## Linear Algebra and Its Applications, Review Exercise 2.22

Review exercise 2.22. a) Given

$A = \begin{bmatrix} 1&2&0&3 \\ 0&0&0&0 \\ 2&4&0&1 \end{bmatrix} \qquad b = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}$

what conditions must $b$ satisfy in order for $Ax = b$ to have a solution?

b) Find a basis for the nullspace of $A$.

c) Find the general solution for $Ax = b$ for those cases when a solution exists.

d) Find a basis for the column space of $A$?

e) What is the rank of $A^T$?

$\begin{bmatrix} 1&2&0&3 \\ 0&0&0&0 \\ 2&4&0&1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}$

from the second row we have

$0 \cdot x_1 + 0 \cdot x_2 + 0 \cdot x_3 + 0 \cdot x_4 = b_2$

This can be true only if $b_2 = 0$.

b) The system $Ax = 0$ corresponds to

$\begin{bmatrix} 1&2&0&3 \\ 0&0&0&0 \\ 2&4&0&1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

Doing Gaussian elimination on $A$ we subtract 2 times the first row from the third:

$\begin{bmatrix} 1&2&0&3 \\ 0&0&0&0 \\ 2&4&0&1 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&2&0&3 \\ 0&0&0&0 \\ 0&0&0&-5 \end{bmatrix}$

Since there are pivots in columns 1 and 4 we have $x_1$ and $x_4$ as basic variables and $x_2$ and $x_3$ as basic variables. (Note that this would be true whether or not we do a row exchange to put the matrix in true echelon form.)

From the third equation we have $x_4 = 0$. Setting $x_2 = 1$ and $x_3 = 0$, from the first equation we have $x_1 + 2x_2 + 3x_4 = x_1 + 2 + 0 = 0$ or $x_1 = -2$. Setting $x_2 = 0$ and $x_3 = 1$, from the first equation we have $x_1 + 2x_2 + 3x_4 = x_1 + 0 + 0 = 0$ or $x_1 = 0$.

The following vectors are thus solutions to $Ax = 0$ and serve as basis vectors for the nullspace of $A$:

$\begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} \qquad \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}$

c) From (a) above, in order for $Ax = b$ to have a solution we must have $b_2 = 0$. The system $Ax = b$ then becomes

$\begin{bmatrix} 1&2&0&3 \\ 0&0&0&0 \\ 2&4&0&1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} b_1 \\ 0 \\ b_3 \end{bmatrix}$

To find a particular solution we set the free variables $x_2 = x_3 = 0$. From the first equation we have $x_1 + 2x_2 + 3x_4 = x_1 + 3x_4 = b_1$ and from the third equation we have $2x_1 + 4x_2 + x_4 = 2x_1 + x_4 = b_3$. Subtracting twice the first equation from the third equation we obtain $-5x_4 = b_3-2b_1$ or $x_4 = \frac{1}{5}(2b_1-b_3)$.

Substituting the values of $x_4$ into the first equation we have

$x_1 + 3x_4 = x_1 + \frac{3}{5}(2b_1-b_3)$

$= x_1 + \frac{6}{5}b_1 - \frac{3}{5}b_3 = b_1$

or

$x_1 = b_1 -\frac{6}{5}b_1 + \frac{3}{5}b_3$

$= -\frac{1}{5}b_1 + \frac{3}{5}b_3 = \frac{1}{5}(-b_1+3b_3)$.

The particular solution to $Ax = b$ is therefore

$\frac{1}{5} \begin{bmatrix} -b_1+3b_3 \\ 0 \\ 0 \\ 2b_1-b_3 \end{bmatrix}$

Combining the particular solution with the homogeneous solution from (b) above gives the general solution

$x = \frac{1}{5} \begin{bmatrix} -b_1+3b_3 \\ 0 \\ 0 \\ 2b_1-b_3 \end{bmatrix} + c_1 \begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + c_2 \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}$

where $c_1$ and $c_2$ are arbitrary scalars.

d) From (b) above we have pivots in columns 1 and 4 of the echelon form of $A$. Columns 1 and 4 of $A$ therefore serve as a basis for the column space of $A$:

$\begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} \qquad \begin{bmatrix} 3 \\ 0 \\ 1 \end{bmatrix}$

e) The rank of $A^T$ is the same as the rank of $A$, namely 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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