## 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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