## Linear Algebra and Its Applications, exercise 1.4.5

Exercise 1.4.5. Multiply the following matrices:

$Ax = \begin{bmatrix} 3&-6&0 \\ 0&2&-2 \\ 1&-1&-1 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix}$

Considering the matrix A as a system of equations, find a solution to the system Ax = 0, for which the right-hand side of all three equations is zero. Specify whether there is just one solution or many; if the latter find another solution.

$Ax = \begin{bmatrix} 3&-6&0 \\ 0&2&-2 \\ 1&-1&-1 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 0 \\ 1 \end{bmatrix} \cdot 2 + \begin{bmatrix} -6 \\ 2 \\ -1 \end{bmatrix} \cdot 1 + \begin{bmatrix} 0 \\ -2 \\ -1 \end{bmatrix} \cdot 1 = \begin{bmatrix} 6 - 6 + 0 \\ 0 + 2 - 2 \\ 2 - 1 - 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

So we have (2, 1, 1) as a solution vector to the system of equations.

Going back to the equation

$\begin{bmatrix} 3 \\ 0 \\ 1 \end{bmatrix} \cdot 2 + \begin{bmatrix} -6 \\ 2 \\ -1 \end{bmatrix} \cdot 1 + \begin{bmatrix} 0 \\ -2 \\ -1 \end{bmatrix} \cdot 1 = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

we see that we can multiply both sides of the equation by a constant c:

$\begin{bmatrix} 3 \\ 0 \\ 1 \end{bmatrix} \cdot 2c + \begin{bmatrix} -6 \\ 2 \\ -1 \end{bmatrix} \cdot c + \begin{bmatrix} 0 \\ -2 \\ -1 \end{bmatrix} \cdot c = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

So (2c, c, c) is also a solution for any c; for c = 2 we have the solution (4, 2, 2).

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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