## Linear Algebra and Its Applications, Exercise 2.2.5

Exercise 2.2.5. Consider the system of linear equations represented by the following matrix: $A = \begin{bmatrix} 0&0 \\ 1&2 \\ 4&8 \\ 0&0 \end{bmatrix}$

(This is the transpose of the matrix from exercise 2.2.4.) Find the echelon matrix $U$, a set of basic variables, a set of free variables, and the general solution to $Ax = 0$. Then use elimination to find when the system $Ax = b$ has a solution, and express that solution as the sum of a particular solution and the general solution to $Ax = 0$. Finally, find the rank of $A$.

Answer: We must do a row exchange to exchange the first and second rows. This is equivalent to multiplying by a permutation matrix $P$ as follows: $PA = \begin{bmatrix} 0&1&0&0 \\ 1&0&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} 0&0 \\ 1&2 \\ 4&8 \\ 0&0 \end{bmatrix} = \begin{bmatrix} 1&2 \\ 0&0 \\ 4&8 \\ 0&0 \end{bmatrix}$

We then perform elimination on $PA$ by subtracting 4 times the first row from the third row (i.e., using the multiplier $l_{31} = 4$): $\begin{bmatrix} 1&2 \\ 0&0 \\ 4&8 \\ 0&0 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&2 \\ 0&0 \\ 0&0 \\ 0&0 \end{bmatrix}$

This completes elimination, and leaves us with the echelon matrix $U = \begin{bmatrix} 1&2 \\ 0&0 \\ 0&0 \\ 0&0 \end{bmatrix}$

and the factorization $PA = LU$: $\begin{bmatrix} 0&1&0&0 \\ 1&0&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} 0&0 \\ 1&2 \\ 4&8 \\ 0&0 \end{bmatrix} = \begin{bmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 4&0&1&0 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} 1&2 \\ 0&0 \\ 0&0 \\ 0&0 \end{bmatrix}$

We now solve for $Ax = \begin{bmatrix} 0&0 \\ 1&2 \\ 4&8 \\ 0&0 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = 0$

which is equivalent to $Ux = \begin{bmatrix} 1&2 \\ 0&0 \\ 0&0 \\ 0&0 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = 0$

The pivot in $U$ is in column 1, so the (only) basic variable is $u$ and the free variable is $v$.

From the first row of $U$ we have $u + 2v = 0$ and thus $u = -2v$. We then have $x = \begin{bmatrix} -2v \\ v \end{bmatrix} = v \begin{bmatrix} -2 \\ 1 \end{bmatrix}$

In considering the inhomogeneous system $Ax = b$ or $\begin{bmatrix} 0&0 \\ 1&2 \\ 4&8 \\ 0&0 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \\ b_4 \end{bmatrix}$

the elimination sequence above (including the initial row exchange) would produce the system $Ux = c$ or $\begin{bmatrix} 1&2 \\ 0&0 \\ 0&0 \\ 0&0 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} b_2 \\ b_1 \\ b_3 - 4b_2 \\ b_4 \end{bmatrix} \Rightarrow \begin{bmatrix} u+2v \\ 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} b_2 \\ b_1 \\ b_3 - 4b_2 \\ b_4 \end{bmatrix}$

From the second and fourth equations we must have $b_1 = b_4 = 0$.  From the third equation we must have $b_3 - 4b_2 = 0$ or $b_3 = 4b_2$. Thus for $Ax = b$ to have a solution the vector $b$ must have the form $(0, b_2, 4b_2, 0)$.

Taking $b = (0, b_2, 4b_2, 0)$ produces the system $\begin{bmatrix} 0&0 \\ 1&2 \\ 4&8 \\ 0&0 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} 0 \\ b_2 \\ 4b_2 \\ 0 \end{bmatrix}$

which after row exchange and elimination becomes $\begin{bmatrix} 1&2 \\ 0&0 \\ 0&0 \\ 0&0 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} b_2 \\ 0 \\ 0 \\ 0 \end{bmatrix}$

The first equation gives $u + 2v = b_2$ or $u = b_2 - 2v$. Setting the free variable $v$ to zero produces the particular solution $x = (u, v) = (b_2, 0)$ to the system $\begin{bmatrix} 0&0 \\ 1&2 \\ 4&8 \\ 0&0 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} 0 \\ b_2 \\ 4b_2 \\ 0 \end{bmatrix}$

We can combine the particular solution to this system with the solution to $Ax = 0$ to produce the general solution $x$ for the system $Ax = b$ $x = \begin{bmatrix} b_2 \\ 0 \end{bmatrix} + v \begin{bmatrix} -2 \\ 1 \end{bmatrix}$

We can check this solution as follows: $Ax = \begin{bmatrix} 0&0 \\ 1&2 \\ 4&8 \\ 0&0 \end{bmatrix} \left( \begin{bmatrix} b_2 \\ 0 \end{bmatrix} + v \begin{bmatrix} -2 \\ 1 \end{bmatrix} \right) = \begin{bmatrix} 0&0 \\ 1&2 \\ 4&8 \\ 0&0 \end{bmatrix} \begin{bmatrix} b_2 - 2v \\ v \end{bmatrix}$ $= \begin{bmatrix} 0 \\ (b_2 - 2v) + 2v \\ 4(b_2 - 2v) + 8v \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ b_2 - 2v + 2v \\ 4b_2 - 8v + 8v \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ b_2 \\ 4b_2 \\ 0 \end{bmatrix}$

The rank of $A$ is 1, the number of basic variables (or pivots).

UPDATE: Corrected a typo found by Argiris.

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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### 2 Responses to Linear Algebra and Its Applications, Exercise 2.2.5

1. Argiris says:

In the last b side 3rd row u need 4b2 and not 4b2 -2

• hecker says:

Thank you for finding this error! I have updated the post to correct the problem.