Linear Algebra and Its Applications, Review Exercise 1.13

Review exercise 1.13. Given the following:

A = LU = \begin{bmatrix} 1&0&0 \\ 4&1&0 \\ 1&0&1 \end{bmatrix} \begin{bmatrix} 2&2&4 \\ 0&1&3 \\ 0&0&1 \end{bmatrix} \qquad b = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}

use the triangular systems Lc = b and Ux = c to find a solution to Ax = b

Answer: We have

Lc = b \rightarrow \begin{bmatrix} 1&0&0 \\ 4&1&0 \\ 1&0&1 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \rightarrow \setlength\arraycolsep{0.2em}\begin{array}{rcrcrcr} c_1&&&&&=&0 \\ 4c_1&+&c_2&&&=&0 \\ c_1&&&+&c_3&=&1 \end{array}

Since c_1 = 0 and 4c_1 + c_2 = 0 we have c_2 = 0 also. Since c_1 + c_3 = 1 we have c_3 = 1 so that

c = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}

We then have

Ux = c \rightarrow \begin{bmatrix} 2&2&4 \\ 0&1&3 \\ 0&0&1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \rightarrow \setlength\arraycolsep{0.2em}\begin{array}{rcrcrcr} 2x_1&+&2x_2&+&4x_3&=&0 \\ &&x_2&+&3x_3&=&0 \\ &&&&x_3&=&1 \end{array}

Since x_3 = 1 and x_2 + 3x_3 = 0 we have x_2 = -3. We then have 2x_1 + 2x_2 + 4x_3 = 2x_1 - 2 = 0 so that x_1 = 1 and

x = \begin{bmatrix} 1 \\ -3 \\ 1 \end{bmatrix}

Note that b is equal to the last column of the 3 by 3 identity matrix I. Since A is nonsingular we know that A^{-1} exists, and since we have Ax = b we see that the solution x is equal to the last column of A^{-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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