Chapter 8.3, Problem 12E

To determine

To show:that B is the inverse of A

Explanation of Solution

Given information: $A=\left[\begin{array}{cccc}-1& 1& 0& -1\\ 1& -1& 1& 0\\ -1& 1& 2& 0\\ 0& -1& 1& 1\end{array}\right]$; $B=\frac{1}{3}\left[\begin{array}{cccc}-3& 1& 1& -3\\ -3& -1& 2& -3\\ 0& 1& 1& 0\\ -3& -2& 1& 0\end{array}\right]$

Concept Involved:

Definition of Matrix Multiplication:

If $A=\left[a_{ij}\right]$ is an $m\times n$ matrix and $A=\left[b_{ij}\right]$ is an $n\times p$ matrix, then the product AB is an $m\times p$ matrix given by $A=\left[c_{ij}\right]$ , where $c_{ij}=a_{i1}{b}_{1j}+a_{i2}{b}_{2j}+a_{i3}{b}_{3j}+\mathrm{....}+a_{in}{b}_{nj}$ .

The definition of matrix multiplication uses a row-by-column multiplication, where the entry in the $i^{th}$ row and $j^{th}$ column of the product AB is obtained by multiplying the entries in the $i^{th}$ row of A by the corresponding entries in the $j^{th}$ column of B and then adding the results.

So, for the product of two matrices to be defined, the number of columns of the first matrix must equal the number of rows of the second matrix. That is, the middletwo indices must be the same.

The outside two indices give the dimension of the product as show below

  $\begin{array}{l}\underset{\underset{\uparrow }{m}\times \underset{\uparrow }{n}}{A}\underset{\begin{array}{l}\\ \\ Equal\end{array}}{\times }\underset{\underset{\uparrow }{n}\times \underset{\uparrow }{p}}{B}=\underset{m\times p}{AB}\\ DimensionofAB\end{array}$

Definition of the Inverse of a Square Matrix:

Let A be $n\times n$ matrix and let $I_n$ be the $n\times n$ identity matrix.

If there exists a matrix $A^{-1}$ such that $A{A}^{-1}=I_n=A^{-1}A$ then $A^{-1}$ is the inverse of A.

The symbol $A^{-1}$ is read as “ $A$ inverse.”

To show that B is the inverse of A, we need to show that AB = I = BA.

Calculation:

Finding the product AB

  $\left[\begin{array}{cccc}-1& 1& 0& -1\\ 1& -1& 1& 0\\ -1& 1& 2& 0\\ 0& -1& 1& 1\end{array}\right]\cdot \frac{1}{3}\left[\begin{array}{cccc}-3& 1& 1& -3\\ -3& -1& 2& -3\\ 0& 1& 1& 0\\ -3& -2& 1& 0\end{array}\right]$

Multiply the elements of each row of the first matrix by the elements of each column in the second matrix and add the products

Simplifying each element of the matrix further

  $AB=\frac{1}{3}\left[\begin{array}{cccc}3& 0& 0& 0\\ 0& 3& 0& 0\\ 0& 0& 3& 0\\ 0& 0& 0& 3\end{array}\right]$

Multiplying the 1/3 with each element inside the matrix

  $AB=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]$

Finding the product BA

  $\frac{1}{3}\left[\begin{array}{cccc}-3& 1& 1& -3\\ -3& -1& 2& -3\\ 0& 1& 1& 0\\ -3& -2& 1& 0\end{array}\right]\cdot \left[\begin{array}{cccc}-1& 1& 0& -1\\ 1& -1& 1& 0\\ -1& 1& 2& 0\\ 0& -1& 1& 1\end{array}\right]$

Multiply the elements of each row of the first matrix by the elements of each column in the second matrix and add the products

Simplifying each element of the matrix further

  $BA=\frac{1}{3}\left[\begin{array}{cccc}3& 0& 0& 0\\ 0& 3& 0& 0\\ 0& 0& 3& 0\\ 0& 0& 0& 3\end{array}\right]$

Multiplying the 1/3 with each element inside the matrix

  $BA=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]$

Conclusion:

Since AB is equal to BA, A is an inverse of B.