Linear Algebra and Its Applications, Review Exercise 2.15

Review exercise 2.15. For each of the following, find a matrix A for which

i) there is either no solution or one solution to Ax =b depending on b

ii) there are an infinite number of solutions to Ax = b for all b

iii) there is either no solution or an infinite number of solutions to Ax =b depending on b

iv) there is exactly one solution for Ax = b for any b

Answer: i) The following system of two equations in two unknowns

\begin{array}{rcrcr} x_1&&&=&b_1 \\ &&x_2&=&b_2 \end{array}

obviously has a solution no matter the value of b. However if we add a third equation as follows

\begin{array}{rcrcr} x_1&&&=&b_1 \\ &&x_2&=&b_2 \\ x_1&+&x_2&=&b_3 \end{array}

then the resulting system may or may not have a solution depending on the value of b. In particular, in order for the system to have a solution we must have b_3 = b_1 + b_2 in which case the single solution is x = (b_1, b_2).

The system above corresponds to Ax = b where

A = \begin{bmatrix} 1&0 \\ 0&1 \\ 1&1 \end{bmatrix}

ii) In order for a system to have an infinite number of solutions we can specify more unknowns than there are equations, so that some variables are free variables that can take on any value. For example, if we start again with the system of two equations in two unknowns

\begin{array}{rcrcr} x_1&&&=&b_1 \\ &&x_2&=&b_2 \end{array}

we can add another unknown to obtain the following system of two equations in three unknowns:

\begin{array}{rcrcrcr} x_1&&&+&x_3&=&b_1 \\ &&x_2&+&x_3&=&b_2 \end{array}

This system has the general solution x = (b_1-c, b_2-c, c) for any b with c being an arbitrary value. Since c can take on any value the number of solutions to the system is thus infinite.

The system above corresponds to Ax = b where

A = \begin{bmatrix} 1&0&1 \\ 0&1&1 \end{bmatrix}

iii) The system of two equations in three unknowns in (ii) above has an infinite number of solutions. To allow for the possibility of having either an infinite number of solutions or no solution at all, we can follow the example of (i) above and add a third equation which might or might not be satisfiable depending on the value of b (and in particular the  value of b_3).

For example, consider the following system of equations:

\begin{array}{rcrcrcr} x_1&&&+&x_3&=&b_1 \\ &&x_2&+&x_3&=&b_2 \\ x_1&-&x_2&&&=&b_3 \end{array}

From (ii) above we know that x = (b_1 - c, b_2 - c, c) (where c is an arbitrary constant) is a solution to the first two equations. For this to also be a solution to the third equation we must have

x_1 - x_2 = (b_1-c) - (b_2-c) = b_1 - b_2 = b_3

So the system has an infinite number of solutions if b_3 = b_1 - b_2 but has no solution otherwise.

The system above corresponds to Ax = b where

A = \begin{bmatrix} 1&0&1 \\ 0&1&1 \\ 1&-1&0 \end{bmatrix}

iv) Consider taking the system of three equations in three unknowns from (iii) above and changing the third equation as follows:

\begin{array}{rcrcrcr} x_1&&&+&x_3&=&b_1 \\ &&x_2&+&x_3&=&b_2 \\ &&&&x_3&=&b_3 \end{array}

From the third equation we have x_3 = b_3. From the second equation we have x_2 + x_3 = x_2+b_3 = b_2 or x_2 = b_2 - b_3. From the first equation we have x_1 + x_3 = x_1+b_3 = b_1 or x_1 = b_1 - b_3. So the system has the single solution x = (b_1-b_3, b_2-b_3, b_3).

This corresponds to Ax = b where

A = \begin{bmatrix} 1&0&1 \\ 0&1&1 \\ 0&0&1 \end{bmatrix}

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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