2
$\begingroup$

Find the inverseof the following matrix:

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

My attempt : I think the inverse will not exist because $\det A=0 $

Is it True ?

Any hints/solution will be appreciated.

$\endgroup$
1
  • 1
    $\begingroup$ Columns are independent so the inverse exists. $\endgroup$
    – Widawensen
    Commented Jan 7, 2019 at 15:17

5 Answers 5

4
$\begingroup$

Note that the given matrix is a so-called permutation matrix, the inverse of it is the permutation matrix associated to the inverse permutation.


Later edit: The matrix $A$ works on a "generic vector" $x$ with components $x_1,\dots, x_5$ as below by mapping it into a vector with components $x_{\sigma(1)}, \dots,x_{\sigma(5)}$ for a particular permutation $\sigma$. Explicitly:

$$ A \begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{bmatrix} = \begin{bmatrix}x_2\\x_1\\x_5\\x_3\\x_4\end{bmatrix} $$ The corresponding permutation $\sigma$ (with the inverse $\tau$) is of course $$ \sigma = \binom{12345}{21534}=(12)(354)\ , \qquad\text{ with } \tau:=\sigma^{-1}= \binom{12345}{21453}=(12)(345)\ . $$ Now associate in the natural way the permutation matrix for $\tau$...

Note: It is important to see at a glance, that this matrix is a permutation matrix. (Exactly one $1$ entry on each row and/or column.) This is part of a structure. (One has a mapping / a representation of the symmetric group into the general linear group of the corresponding dimension over $\Bbb Z$, or $\Bbb Q$ if we prefer a field.) "Other answers" like writing $A$ as a $(2+3)\times(2+3)$ block matrix, then doing the work in smaller pieces, may also lead to a good end, but for my taste, if i see a structural answer, and a computational one (Gauss scheme, block split, adjoint matrix, etc...) the structural one wins for me.

$\endgroup$
1
  • $\begingroup$ can u elaborate ur answer $\endgroup$
    – jasmine
    Commented Jan 7, 2019 at 11:47
2
$\begingroup$

This matrix is a permutation matrix and also orthogonal. What do you know about inverses of orthogonal matrices?

$\endgroup$
1
$\begingroup$

Interchange rows three times ($R_1-R_2, R_3-R_4,R_4-R_5$): $$\det(A)= \begin{vmatrix} 0 & \color{blue}1 & 0 & 0 &0 \\ \color{red}1 & 0 & 0 & 0 &0 \\ 0 & 0 & 0 & 0&\color{green}1 \\ 0 & 0 & \color{brown}1 & 0 &0 \\0&0&0&\color{purple}1&0 \end{vmatrix}=(-1)^3\begin{vmatrix} \color{red}1 & 0 & 0 & 0 &0 \\ 0 & \color{blue}1 & 0 & 0 &0 \\ 0 & 0 & \color{brown}1 & 0& 0 \\ 0 & 0 & 0 & \color{purple}1&0 \\0&0&0&0&\color{green}1 \end{vmatrix}=-1\ne 0.$$ Consider the expanded matrix and perform the same row interchanges: $$\begin{vmatrix} 0 & \color{blue}1 & 0 & 0 &0 \\ \color{red}1 & 0 & 0 & 0 &0 \\ 0 & 0 & 0 & 0&\color{green}1 \\ 0 & 0 & \color{brown}1 & 0 &0 \\0&0&0&\color{purple}1&0 \end{vmatrix}\begin{vmatrix} 1 & 0 & 0 & 0 &0 \\ 0 & 1 & 0 & 0 &0 \\ 0 & 0 & 1 & 0&0 \\ 0 & 0 & 0 & 1 &0 \\0&0&0&0&0 \end{vmatrix} \Rightarrow \begin{vmatrix} \color{red}1 & 0 & 0 & 0 &0 \\ 0 & \color{blue}1 & 0 & 0 &0 \\ 0 & 0 & \color{brown}1 & 0& 0 \\ 0 & 0 & 0 & \color{purple}1&0 \\0&0&0&0&\color{green}1 \end{vmatrix}\begin{vmatrix} 0 & 1 & 0 & 0 &0 \\ 1 & 0 & 0 & 0 &0 \\ 0 & 0 & 0 & 1&0 \\ 0 & 0 & 0 & 0 &1 \\ 0 & 0 & 1 & 0 &0 \end{vmatrix} \ \ \ \ $$ Hence, the inverse matrix is: $$A^{-1}=\begin{vmatrix} 0 & 1 & 0 & 0 &0 \\ 1 & 0 & 0 & 0 &0 \\ 0 & 0 & 0 & 1&0 \\ 0 & 0 & 0 & 0 &1 \\ 0 & 0 & 1 & 0 &0 \end{vmatrix}$$

$\endgroup$
1
$\begingroup$

The determinant of the matrix is not $0$, the determinant is $1$.

Hint:

Take a look at $A^2, A^3,A^4,\dots$.

$\endgroup$
2
  • $\begingroup$ eigenvalue are $0$...i thinks det $A =0$ @Xum $\endgroup$
    – jasmine
    Commented Jan 7, 2019 at 11:35
  • $\begingroup$ @jasmine None of the eigenvalues of the matrix is $0$. And the determinant is $1$. $\endgroup$
    – 5xum
    Commented Jan 7, 2019 at 11:43
1
$\begingroup$

Note that this matrix is obtained from the identity matrix by exchanging columns (or rows), and that columns (and rows) are still linearly independent, so immediately we have that the determinant is not $0$. Even more precisely, we immediately have that the determinant is $1$ or $-1$. (Immediately= without doing any calculation)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .