Linear Algebra and Its Applications, Review Exercise 2.7

Review exercise 2.7. Find the most general solution to the following system of linear equations:

\begin{array}{rcrcrcr} u&+&v&+&w&=&1 \\ u&&&-&w&=&2 \end{array}

Answer: This system corresponds to the system Ax = b where

A = \begin{bmatrix} 1&1&1 \\ 1&0&-1 \end{bmatrix} \qquad x = \begin{bmatrix} u \\ v \\ w \end{bmatrix} \qquad b = \begin{bmatrix} 1 \\ 2 \end{bmatrix}

The general solution to this system is a combination of a particular solution to Ax = b plus the homogeneous solution to Ax = 0. We first solve the homogeneous system.

We start by multiplying the first row of A by 1 and subtracting it from the second row:

\begin{bmatrix} 1&1&1 \\ 1&0&-1 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&1&1 \\ 0&-1&-2 \end{bmatrix}

The resulting matrix is in echelon form with pivots in columns 1 and 2, u and v as basic variables, and w as a free variable. Setting w to 1 and solving the homogeneous system, from the second row we have -v - 2w = -v - 2 = 0 or v = -2. From the first row we then have u+v+w = u - 2 + 1 = u - 1 = 0 or u=1. So (1, -2, 1) is a solution for the homogeneous system, as is any multiple of that vector.

To find a particular solution we set w = 0 and go back to the original system. From the second equation we have u - w = u - 0 = 2 or u = 2. From the first equation we have u + v + w = 2 + v + 0 = 1 or v = -1. So (2, -1, 0) is a particular solution.

The general solution to the original system is thus all vectors of the form

\begin{bmatrix} 2 \\ -1 \\ 0 \end{bmatrix} + w \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix}

where w can take on any value.

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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