Chapter 16.7, Problem 1OE

To determine

  $a_1$ =?

  $\left|\begin{array}{ccc}0& 1& 2\\ 1& -1& -3\\ -2& 4& 5\end{array}\right|=a_1\left|\begin{array}{cc}-1& -3\\ b_3& 5\end{array}\right|-b_1\left|\begin{array}{cc}{a}_2& -3\\ -2& 5\end{array}\right|+2C_1$

Answer to Problem 1OE

0

Explanation of Solution

Given: $\left|\begin{array}{ccc}0& 1& 2\\ 1& -1& -3\\ -2& 4& 5\end{array}\right|=a_1\left|\begin{array}{cc}-1& -3\\ b_3& 5\end{array}\right|-b_1\left|\begin{array}{cc}{a}_2& -3\\ -2& 5\end{array}\right|+2C_1$

Formula used:

Expansion of a given 3 by 3 determinant can be given as:

  $\left|\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right|=a.\left|\begin{array}{cc}e& f\\ h& i\end{array}\right|-b.\left|\begin{array}{cc}d& f\\ g& i\end{array}\right|+c.\left|\begin{array}{cc}d& e\\ g& h\end{array}\right|$

Comparing the formula of expansion as stated above with the given determinant expansion in the form of minors, the answer is:

  $a_1=0$