Answer to Question #206417 in Linear Algebra for Elissa

Question

Find the determinant and inverse of:

  a)

3 0

5 9

  b)  

−3 7 9

1 1 3

4 9 3

Answer 1

a)

$$\det A=\begin{vmatrix} 3 & 0 \\ 5 & 9 \end{vmatrix}=3(9)-0(5)=27\not=0$$

$=>A^{-1}\ exists$

$$A^{-1}=\dfrac{1}{27}\begin{pmatrix} 9 & 0 \\ -5 & 3 \end{pmatrix}=\begin{pmatrix} 1/3 & 0 \\ -5/27 & 1/9 \end{pmatrix}$$

b)

$$\det A=\begin{vmatrix} -3 & 7 & 9 \\ 1 & 1 & 3 \\ 4 & 9 & 3 \\ \end{vmatrix}=-3\begin{vmatrix} 1 & 3 \\ 9 & 3 \end{vmatrix}-7\begin{vmatrix} 1 & 3\\ 4 & 3 \end{vmatrix}+9\begin{vmatrix} 1 & 1 \\ 4 & 9 \end{vmatrix}$$

$$=-3(3-27)-7(3-12)+9(9-4)=180\not=0$$

$=>A^{-1}\ exists$

To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be the inverse matrix.

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

$R_1=-R_1/3$

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

$R_2=R_2-R_1$

$$\begin{pmatrix} 1 & -7/3 & -3 & & -1/3 & 0 & 0 \\ 0 & 10/3 & 6 & & 1/3 & 1 & 0 \\ 4 & 9 & 3 & & 0 & 0 & 1 \\ \end{pmatrix}$$

$R_3=R_3-4R_1$

$$\begin{pmatrix} 1 & -7/3 & -3 & & -1/3 & 0 & 0 \\ 0 & 10/3 & 6 & & 1/3 & 1 & 0 \\ 0 & 55/3 & 15 & & 4/3 & 0 & 1 \\ \end{pmatrix}$$

$R_2=(3/10)R_2$

$$\begin{pmatrix} 1 & -7/3 & -3 & & -1/3 & 0 & 0 \\ 0 & 1 & 9/5 & & 1/10 & 3/10 & 0 \\ 0 & 55/3 & 15 & & 4/3 & 0 & 1 \\ \end{pmatrix}$$

$R_1=R_1+(7/3)R_2$

$$\begin{pmatrix} 1 & 0 & 6/5 & & -1/10 & 7/10 & 0 \\ 0 & 1 & 9/5 & & 1/10 & 3/10 & 0 \\ 0 & 55/3 & 15 & & 4/3 & 0 & 1 \\ \end{pmatrix}$$

$R_3=R_3-(55/3)R_2$

$$\begin{pmatrix} 1 & 0 & 6/5 & & -1/10 & 7/10 & 0 \\ 0 & 1 & 9/5 & & 1/10 & 3/10 & 0 \\ 0 & 0 & -18 & & -1/2 & -11/2 & 1 \\ \end{pmatrix}$$

$R_3=-R_3/18$

$$\begin{pmatrix} 1 & 0 & 6/5 & & -1/10 & 7/10 & 0 \\ 0 & 1 & 9/5 & & 1/10 & 3/10 & 0 \\ 0 & 0 & 1 & & 1/36 & 11/36 & -1/18 \\ \end{pmatrix}$$

$R_1=R_1-(6/5)R_3$

$$\begin{pmatrix} 1 & 0 & 0 & & -2/15 & 1/3 & 1/15 \\ 0 & 1 & 9/5 & & 1/10 & 3/10 & 0 \\ 0 & 0 & 1 & & 1/36 & 11/36 & -1/18 \\ \end{pmatrix}$$

$R_2=R_2-(9/5)R_3$

$$\begin{pmatrix} 1 & 0 & 0 & & -2/15 & 1/3 & 1/15 \\ 0 & 1 & 0 & & 1/20 & -1/4 & 1/10 \\ 0 & 0 & 1 & & 1/36 & 11/36 & -1/18 \\ \end{pmatrix}$$

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

$$A^{-1}=\begin{pmatrix} -2/15 & 1/3 & 1/15 \\ 1/20 & -1/4 & 1/10 \\ 1/36 & 11/36 & -1/18 \\ \end{pmatrix}$$