Chapter 4.6, Problem 1CYP

To determine

To verify: whether $X=\left[\begin{array}{cc}4& -1\\ 2& -2\end{array}\right]$ and $Y=\left[\begin{array}{cc}\frac{1}{3}& -\frac{1}{6}\\ \frac{1}{3}& -\frac{2}{3}\end{array}\right]$ are inverse of each other.

Answer to Problem 1CYP

X and Y are inverse of each other.

Explanation of Solution

Given information:

  $X=\left[\begin{array}{cc}4& -1\\ 2& -2\end{array}\right]$ and $Y=\left[\begin{array}{cc}\frac{1}{3}& -\frac{1}{6}\\ \frac{1}{3}& -\frac{2}{3}\end{array}\right]$ .

Formula used:

Two $n\times n$ matrices are inverse of each other if their product is the identity matrix. We can say that, $A.B=B.A=I$ where A is a matrix.

Calculation:

Consider ,

  $X=\left[\begin{array}{cc}4& -1\\ 2& -2\end{array}\right]$ and $Y=\left[\begin{array}{cc}\frac{1}{3}& -\frac{1}{6}\\ \frac{1}{3}& -\frac{2}{3}\end{array}\right]$

Now ,

  $X.Y=\left[\begin{array}{cc}4& -1\\ 2& -2\end{array}\right].\left[\begin{array}{cc}\frac{1}{3}& -\frac{1}{6}\\ \frac{1}{3}& -\frac{2}{3}\end{array}\right]$

  $\begin{array}{l}=\left[\begin{array}{cc}\left(4\times \frac{1}{3}\right)+\left(-1\times \frac{1}{3}\right)& \left(4\times \frac{-1}{6}\right)+\left(-1\times \frac{-2}{3}\right)\\ \left(2\times \frac{1}{3}\right)+\left(-2\times \frac{1}{3}\right)& \left(2\times \frac{-1}{6}\right)+\left(-2\times \frac{-2}{3}\right)\end{array}\right]\\ =\left[\begin{array}{cc}\frac{4}{3}-\frac{1}{3}& -\frac{4}{6}+\frac{2}{3}\\ \frac{2}{3}-\frac{2}{3}& -\frac{2}{6}+\frac{4}{3}\end{array}\right]\\ =\left[\begin{array}{cc}\frac{4-1}{3}& \frac{-4+4}{6}\\ \frac{2-2}{3}& \frac{-2+8}{6}\end{array}\right]\\ =\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\end{array}$

and

  $Y.X=\left[\begin{array}{cc}\frac{1}{3}& -\frac{1}{6}\\ \frac{1}{3}& -\frac{2}{3}\end{array}\right].\left[\begin{array}{cc}4& -1\\ 2& -2\end{array}\right]$

  $\begin{array}{l}=\left[\begin{array}{cc}\left(\frac{1}{3}\times 4\right)+\left(\frac{1}{3}\times \left(-1\right)\right)& \left(\frac{-1}{6}\times 4\right)+\left(\frac{-2}{3}\times (-1)\right)\\ \left(\frac{1}{3}\times 2\right)+\left(\frac{1}{3}\times (-2)\right)& \left(\frac{-1}{6}\times 2\right)+\left(\frac{-2}{3}\times (-2)\right)\end{array}\right]\\ =\left[\begin{array}{cc}\frac{4}{3}-\frac{1}{3}& -\frac{4}{6}+\frac{2}{3}\\ \frac{2}{3}-\frac{2}{3}& -\frac{2}{6}+\frac{4}{3}\end{array}\right]\\ =\left[\begin{array}{cc}\frac{4-1}{3}& \frac{-4+4}{6}\\ \frac{2-2}{3}& \frac{-2+8}{6}\end{array}\right]\\ =\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\end{array}$

This implies $X.Y=Y.X$ .

Since, $X.Y=Y.X=I$ , X and Y are inverses.