Chapter 11.3, Problem 63AYU

To determine

To show: When the entire second row of a 3 by 3 determinant is multiplied by a number k , $k\ne 0$ the value of the new determinant is $k$ times the value of the original determinant.

Explanation of Solution

Given information:

The determinant of a 3 by 3 matrix.

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 $\left|A\right|=ps-qr$ .

Proof:

Consider a 3 by 3 matrix say, $\left(\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right)$ .

Now, denote the determinant of the above matrix as, $D_1=\left|\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right|$ .

Compute it,

  $\begin{array}{c}{D}_1=a\left|\begin{array}{cc}e& f\\ h& i\end{array}\right|-b\left|\begin{array}{cc}d& f\\ g& i\end{array}\right|+c\left|\begin{array}{cc}d& e\\ g& h\end{array}\right|\\ =a\left(ei-fh\right)-b\left(di-fg\right)+c\left(dh-eg\right)\\ =aei-afh-bdi+bfg+cdh-ceg\end{array}$

Now, multiply the entire second row by a number k , $k\ne 0$ of the original matrix, so new matrix is $\left(\begin{array}{ccc}a& b& c\\ kd& ke& kf\\ g& h& i\end{array}\right)$

Now, denote the determinant of the above matrix as, $D_2=\left|\begin{array}{ccc}a& b& c\\ kd& ke& kf\\ g& h& i\end{array}\right|$ .

Compute it,

  $\begin{array}{c}{D}_2=a\left|\begin{array}{cc}ke& kf\\ h& i\end{array}\right|-b\left|\begin{array}{cc}kd& kf\\ g& i\end{array}\right|+c\left|\begin{array}{cc}kd& ke\\ g& h\end{array}\right|\\ =a\left(kei-kfh\right)-b\left(kdi-kfg\right)+c\left(kdh-keg\right)\\ =kaei-kafh-kbdi+kbfg+kcdh-kceg\\ =k\left(aei-afh-bdi+bfg+cdh-ceg\right)\end{array}$

Observe that, $D_2=k{D}_1$ .

Hence, it is shown that when the entire second row of a 3 by 3 determinant is multiplied by a number k , $k\ne 0$ the value of the new determinant is $k$ times the value of the original determinant.