $\setlength\arraycolsep{0.2em}\begin{array}{rcrcl} 1 \cdot x&+&y&=&1 \\ &&(1 - \frac{1}{1})y&=&1 - \frac{1}{1} \end{array} \rightarrow \begin{array}{rcrcl} x&+&y&=&1 \\ &&0&=&0 \end{array}$

This system has an infinite number of solutions, namely any value of $(x, y)$ for which $y = -x + 1$.

If $k \ne 0$ and $k \ne 1$ then we can solve for $y$ as follows: $y = (1 - \frac{1}{k}) / (k - \frac{1}{k}) = k (1 - \frac{1}{k}) / k(k - \frac{1}{k})$ $= (k - 1) / (k^2 - 1) = (k - 1) / (k - 1)(k + 1) = \frac{1}{k+1}$

We then solve for $x$ as follows: $kx = 1 - y = 1 - \frac{1}{k+1} = \frac{k+1}{k+1} - \frac{1}{k+1}$ $= \frac{k+1-1}{k+1} = \frac{k}{k+1}$ $\rightarrow x = \frac{1}{k} \frac{k}{k+1} = \frac{1}{k+1}$

So for $k \ne 0$ and $k \ne 1$ the system has the unique solution $(\frac{1}{k+1}, \frac{1}{k+1})$. For example, for $k = 2$ the unique solution is $(\frac{1}{2+1}, \frac{1}{2+1}) = (\frac{1}{3}, \frac{1}{3})$.

Note that the solution for $k = 0$ was $(1, 1)$ which matches that given by our formula: $(\frac{1}{0+1}, \frac{1}{0+1}) = (1, 1)$. Also note that there is no value of $k$ for which no solution exists.

Also note that there is no solution for $k = -1$, for which the solution $(\frac{1}{k+1}, \frac{1}{k+1})$ would require dividing by zero. In this case the system of equations is $\setlength\arraycolsep{0.2em}\begin{array}{rcrcr} -x&+&y&=&1 \\ x&-&y&=&1 \end{array} \rightarrow \begin{array}{rcrcr} x&-&y&=&-1 \\ x&-&y&=&1 \end{array}$

and produces the contradiction $-1 = 1$.

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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### 2 Responses to Linear Algebra and Its Applications, Review Exercise 1.16

1. Steve says:

I agree with the answers for one solution and infinitely many solutions, but I do not agree with your statement “that there is no value of k for which no solution exists.” If one sets k equal to -1, then x and y are undefined, for you are dividing by zero. Wouldn’t k = -1 produce no solution for the system?

• hecker says:

You are absolutely correct; thanks for catching that! I’ve corrected the answer.