## Linear Algebra and Its Applications, Exercise 2.2.9

Exercise 2.2.9. Consider the following $A$ and $b$: $A = \begin{bmatrix} 1&2&0&3 \\ 2&4&0&7 \end{bmatrix} \qquad b = \begin{bmatrix} b_1 \\ b_2 \end{bmatrix}$

For what values of $b_1$ and $b_2$ does the system $Ax = b$ have a solution? Also, determine the nullspace of $A$ and provide two examples of vectors within it, and find the general solution to $Ax = b$.

Answer: We perform elimination by subtracting 2 times the first row from the third row: $\begin{bmatrix} 1&2&0&3 \\ 2&4&0&7 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&2&0&3 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 - 2b_1 \end{bmatrix}$

This completes elimination and produces the system $Ux = c$: $\begin{bmatrix} 1&2&0&3 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 - 2b_1 \end{bmatrix}$

Since the pivots of $U$ are in columns 1 and 4, the basic variables are $u$ and $y$ and the free variables are $v$ and $w$. There are no constraints on the values of $b_1$ and $b_2$.

To find the general solution to $Ax = b$ we start with the system $Ax = 0$, which is equivalent to $Ux = 0$: $\begin{bmatrix} 1&2&0&3 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

From the second equation we must have $y = 0$. Substituting the value of $y$ into the first equation we have $u + 2v = 0$ or $u = -2v$. The general solution to $Ax = 0$ can then be expressed in terms of the free variables $v$ and $w$ as follows: $x_{homogeneous} = v \begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + w \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}$

By setting $v = 1$ and $w = 0$ we obtain one vector $(-2, 1, 0, 0)$ in the nullspace of $A$ and by setting $v = 0$ and $w = 1$ we obtain another vector $(0, 0, 1, 0)$ also in the nullspace.

We can obtain a particular solution to $Ax = b$ by going back to the system $Ux = c$ derived above: $\begin{bmatrix} 1&2&0&3 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 - 2b_1 \end{bmatrix}$

From the second equation we have $y = b_2 - 2b_1$. Substituting for $y$ into the first equation we have $u + 2v + 3(b_2 - 2b_1) = u + 2v +3b_2 - 6b_1 = b_1$ or $u = -2v + 7b_1 - 3b_2$. Setting the free variables $v$ and $w$ to zero gives the particular solution $x_{particular} = \begin{bmatrix} 7b_1 - 3b_2 \\ 0 \\ 0 \\ -2b_1 + b_2 \end{bmatrix}$

The general solution to $Ax = b$ is then $x = x_{particular} + x_{homogeneous} = \begin{bmatrix} 7b_1 - 3b_2 \\ 0 \\ 0 \\ -2b_1 + b_2 \end{bmatrix} + v \begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + w \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}$

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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