recategorized by
4,549 views

3 Answers

Best answer
27 votes
27 votes
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,

 $A^{-1}= \begin{bmatrix} \frac{1}{2} & \frac{-1}{2} & \frac{1}{2}\\ 0&0 &1 \\ \frac{1}{2}&\frac {1}{2} &\frac{-1}{2} \end{bmatrix}$
edited by
20 votes
20 votes
A=$\begin{bmatrix} 1 & 0 & 1 \\ -1 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix}$

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

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

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

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

$adjA=B^{T}$

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

$|A|=-2$

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

$|A|^{-1}=\begin{bmatrix} \frac{1}{2} & \frac{-1}{2} &\frac{1}{2} \\ 0& 0 & 1\\ \frac{1}{2}&\frac{1}{2} & \frac{-1}{2} \end{bmatrix}$
9 votes
9 votes
Answer:

$
\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}
$

Related questions

18 votes
18 votes
4 answers
1
Kathleen asked Oct 4, 2014
5,541 views
The rank of matrix $\begin{bmatrix} 0 & 0 & -3 \\ 9 & 3 & 5 \\ 3 & 1 & 1 \end{bmatrix}$ is:$0$$1$$2$$3$
26 votes
26 votes
3 answers
2
Kathleen asked Oct 4, 2014
7,735 views
Let $A$ and $B$ be real symmetric matrices of size $n \times n$. Then which one of the following is true?$AA'=I$$A=A^{-1}$$AB=BA$$(AB)'=BA$
33 votes
33 votes
4 answers
3
Kathleen asked Oct 5, 2014
8,847 views
An instance of a relational scheme $R(A, B, C)$ has distinct values for attribute $A$. Can you conclude that $A$ is a candidate key for $R?$