844 views
Find the inverse of the matrix $\begin{bmatrix} 1 & 0 & 1 \\ -1 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix}$
asked | 844 views
+3

This might help .......

An Easy Procedure.

Using Eigen values, the characteristic equation we get is -

$-\lambda ^{3} + 2\lambda ^{2} -2 =0$

Using Cayley-Hamilton Theorem-

$-A^{3}+2A^{2}-2I=0$

So, $A^{-1}=\frac{1}{2}(2A-A^{2})$

Solving that we get,

$A^{-1}= \begin{bmatrix} \frac{1}{2} & \frac{-1}{2} & \frac{1}{2}\\ 0&0 &1 \\ \frac{1}{2}&\frac {1}{2} &\frac{-1}{2} \end{bmatrix}$
answered by Junior (655 points)
edited
0
Can u please show how u got characteristic equation. I am getting characteristic equation of degree 1. I know its wrong, so please guide me how to find characteristic equation
0

+1
A=$\begin{bmatrix} 1 & 0 & 1 \\ -1 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix}$

$A_{11} = -1 , A_{12}=0 , A_{13}=-1$

$A_{21} = 1 , A_{22}=0 , A_{23}=-1$

$A_{31} = -1 , A_{32}=-2 , A_{33}=1$

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

$adjA=B^{T}$

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

$|A|=-2$

$|A|^{-1}=\frac{adjA}{|A|}$

$|A|^{-1}=\begin{bmatrix} \frac{1}{2} & \frac{-1}{2} &\frac{1}{2} \\ 0& 0 & 1\\ \frac{1}{2}&\frac{1}{2} & \frac{-1}{2} \end{bmatrix}$
answered by Boss (40.9k points)

$\begin{bmatrix} \frac{1}{2} & \frac{-1}{2} & \frac{1}{2}\\ 0 & 0 & 1\\ \frac{1}{2} & \frac{1}{2} & \frac{-1}{2} \end{bmatrix}$
answered by Boss (34.1k points)

1
2