## 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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