## 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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