## 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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