Chapter 8.3, Problem 7E

To determine

Toshow: that B is the inverse of A.

Answer to Problem 7E

  $\text{AB=I=BA}$

So, B is the inverse of A.

Explanation of Solution

Given:

  $A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$ and $B=\left[\begin{array}{cc}-2& 1\\ \frac{3}{2}& -\frac{1}{2}\end{array}\right]$ .

Concept used:

If an $\text{m×n}$ matrix $\text{A=}\left[{\text{a}}_{\text{ij}}\right]$ and $\text{n×p}$ matrix $\text{B=}\left[{\text{b}}_{\text{ij}}\right]$ are there, then the matrix product AB is an $\text{m×p}$ matrix $\text{C=}\left[{\text{c}}_{\text{ij}}\right]=$ with $\text{ij}$ entry given by the formula

  ${\text{c}}_{\text{ij}}{\text{=a}}_{\text{i1}}{\text{b}}_{\text{1j}}{\text{+a}}_{\text{i2}}{\text{b}}_{\text{2j}}\text{+}\mathrm{........}{\text{+a}}_{\text{in}}{\text{b}}_{\text{nj}}$ .

Calculation:

Here,

  $A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$ and $B=\left[\begin{array}{cc}-2& 1\\ \frac{3}{2}& -\frac{1}{2}\end{array}\right]$

To show that B is the inverse of A, first find out the product of AB and BA.

If an $\text{m×n}$ matrix $\text{A=}\left[{\text{a}}_{\text{ij}}\right]$ and $\text{n×p}$ matrix $\text{B=}\left[{\text{b}}_{\text{ij}}\right]$ are there, then the matrix product AB is an $\text{m×p}$ matrix $\text{C=}\left[{\text{c}}_{\text{ij}}\right]=$ with $\text{ij}$ entry given by the formula

  ${\text{c}}_{\text{ij}}{\text{=a}}_{\text{i1}}{\text{b}}_{\text{1j}}{\text{+a}}_{\text{i2}}{\text{b}}_{\text{2j}}\text{+}\mathrm{........}{\text{+a}}_{\text{in}}{\text{b}}_{\text{nj}}$

Here A and B both are $\text{2}\times \text{2}$ square matrices, so the number of columns of A equals to the numbers of row of B. thus find the product AB, which is a $\text{2}\times \text{2}$ matrix, as follows

  $\text{AB=}\left[\begin{array}{cc}\text{1}& \text{2}\\ \text{3}& \text{4}\end{array}\right]\left[\begin{array}{cc}\text{-2}& \text{1}\\ \frac{\text{3}}{\text{2}}& \text{-}\frac{\text{1}}{\text{2}}\end{array}\right]$

  $\Rightarrow \left[\begin{array}{cc}\left(\text{1}\right)\left(\text{-2}\right)\text{+}\left(\text{2}\right)\left(\frac{\text{3}}{\text{2}}\right)& \left(\text{1}\right)\left(\text{1}\right)\text{+}\left(\text{2}\right)\left(\text{-}\frac{\text{1}}{\text{2}}\right)\\ \left(\text{3}\right)\left(\text{-2}\right)\text{+}\left(\text{4}\right)\left(\frac{\text{3}}{\text{2}}\right)& \left(\text{3}\right)\left(\text{1}\right)\text{+}\left(\text{4}\right)\left(\text{-}\frac{\text{1}}{\text{2}}\right)\end{array}\right]$

  $\Rightarrow \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

  $\text{AB=I}$ .

Similarly, the product of BA is

  $\text{BA=}\left[\begin{array}{cc}\text{-2}& \text{1}\\ \frac{\text{3}}{\text{2}}& \text{-}\frac{\text{1}}{\text{2}}\end{array}\right]\left[\begin{array}{cc}\text{1}& \text{2}\\ \text{3}& \text{4}\end{array}\right]$

  $\Rightarrow \left[\begin{array}{cc}\text{+}\left(\text{1}\right)\left(\text{3}\right)& \left(\text{-2}\right)\left(\text{2}\right)\text{+}\left(\text{1}\right)\left(\text{4}\right)\\ \left(\frac{\text{3}}{\text{2}}\right)\left(\text{1}\right)\text{+}\left(\text{-}\frac{\text{1}}{\text{2}}\right)\left(\text{3}\right)& \left(\frac{\text{3}}{\text{2}}\right)\left(\text{2}\right)\text{+}\left(\text{-}\frac{\text{1}}{\text{2}}\right)\left(\text{4}\right)\end{array}\right]$

  $\Rightarrow \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

  $\Rightarrow \text{I}$

Here that an $\text{n×n}$ matrix ${\text{A}}^{-1}$ such that ${\text{AA}}^{\text{-1}}{\text{=I}}_{\text{n}}{\text{=A}}^{\text{-1}}\text{A}$ then ${\text{A}}^{-1}$ is called the Inverse matrix.

Hence, $\text{AB=I=BA}$

So, B is the inverse of A.