Linear Algebra and Its Applications, Exercise 1.5.11

Exercise 1.5.11. We have a system LUx = b with values for L, U, and b as follows:

$\begin{bmatrix} 1&0&0 \\ 1&1&0 \\ 1&0&1 \end{bmatrix} \begin{bmatrix} 2&4&4 \\ 0&1&2 \\ 0&0&1 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \\ 2 \end{bmatrix}$

Solve for x without multiplying L and U to find A.

Answer: We can take advantage of the equations Lc = b and Ux = c to first solve for c and then for x. From Lc = b we have

$\begin{bmatrix} 1&0&0 \\ 1&1&0 \\ 1&0&1 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \\ 2 \end{bmatrix}$

Solving for c we have

$c_1 = 2$
$c_1+c_2 = 0 \Rightarrow c_2 = -2$
$c_1+c_3 = 2 \Rightarrow c_3 = 0$

From Ux = c we then have

$\begin{bmatrix} 2&4&4 \\ 0&1&2 \\ 0&0&1 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} 2 \\ -2 \\ 0 \end{bmatrix}$

Solving for u, v, and w we have

$w = 0$
$v+2w = -2 \Rightarrow v = -2$
$2u+4v+4w = 2 \Rightarrow 2u - 8 = 2 \Rightarrow u = 5$

and the solution to the system above is

$x =\begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} 5 \\ -2 \\ 0 \end{bmatrix}$

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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