Chapter 8.4, Problem 77E

To determine

To verify: The equation,

  $\begin{array}{cc}w& x\\ y& z\end{array}=\begin{array}{cc}w& x+cw\\ y& z+cy\end{array}$

With help of determinants.

Explanation of Solution

Given information:

The equation,

  $\begin{array}{cc}w& x\\ y& z\end{array}=\begin{array}{cc}w& x+cw\\ y& z+cy\end{array}$

Formula used:

For the matrix of order $2\times 2$ ,

  $A=\left[\begin{array}{ll}p\hfill & q\hfill \\ r\hfill & s\hfill \end{array}\right]$

The determinant is given as $\left|A\right|=ps-qr$ .

Calculation:

Consider the equation,

  $\begin{array}{cc}w& x\\ y& z\end{array}=\begin{array}{cc}w& x+cw\\ y& z+cy\end{array}$

Recall that for the matrix of order $2\times 2$ ,

  $A=\left[\begin{array}{ll}p\hfill & q\hfill \\ r\hfill & s\hfill \end{array}\right]$

The determinant is given as $\left|A\right|=ps-qr$ .

Therefore, determinant of the matrix on the left hand side is given by,

  $\begin{array}{c}\begin{array}{cc}w& x\\ y& z\end{array}=\left(w\right)\left(z\right)-\left(x\right)\left(y\right)\\ =wz-xy\end{array}$

Therefore, determinant of the matrix on the right hand side is given by,

  $\begin{array}{c}\begin{array}{cc}w& x+cw\\ y& z+cy\end{array}=\left(w\right)\left(z+cy\right)-\left(x+cw\right)\left(y\right)\\ =wz+cwy-\left(xy+cwy\right)\\ =wz+cwy-xy-cwy\\ =wz-xy\end{array}$

Since, the left hand side is equal to right hand side, the equation is verified.

Thus, the equation, $\begin{array}{cc}w& x\\ y& z\end{array}=\begin{array}{cc}w& x+cw\\ y& z+cy\end{array}$ is verified with help of determinants.