## Linear Algebra and Its Applications, Exercise 2.2.4

Exercise 2.2.4. Consider the system of linear equations represented by the following matrix: $A = \begin{bmatrix} 0&1&4&0 \\ 0&2&8&0 \end{bmatrix}$

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 perform elimination on $A$ by subtracting 2 times the first row from the second row (i.e., using the multiplier $l_{21} = 2$): $\begin{bmatrix} 0&1&4&0 \\ 0&2&8&0 \end{bmatrix} \Rightarrow \begin{bmatrix} 0&1&4&0 \\ 0&0&0&0 \end{bmatrix}$

This completes elimination, and leaves us with the echelon matrix $U = \begin{bmatrix} 0&1&4&0 \\ 0&0&0&0 \end{bmatrix}$

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

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

which is equivalent to $Ux = \begin{bmatrix} 0&1&4&0 \\ 0&0&0&0 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = 0$

The pivot in $U$ is in column 2, so the (only) basic variable is $v$ and the free variables are $u$, $w$ and $y$.

From the first row of $U$ we have $v + 4w = 0$ and thus $v = -4w$. We then have $x = \begin{bmatrix} u \\ -4w \\ w \\ y \end{bmatrix} = u \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + w \begin{bmatrix} 0 \\ -4 \\ 1 \\ 0 \end{bmatrix} + y \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}$

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

the elimination sequence above would produce the system $Ux = c$ or $\begin{bmatrix} 0&1&4&0 \\ 0&0&0&0 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 - 2b_1 \end{bmatrix}$

From the second equation we must have $b_2 - 2b_1 = 0$ or $b_2 = 2b_1$. Thus for $Ax = b$ to have a solution the vector $b$ must lie on the line passing through the origin and the point $(1, 2)$ so that $b = (b_1, 2b_1)$.

Taking $b = (b_1, 2b_2)$ produces the system $\begin{bmatrix} 0&1&4&0 \\ 0&2&8&0 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = \begin{bmatrix} b_1 \\ 2b_1 \end{bmatrix}$

which after elimination becomes $\begin{bmatrix} 0&1&4&0 \\ 0&0&0&0 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = \begin{bmatrix} b_1 \\ 0 \end{bmatrix}$

The first equation gives $v + 4w = b_1$ or $v = b_1 - 4w$. Setting the free variables $u$, $w$, and $y$ all to zero produces the particular solution $x = (0, b_1, 0, 0)$ to the system $\begin{bmatrix} 0&1&4&0 \\ 0&2&8&0 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = \begin{bmatrix} b_1 \\ 2b_1 \end{bmatrix}$

We can combine the particular solution to this system with the solution to $Ax = 0$ to produce the general solution for the system $x = \begin{bmatrix} 0 \\ b_1 \\ 0 \\ 0 \end{bmatrix} + u \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + w \begin{bmatrix} 0 \\ -4 \\ 1 \\ 0 \end{bmatrix} + y \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}$

We can check this solution as follows: $Ax = \begin{bmatrix} 0&1&4&0 \\ 0&2&8&0 \end{bmatrix} \left( \begin{bmatrix} 0 \\ b_1 \\ 0 \\ 0 \end{bmatrix} + u \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + w \begin{bmatrix} 0 \\ -4 \\ 1 \\ 0 \end{bmatrix} + y \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} \right)$ $= \begin{bmatrix} 0&1&4&0 \\ 0&2&8&0 \end{bmatrix} \begin{bmatrix} u \\ b_1 - 4w \\ w \\ y \end{bmatrix} = \begin{bmatrix} (b_1-4w) + 4w \\ 2(b_1-4w) + 8w \end{bmatrix}$ $= \begin{bmatrix} b_1 - 4w + 4w \\ 2b_1 - 8w + 8w \end{bmatrix} = \begin{bmatrix} b_1 \\ 2b_1 \end{bmatrix}$

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

UPDATE: I expanded the answer to conform to the presentation in the answer to exercise 2.2.5.

UPDATE 2: I corrected the value for $U$ used in various places. Thanks go to Zoi for catching this error!

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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### 4 Responses to Linear Algebra and Its Applications, Exercise 2.2.4

1. Zoi says:

Hi! You have a mistake – some times you add an extra 1 in U.

• hecker says:
2. Jeff Berhow says:
• hecker says: