Q27. - Algebra 2

Question

Evaluate each determinant using expansion by minors.

 

$\left|\begin{array}{ccc}3& 1& 2\\ 0& 6& 4\\ 2& 5& 1\end{array}\right|$

Step-by-Step Solution

Answer

The value of the determinant $\left|\begin{array}{ccc}3& 1& 2\\ 0& 6& 4\\ 2& 5& 1\end{array}\right|=-58$.

Step 1 - Decide row to use for expansion

In this case, use first row for the expansion by minors.

 

$\left|\begin{array}{ccc}3& 1& 2\\ 0& 6& 4\\ 2& 5& 1\end{array}\right|=3\left|\begin{array}{cc}6& 4\\ 5& 1\end{array}\right|-\left(1\right)\left|\begin{array}{cc}0& 4\\ 2& 1\end{array}\right|+2\left|\begin{array}{cc}0& 6\\ 2& 5\end{array}\right|$

Step 2 - Evaluate 2 × 2 determinants

The determinant of second order matrix is found by calculating the difference of the product of the two diagonals, that is., $\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|=ad-bc$ . Apply this definition to find the $2\times 2$ determinant.

$\begin{array}{c}\left|\begin{array}{ccc}3& 1& 2\\ 0& 6& 4\\ 2& 5& 1\end{array}\right|=3\left|\begin{array}{cc}6& 4\\ 5& 1\end{array}\right|-\left(1\right)\left|\begin{array}{cc}0& 4\\ 2& 1\end{array}\right|+2\left|\begin{array}{cc}0& 6\\ 2& 5\end{array}\right|\\ =3\left(6-20\right)-\left(1\right)\left(0-8\right)+2\left(0-12\right)\end{array}$

Step 3 - Multiply and simplify

Further simplify the values.

 

 $\begin{array}{c}\left|\begin{array}{ccc}3& 1& 2\\ 0& 6& 4\\ 2& 5& 1\end{array}\right|=3\left(6-20\right)-\left(1\right)\left(0-8\right)+2\left(0-12\right)\\ =3\left(-14\right)-\left(-8\right)+2\left(-12\right)\\ =-42+8-24\\ =-58\end{array}$