Answer to Question #190381 in Linear Algebra for Regomoditswe Dibob

Question

Consider the given matrix B =   2 2 0 1 0 1 0 1 1   . Find detB and use it to determine whether or not B is invertible, and if so, find B −1 . (Hint: Use the matrix equation BX = I)

Answer 1

$B=\begin{pmatrix} 2 & 2 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix}$

$\text{det}B=\begin{vmatrix} 2 & 2 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{vmatrix}$

$=(-1)^{1+1}\cdot 2\cdot\begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix}+(-1)^{2+1}\cdot 1\cdot\begin{vmatrix} 2 & 0 \\ 1 & 1 \end{vmatrix}$

$+(-1)^{3+1}\cdot 0\cdot\begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix}=$

$=2(0-1)-(2-0)=-4\not=0$

Therefore the matrix $B$ is invertible and $B^{-1}$ exists.

Augment the matrix with the identity matrix:

$\begin{bmatrix} 2 & 2 & 0 & & 1 & 0 & 0 \\ 1 & 0 & 1 & & 0 & 1 & 0 \\ 0 & 1 & 1 & & 0 & 0 & 1 \\ \end{bmatrix}$

$R_1=\dfrac{R_1}{2}$

$\begin{bmatrix} 1 & 1 & 0 & & 1/2 & 0 & 0 \\ 1 & 0 & 1 & & 0 & 1 & 0 \\ 0 & 1 & 1 & & 0 & 0 & 1 \\ \end{bmatrix}$

$R_2=R_2-R_1$

$\begin{bmatrix} 1 & 1 & 0 & & 1/2 & 0 & 0 \\ 0 & -1 & 1 & & -1/2 & 1 & 0 \\ 0 & 1 & 1 & & 0 & 0 & 1 \\ \end{bmatrix}$

$R_2=-R_2$

$\begin{bmatrix} 1 & 1 & 0 & & 1/2 & 0 & 0 \\ 0 & 1 & -1 & & 1/2 & -1 & 0 \\ 0 & 1 & 1 & & 0 & 0 & 1 \\ \end{bmatrix}$

$R_1=R_1-R_2$

$\begin{bmatrix} 1 & 0 & 1 & & 0 & 1 & 0 \\ 0 & 1 & -1 & & 1/2 & -1 & 0 \\ 0 & 1 & 1 & & 0 & 0 & 1 \\ \end{bmatrix}$

$R_3=R_3-R_2$

$\begin{bmatrix} 1 & 0 & 1 & & 0 & 1 & 0 \\ 0 & 1 & -1 & & 1/2 & -1 & 0 \\ 0 & 0 & 2 & & -1/2 & 1 & 1 \\ \end{bmatrix}$

$R_3=R_3/2$

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

$R_1=R_1-R_3$

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

$R_2=R_2+R_3$

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

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

$B^{-1}=\begin{pmatrix} 1/4 & 1/2 & -1/2 \\ 1/4 & -1/2 & 1/2 \\ -1/4 & 1/2 & 1/2 \end{pmatrix}$