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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