Chapter 4.6, Problem 4CYU

To determine

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

Answer to Problem 4CYU

H and J are not inverse of each other.

Explanation of Solution

Given information:

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

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 ,

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

Now ,

  $H.J=\left[\begin{array}{cc}-3& -1\\ -4& -2\end{array}\right].\left[\begin{array}{cc}-1& 2\\ 3& -4\end{array}\right]$

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

and

  $J.H=\left[\begin{array}{cc}-1& 2\\ 3& -4\end{array}\right].\left[\begin{array}{cc}-3& -1\\ -4& -2\end{array}\right]$

  $\begin{array}{l}=\left[\begin{array}{cc}\left((-1)\times (-3)\right)+\left(2\times \left(-4\right)\right)& \left((-1)\times (-1)\right)+\left(2\times (-2)\right)\\ \left(3\times (-3)\right)+\left((-4)\times (-4)\right)& \left(3\times (-1)\right)+\left((-4)\times (-2)\right)\end{array}\right]\\ =\left[\begin{array}{cc}3-8& 1-4\\ -9+16& -3+8\end{array}\right]\\ =\left[\begin{array}{cc}-5& -3\\ 7& 5\end{array}\right]\end{array}$

This implies $H.J\ne J.H$ .

Since, $H.J\ne J.H\ne I$ , H and J are not inverses.