Answer to Question #287797 in Linear Algebra for Zeki

Question

find the inverse of this matrix

A= 1 3 5 4

7 2 9 6

Answer 1

$$A=\begin{bmatrix} 1 & 3 & 5 \\ 4 & 7 & 2 \\ 9 & 6 & 8 \end{bmatrix}$$

$$\begin{bmatrix} 1 & 3 & 5 & & 1 & 0 & 0 \\ 4 & 7 & 2 & & 0 & 1 & 0 \\ 9 & 6 & 8 & & 0 & 0 & 1 \end{bmatrix}$$

$R_2=R_2-4R_1$

$$\begin{bmatrix} 1 & 3 & 5 & & 1 & 0 & 0 \\ 0 & -5 & -18 & & -4 & 1 & 0 \\ 9 & 6 & 8 & & 0 & 0 & 1 \end{bmatrix}$$

$R_3=R_3-9R_1$

$$\begin{bmatrix} 1 & 3 & 5 & & 1 & 0 & 0 \\ 0 & -5 & -18 & & -4 & 1 & 0 \\ 0 & -21 & -37 & & -9 & 0 & 1 \end{bmatrix}$$

$R_2=-R_2/5$

$$\begin{bmatrix} 1 & 3 & 5 & & 1 & 0 & 0 \\ 0 & 1 & 18/5 & & 4/5 & -1/5 & 0 \\ 0 & -21 & -37 & & -9 & 0 & 1 \end{bmatrix}$$

$R_1=R_1-3R_2$

$$\begin{bmatrix} 1 & 0 & -29/5 & & -7/5 & 3/5 & 0 \\ 0 & 1 & 18/5 & & 4/5 & -1/5 & 0 \\ 0 & -21 & -37 & & -9 & 0 & 1 \end{bmatrix}$$

$R_3=R_3+21R_2$

$$\begin{bmatrix} 1 & 0 & -29/5 & & -7/5 & 3/5 & 0 \\ 0 & 1 & 18/5 & & 4/5 & -1/5 & 0 \\ 0 & 0 & 193/5 & & 39/5 & -21/5 & 1 \end{bmatrix}$$

$R_3=5R_3/193$

$$\begin{bmatrix} 1 & 0 & -29/5 & & -7/5 & 3/5 & 0 \\ 0 & 1 & 18/5 & & 4/5 & -1/5 & 0 \\ 0 & 0 & 1 & & 39/193 & -21/193 & 5/193 \end{bmatrix}$$

$R_1=R_1+29R_3/5$

$$\begin{bmatrix} 1 & 0 & 0 & & -44/193 & -6/193 & 29/193 \\ 0 & 1 & 18/5 & & 4/5 & -1/5 & 0 \\ 0 & 0 & 1 & & 39/193 & -21/193 & 5/193 \end{bmatrix}$$

$R_2=R_2-18R_3/5$

$$\begin{bmatrix} 1 & 0 & 0 & & -44/193 & -6/193 & 29/193 \\ 0 & 1 & 0 & & 14/193 & 37/193 & -18/193 \\ 0 & 0 & 1 & & 39/193 & -21/193 & 5/193 \end{bmatrix}$$

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

$$A^{-1}=\begin{bmatrix} -44/193 & -6/193 & 29/193 \\ 14/193 & 37/193 & -18/193 \\ 39/193 & -21/193 & 5/193 \end{bmatrix}$$