## Linear Algebra and Its Applications, exercise 1.2.6

Exercise 1.2.6. Start with equation (4) from p. 8 of the text:

$u \begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix} + v \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix} + w \begin{bmatrix}1 \\ 3 \\ 4\end{bmatrix} = b$

and assume that b = (2, 5, 7) (a value of b for which solutions can be found for u, v, and w). The text mentions two solutions, (1, 0, 1) and (4, -1, -1). What is another solution (u, v, w)?

$u \begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix} + v \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix} + w \begin{bmatrix}1 \\ 3 \\ 4\end{bmatrix} = \begin{bmatrix}2 \\ 5 \\ 7\end{bmatrix}$

As noted in the text, the equation

$u \begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix} + v \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix} + w \begin{bmatrix}1 \\ 3 \\ 4\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}$

has the solution (3, -1, -2):

$3 \begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix} - \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix} - 2 \begin{bmatrix}1 \\ 3 \\ 4\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}$

We can then multiply each side of the equation by any constant c to obtain:

$3c \begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix} - c \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix} - 2c \begin{bmatrix}1 \\ 3 \\ 4\end{bmatrix} = c \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}$

Since (1, 0, 1) is a solution to the original equation we have:

$\begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix} + 0 \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix} + \begin{bmatrix}1 \\ 3 \\ 4\end{bmatrix} = \begin{bmatrix}2 \\ 5 \\ 7\end{bmatrix}$

Adding this equation to the previous one we obtain:

$(3c + 1) \begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix} + (-c) \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix} + (-2c + 1) \begin{bmatrix}1 \\ 3 \\ 4\end{bmatrix} = \begin{bmatrix}2 \\ 5 \\ 7\end{bmatrix}$

Thus there are an infinite number of solutions of the form (3c + 1, -c, -2c + 1) where c is any real number. Setting c = 0 and c = 1 respectively gives the solutions (1, 0, 1) and (4, -1, -1) previously noted, while setting c = -1 gives the additional solution (-2, 1, 3) and setting c = 2 the additional solution (7, -2, -3).

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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