## 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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