## 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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