Answer to Question #192268 in Linear Algebra for prince

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

det(B)=

$\begin{vmatrix} 2& 2 & 0\\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{vmatrix}$ = -2 -2 = -4 $\ne0$ means inverse exists

 BX = I

x = B^-1I

x= $\left[\begin{array}{ccc|ccc}2 & 2 & 0 & 1 & 0 & 0\\1 & 0 & 1 & 0 & 1 & 0\\0 & 1 & 1 & 0 & 0 & 1\end{array}\right]$

Divide row 1 by 2 $: R_{1} = \frac{R_{1}}{2}$

$\left[\begin{array}{ccc|ccc}1 & 1 & 0 & \frac{1}{2} & 0 & 0\\1 & 0 & 1 & 0 & 1 & 0\\0 & 1 & 1 & 0 & 0 & 1\end{array}\right]$

Subtract row 1 from row 2 :$R_{2} = R_{2} - R_{1}$

$\left[\begin{array}{ccc|ccc}1 & 1 & 0 & \frac{1}{2} & 0 & 0\\0 & -1 & 1 & - \frac{1}{2} & 1 & 0\\0 & 1 & 1 & 0 & 0 & 1\end{array}\right]$

$R_{2} = - R_{2}$

$\left[\begin{array}{ccc|ccc}1 & 1 & 0 & \frac{1}{2} & 0 & 0\\0 & 1 & -1 & \frac{1}{2} & -1 & 0\\0 & 1 & 1 & 0 & 0 & 1\end{array}\right]$

$R_{1} = R_{1} - R_{2}$

$\left[\begin{array}{ccc|ccc}1 & 0 & 1 & 0 & 1 & 0\\0 & 1 & -1 & \frac{1}{2} & -1 & 0\\0 & 1 & 1 & 0 & 0 & 1\end{array}\right]$

$R_{3} = R_{3} - R_{2}$

$\left[\begin{array}{ccc|ccc}1 & 0 & 1 & 0 & 1 & 0\\0 & 1 & -1 & \frac{1}{2} & -1 & 0\\0 & 0 & 2 & - \frac{1}{2} & 1 & 1\end{array}\right]$

$R_{3} = \frac{R_{3}}{2}$

$\left[\begin{array}{ccc|ccc}1 & 0 & 1 & 0 & 1 & 0\\0 & 1 & -1 & \frac{1}{2} & -1 & 0\\0 & 0 & 1 & - \frac{1}{4} & \frac{1}{2} & \frac{1}{2}\end{array}\right]$

$R_{1} = R_{1} - R_{3}$

$\left[\begin{array}{ccc|ccc}1 & 0 & 0 & \frac{1}{4} & \frac{1}{2} & - \frac{1}{2}\\0 & 1 & -1 & \frac{1}{2} & -1 & 0\\0 & 0 & 1 & - \frac{1}{4} & \frac{1}{2} & \frac{1}{2}\end{array}\right]$

$R_{2} = R_{2} + R_{3}$

$\left[\begin{array}{ccc|ccc}1 & 0 & 0 & \frac{1}{4} & \frac{1}{2} & - \frac{1}{2}\\0 & 1 & 0 & \frac{1}{4} & - \frac{1}{2} & \frac{1}{2}\\0 & 0 & 1 & - \frac{1}{4} & \frac{1}{2} & \frac{1}{2}\end{array}\right]$

$ANSWER$ =

$\left[\begin{array}{ccc}\frac{1}{4} & \frac{1}{2} & - \frac{1}{2}\\\frac{1}{4} & - \frac{1}{2} & \frac{1}{2}\\- \frac{1}{4} & \frac{1}{2} & \frac{1}{2}\end{array}\right] \\ = \left[\begin{array}{ccc}0.25 & 0.5 & -0.5\\0.25 & -0.5 & 0.5\\-0.25 & 0.5 & 0.5\end{array}\right]$ = $B^{-1}$