## 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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