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 $Ax = b$

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

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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