Answer to Question #350279 in Linear Algebra for Busi

Question

Determine the inverse of A, and show that A

−1A = I.

A =

(2 1 0

2 −1 1

3 −2 4)

Answer 1

$$A=\begin{pmatrix} 2 & 1 & 0\\ 2 & -1 & 1\\ 3 & -2 & 4\\ \end{pmatrix}$$

Augment the matrix with the identity matrix:

$$\begin{pmatrix} 2 & 1 & 0 & & 1 & 0 & 0\\ 2 & -1 & 1 & & 0 & 1 & 0\\ 3 & -2 & 4 & & 0 & 0 & 1\\ \end{pmatrix}$$

$R_1=R_1/2$

$$\begin{pmatrix} 1 & 1/2 & 0 & & 1/2 & 0 & 0\\ 2 & -1 & 1 & & 0 & 1 & 0\\ 3 & -2 & 4 & & 0 & 0 & 1\\ \end{pmatrix}$$

$R_2=R_2-2R_1$

$$\begin{pmatrix} 1 & 1/2 & 0 & & 1/2 & 0 & 0\\ 0 & -2 & 1 & & -1 & 1 & 0\\ 3 & -2 & 4 & & 0 & 0 & 1\\ \end{pmatrix}$$

$R_3=R_3-3R_1$

$$\begin{pmatrix} 1 & 1/2 & 0 & & 1/2 & 0 & 0\\ 0 & -2 & 1 & & -1 & 1 & 0\\ 0 & -7/2 & 4 & & -3/2 & 0 & 1\\ \end{pmatrix}$$

$R_2=-R_2/2$

$$\begin{pmatrix} 1 & 1/2 & 0 & & 1/2 & 0 & 0\\ 0 & 1 & -1/2 & & 1/2 & -1/2 & 0\\ 0 & -7/2 & 4 & & -3/2 & 0 & 1\\ \end{pmatrix}$$

$R_1=R_1-R_2/2$

$$\begin{pmatrix} 1 & 0 & 1/4 & & 1/4 & 1/4 & 0\\ 0 & 1 & -1/2 & & 1/2 & -1/2 & 0\\ 0 & -7/2 & 4 & & -3/2 & 0 & 1\\ \end{pmatrix}$$

$R_3=R_3+7R_2/2$

$$\begin{pmatrix} 1 & 0 & 1/4 & & 1/4 & 1/4 & 0\\ 0 & 1 & -1/2 & & 1/2 & -1/2 & 0\\ 0 & 0 & 9/4 & & 1/4 & -7/4 & 1\\ \end{pmatrix}$$

$R_3=4R_3/9$

$$\begin{pmatrix} 1 & 0 & 1/4 & & 1/4 & 1/4 & 0\\ 0 & 1 & -1/2 & & 1/2 & -1/2 & 0\\ 0 & 0 & 1 & & 1/9 & -7/9 & 4/9\\ \end{pmatrix}$$

$R_1=R_1-R_3/4$

$$\begin{pmatrix} 1 & 0 & 0 & & 2/9 & 4/9 & -1/9\\ 0 & 1 & -1/2 & & 1/2 & -1/2 & 0\\ 0 & 0 & 1 & & 1/9 & -7/9 & 4/9\\ \end{pmatrix}$$

$R_2=R_2+R_3/2$

$$\begin{pmatrix} 1 & 0 & 0 & & 2/9 & 4/9 & -1/9\\ 0 & 1 & 0 & & 5/9 & -8/9 & 2/9\\ 0 & 0 & 1 & & 1/9 & -7/9 & 4/9\\ \end{pmatrix}$$

We are done. On the left is the identity matrix. On the right is the inverse matrix.

$$A^{-1}=\begin{pmatrix} 2/9 & 4/9 & -1/9\\ 5/9 & -8/9 & 2/9\\ 1/9 & -7/9 & 4/9\\ \end{pmatrix}$$

Check

$$A^{-1}A=\begin{pmatrix} 2/9 & 4/9 & -1/9\\ 5/9 & -8/9 & 2/9\\ 1/9 & -7/9 & 4/9\\ \end{pmatrix}\begin{pmatrix} 2 & 1 & 0\\ 2 & -1 & 1\\ 3 & -2 & 4\\ \end{pmatrix}$$

$$=\begin{pmatrix} \dfrac{4+8-3}{9} & \dfrac{2-4+2}{9} & \dfrac{0+4-4}{9}\\ \\ \dfrac{10-16+6}{9} & \dfrac{5+8-4}{9} & \dfrac{0-8+8}{9}\\ \\ \dfrac{2-14+12}{9} & \dfrac{1+7-8}{9} & \dfrac{0-7+16}{9}\\ \\ \end{pmatrix}$$

$$=\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{pmatrix}=I$$