## Linear Algebra and Its Applications, Exercise 1.5.13

Exercise 1.5.13. Given the systems of equations $\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcr}u&+&4v&+&2w&=&-2 \\ -2u&-&8v&+&3w&=&32 \\ &&v&+&w&=&1 \end{array}$

and $\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcr}&&v&+&w&=&0 \\ u&+&v&&&=&0 \\ u&+&v&+&w&=&1 \end{array}$

solve both by elimination. Do row exchanges where necessary, and specify any permutation matrices required.

Answer: The first system of equations can be expressed as follows: $\begin{bmatrix} 1&4&2 \\ -2&-8&3 \\ 0&1&1 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} -2 \\ 32 \\ 1 \end{bmatrix}$

In the first step of elimination we subtract -2 times row 1 from row 2: $\begin{bmatrix} 1&4&2 \\ 0&0&7 \\ 0&1&1 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} -2 \\ 28 \\ 1 \end{bmatrix}$

We then need to exchange rows 2 and 3, so multiply both sides by the appropriate permutation matrix: $\begin{bmatrix} 1&0&0 \\ 0&0&1 \\ 0&1&0 \end{bmatrix} \begin{bmatrix} 1&4&2 \\ 0&0&7 \\ 0&1&1 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} 1&0&0 \\ 0&0&1 \\ 0&1&0 \end{bmatrix} \begin{bmatrix} -2 \\ 28 \\ 1 \end{bmatrix}$

which produces the following: $\begin{bmatrix} 1&4&2 \\ 0&1&1 \\ 0&0&7 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} -2 \\ 1 \\ 28 \end{bmatrix}$

We can then solve for u, v, and w: $7w = 28 \Rightarrow w = 4$ $v + w = 1 \Rightarrow v = -3$ $u + 4v + 2w = -2 \Rightarrow u - 12 + 8 = -2 \Rightarrow u = 2$

So the solution is (2, -3, 4).

The second system of equations can be expressed as follows: $\begin{bmatrix} 0&1&1 \\ 1&1&0 \\ 1&1&1 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$

Since we have a zero pivot in row 1, we must immediately do a row exchange: $\begin{bmatrix} 0&1&0 \\ 1&0&0 \\ 0&0&1 \end{bmatrix} \begin{bmatrix} 0&1&1 \\ 1&1&0 \\ 1&1&1 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} 0&1&0 \\ 1&0&0 \\ 0&0&1 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$

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

We can then multiply row 1 by 1 and subtract from row 3: $\begin{bmatrix} 1&1&0 \\ 0&1&1 \\ 0&0&1 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$

This completes elimination, so we solve for u, v, and w: $w = 1$ $v + w = 0 \Rightarrow v = -1$ $u + v = 0 \Rightarrow u = 1$

So the solution is (1, -1, 1).

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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