Answer to Question #203386 in Linear Algebra for umme habiba sumi

Question

Find the inverse of A = ( 1,3,0) ( 0,4,-6) ( -1,5,7) .

Answer 1

$$A=\begin{pmatrix} 1 & 3 & 0 \\ 0 & 4 & -6 \\ -1 & 5 & 7\\ \end{pmatrix}$$

Augment the matrix with the identity matrix:

$$\begin{pmatrix} 1 & 3 & 0 & & 1 & 0 & 0 \\ 0 & 4 & -6 & & 0 & 1 & 0 \\ -1 & 5 & 7 & & 0 & 0 & 1\\ \end{pmatrix}$$

$R_3=R_3+R_1$

$$\begin{pmatrix} 1 & 3 & 0 & & 1 & 0 & 0 \\ 0 & 4 & -6 & & 0 & 1 & 0 \\ 0 & 8 & 7 & & 1 & 0 & 1\\ \end{pmatrix}$$

$R_2=R_2/4$

$$\begin{pmatrix} 1 & 3 & 0 & & 1 & 0 & 0 \\ 0 & 1 & -3/2 & & 0 & 1/4 & 0 \\ 0 & 8 & 7 & & 1 & 0 & 1\\ \end{pmatrix}$$

$R_1=R_1-3R_2$

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

$R_3=R_3-8R_2$

$$\begin{pmatrix} 1 & 0 & 9/2 & & 1 & -3/4 & 0 \\ 0 & 1 & -3/2 & & 0 & 1/4 & 0 \\ 0 & 0 & 19 & & 1 & -2 & 1\\ \end{pmatrix}$$

$R_3=R_3/19$

$$\begin{pmatrix} 1 & 0 & 9/2 & & 1 & -3/4 & 0 \\ 0 & 1 & -3/2 & & 0 & 1/4 & 0 \\ 0 & 0 & 1 & & 1/19 & -2/19 & 1/19\\ \end{pmatrix}$$

$R_1=R_1-(9/2)R_3$

$$\begin{pmatrix} 1 & 0 & 0 & & 29/28 & -21/76 & -9/38 \\ 0 & 1 & -3/2 & & 0 & 1/4 & 0 \\ 0 & 0 & 1 & & 1/19 & -2/19 & 1/19\\ \end{pmatrix}$$

$R_2=R_2+(3/2)R_3$

$$\begin{pmatrix} 1 & 0 & 0 & & 29/38 & -21/76 & -9/38 \\ 0 & 1 & 0 & & 3/38 & 7/76 & 3/38 \\ 0 & 0 & 1 & & 1/19 & -2/19 & 1/19\\ \end{pmatrix}$$

On the left is the identity matrix. On the right is the inverse matrix.

$$\begin{pmatrix} 1 & 3 & 0 \\ 0 & 4 & -6 \\ -1 & 5 & 7\\ \end{pmatrix}\begin{pmatrix} 29/38 & -21/76 & -9/38 \\ 3/38 & 7/76 & 3/38 \\ 1/19 & -2/19 & 1/19\\ \end{pmatrix}$$

$$=\begin{pmatrix} (29+9+0)/38 & (-21+21-0)/76 &(-9+9+0)/38 \\ (-0+12-12)/38 & (-0+28+48)/76 & (-0+12-12)/38 \\ (-29+15+14)/38 & (21+35-56)/76 & (9+15+14)/38\\ \end{pmatrix}$$

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

$$A^{-1}=\begin{pmatrix} 29/38 & -21/76 & -9/38 \\ 3/38 & 7/76 & 3/38 \\ 1/19 & -2/19 & 1/19\\ \end{pmatrix}$$