## Linear Algebra and Its Applications, Review Exercise 1.4

Review exercise 1.4. Solve the following systems of equations using elimination and back substitution:

$\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcr}u&&&+&w&=&4 \\ u&+&v&&&=&3 \\ u&+&v&+&w&=&6 \end{array}$    and    $\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcr}&&v&+&w&=&0 \\ u&&&+&w&=&0 \\ u&+&v&&&=&6 \end{array}$

$\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcr}u&&&+&w&=&4 \\ u&+&v&&&=&3 \\ u&+&v&+&w&=&6 \end{array}$

by subtracting the first equation from the second and third equations to obtain the following system:

$\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcr}u&&&+&w&=&4 \\ &&v&-&w&=&-1 \\ &&v&&&=&2 \end{array}$

We can then subtract the second equation from the third to produce the following system:

$\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcr}u&&&+&w&=&4 \\ &&v&-&w&=&-1 \\ &&&&w&=&3 \end{array}$

We then back-substitute the value of $w$ into the second equation to obtain

$\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcr}u&&&+&w&=&4 \\ &&v&&&=&2 \\ &&&&w&=&3 \end{array}$

and then back-substitute the value of $v$ into the first equation  to obtain

$\setlength\arraycolsep{0.2em}\begin{array}{rcr}u&=&1 \\ v&=&2 \\ w&=&3 \end{array}$

Note that in matrix terms the elimination sequence above corresponds to

$\begin{bmatrix} 1&0&1 \\ 1&1&0 \\ 1&1&1 \end{bmatrix} \rightarrow \begin{bmatrix} 1&0&1 \\ 0&1&-1 \\ 0&1&0 \end{bmatrix} \rightarrow \begin{bmatrix} 1&0&1 \\ 0&1&-1 \\ 0&0&1 \end{bmatrix}$

We next look at the system

$\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcr}&&v&+&w&=&0 \\ u&&&+&w&=&0 \\ u&+&v&&&=&6 \end{array}$

which can be represented in matrix terms as

$\begin{bmatrix} 0&1&1 \\ 1&0&1 \\ 1&1&0 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 6 \end{bmatrix}$

We can exchange the first and third equations to obtain the system

$\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcr} u&+&v&&&=&6 \\ u&&&+&w&=&0 \\ &&v&+&w&=&0 \end{array}$

and then subtract the first equation from the second to obtain

$\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcr} u&+&v&&&=&6 \\ &&-v&+&w&=&-6 \\ &&v&+&w&=&0 \end{array}$

We next add the second equation to the third to obtain

$\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcr} u&+&v&&&=&6 \\ &&-v&+&w&=&-6 \\ &&&&2w&=&-6 \end{array}$

Solving the third equation yields $w = -3$, and back-substituting the value of $w$ into the second equation yields $v = 3$. Finally, back-substituting the value of $v$ into the first equation yields $u = 3$. The solution is therefore

$\setlength\arraycolsep{0.2em}\begin{array}{rcr}u&=&3 \\ v&=&3 \\ w&=&-3 \end{array}$

In matrix terms the sequence above corresponds to first multiplying by a permutation matrix to exchange rows 1 and 3:

$\begin{bmatrix} 0&0&1 \\ 0&1&0 \\ 1&0&0 \end{bmatrix} \begin{bmatrix} 0&1&1 \\ 1&0&1 \\ 1&1&0 \end{bmatrix} = \begin{bmatrix} 1&1&0 \\ 1&0&1 \\ 0&1&1 \end{bmatrix}$

and then doing elimination:

$\begin{bmatrix} 1&1&0 \\ 1&0&1 \\ 0&1&1 \end{bmatrix} \rightarrow \begin{bmatrix} 1&1&0 \\ 0&-1&1 \\ 0&1&1 \end{bmatrix} \rightarrow \begin{bmatrix} 1&1&0 \\ 0&-1&1 \\ 0&0&2 \end{bmatrix}$

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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