Chapter 7.2, Problem 40E

Solution

To calculate: The inverse of the matrix if it exists, or explain that the inverse does not exists.

  $A=\left[a_{ij}\right],a_{ij}=\left|i-j\right|,1\le i\le 3,1\le j\le 3$

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

Given Information:

The matrix is defined as,

  $A=\left[a_{ij}\right],a_{ij}=\left|i-j\right|,1\le i\le 3,1\le j\le 3$

Formula Used:

Suppose that $A$ is general matrix of order 3 and defined as,

  $A=\left[\begin{array}{ccc}{A}_{11}& A_{12}& A_{13}\\ A_{21}& A_{22}& A_{22}\\ A_{31}& A_{32}& A_{33}\end{array}\right]$

The inverse of the matrix is defined as,

  $A^{-1}=\frac{1}{\left|A\right|}\text{adj}\left(A\right)$

Where, $\left|A\right|$ is representing the determinant and $adj\left(A\right)$ is defined as,

  $\begin{array}{c}adjA={\left[\begin{array}{ccc}{A}_{11}& A_{12}& A_{13}\\ A_{21}& A_{22}& A_{23}\\ A_{31}& A_{32}& A_{33}\end{array}\right]}^T\\ =\left[\begin{array}{ccc}{A}_{11}& A_{21}& A_{31}\\ A_{12}& A_{22}& A_{32}\\ A_{13}& A_{23}& A_{33}\end{array}\right]\end{array}$

And, $\left[\begin{array}{ccc}{A}_{11}& A_{12}& A_{13}\\ A_{21}& A_{22}& A_{23}\\ A_{31}& A_{32}& A_{33}\end{array}\right]$ is a cofactor matrix of $A$ .

Calculation:

Consider the given matrix,

  $A=\left[a_{ij}\right],a_{ij}=\left|i-j\right|,1\le i\le 3,1\le j\le 3$

The general form of the matrix is defined as,

  $A=\left[\begin{array}{ccc}{A}_{11}& A_{12}& A_{13}\\ A_{21}& A_{22}& A_{22}\\ A_{31}& A_{32}& A_{33}\end{array}\right]$

So,

  $\begin{array}{c}A=\left[\begin{array}{ccc}\left|1-1\right|& \left|1-2\right|& \left|1-3\right|\\ \left|2-1\right|& \left|2-2\right|& \left|2-3\right|\\ \left|3-1\right|& \left|3-2\right|& \left|3-3\right|\end{array}\right]\\ =\left[\begin{array}{ccc}0& 1& 2\\ 1& 0& 1\\ 2& 1& 0\end{array}\right]\end{array}$

First, find the determinant.

  $\begin{array}{c}\left|A\right|=\left|\begin{array}{ccc}0& 1& 2\\ 1& 0& 1\\ 2& 1& 0\end{array}\right|\\ =0\left(0-1\right)-1\left(0-2\right)+2\left(1-0\right)\\ =0+2+2\\ =4\end{array}$

Now, find the ad-joint matrix.

  $\begin{array}{c}adjA=\left[\begin{array}{ccc}0-1& -\left(0-2\right)& 1-0\\ -\left(1-2\right)& 0-4& -\left(0-2\right)\\ 1-0& -\left(0-2\right)& 0-1\end{array}\right]\\ =\left[\begin{array}{ccc}-1& 2& 1\\ 2& -4& 2\\ 1& 2& -1\end{array}\right]\end{array}$

Put the above the value in the formula of inverse.

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

Hence, the inverse of the matrix is $\frac{1}{4}\left[\begin{array}{ccc}-1& 2& 1\\ 2& -4& 2\\ 1& 2& -1\end{array}\right]$ .