Chapter 7.2, Problem 56E

Solution

To prove: The given matrix multiplication $A{I}_n=I_{n}A=A$ , where $A=\left[a_{ij}\right]$ be an $n\times n$ .

Given Information:

The matrix is defined as,

  $A=\left[a_{ij}\right]$

Calculation:

Consider the given information,

  $A=\left[a_{ij}\right]$

Multiply the inverse matrix with the given matrix.

  $\begin{array}{c}A{I}_n=\left[\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ a_{21}& a_{22}& \cdots & \\ & & \cdots & \\ a_{n1}& a_{n2}& \cdots & a_{nm}\end{array}\right]\cdot \left[\begin{array}{cccc}1& 0& \cdots & 0\\ 0& 1& \cdots & 0\\ & & \cdots & \\ 0& 0& \cdots & 1\end{array}\right]\\ =B\end{array}$

And,

In order to compute the term  $b_{ij}$ ? of the product matrix  B , multiply the row  $i$ from the matrix  A  by the column  $j$  of the matrix  $I_n$ ? (only the element on the  $ij$  position in the column  j  of the matrix  $I_n$ ? is not zero).

  $\begin{array}{c}{b}_{ij}=a_{i1}\cdot 0+a_{i2}\cdot 0+\cdots +a_{ij}\cdot 1+a_{i\left(j+1\right)}\cdot 0+\cdots +a_{in}\cdot 0\\ =a_{ij}\end{array}$

Since  $b_{ij}=a_{ij}$ for any  $i,j=1,n$ , it means that there is defined as:

  $A{I}_n=A$

So, it can be compute as,

  $\begin{array}{c}{I}_{n}A=\left[\begin{array}{cccc}1& 0& \cdots & 0\\ 0& 1& \cdots & 0\\ & & \cdots & \\ 0& 0& \cdots & 1\end{array}\right]\cdot \left[\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ a_{21}& a_{22}& \cdots & \\ & & \cdots & \\ a_{n1}& a_{n2}& \cdots & a_{nm}\end{array}\right]\\ =C\end{array}$

In order to compute the term  $c_{ij}$ ? of the product matrix  C , and multiply the row  i  from the matrix $I_n$ ? by the column  j  of the matrix  A .

  $\begin{array}{c}{c}_{ij}=0\cdot a_{1j}+0\cdot a_{2j}+\cdots +1\cdot a_{ij}+\cdots +0\cdot a_{nj}\\ =a_{ij}\end{array}$

It can be observed that $c_{ij}=a_{ij}$ for any $i,j=1,n$ . This implies that $I_{n}A=A$ .

Therefore, the given statement has proved.