Q28. - Algebra 2

Question

Evaluate each determinant using expansion by minors.

 

$\left|\begin{array}{ccc}7& 3& -4\\ -2& 9& 6\\ 0& 0& 0\end{array}\right|$

Step-by-Step Solution

Answer

The value of the determinant $\left|\begin{array}{ccc}7& 3& -4\\ -2& 9& 6\\ 0& 0& 0\end{array}\right|=0$.

Step 1 - Decide row to use for expansion

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

 

$\left|\begin{array}{ccc}7& 3& -4\\ -2& 9& 6\\ 0& 0& 0\end{array}\right|=0\left|\begin{array}{cc}3& -4\\ 9& 6\end{array}\right|-0\left|\begin{array}{cc}7& -4\\ -2& 6\end{array}\right|+0\left|\begin{array}{cc}7& 3\\ -2& 9\end{array}\right|$

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.

 

$\begin{array}{c}\left|\begin{array}{ccc}7& 3& -4\\ -2& 9& 6\\ 0& 0& 0\end{array}\right|=0\left|\begin{array}{cc}3& -4\\ 9& 6\end{array}\right|-0\left|\begin{array}{cc}7& -4\\ -2& 6\end{array}\right|+0\left|\begin{array}{cc}7& 3\\ -2& 9\end{array}\right|\\ =0\left(18-\left(-36\right)\right)-0\left(42-8\right)+0\left(63-\left(-6\right)\right)\end{array}$

Step 3 - Multiply and simplify

Further simplify the values.

 

 $\begin{array}{c}\left|\begin{array}{ccc}7& 3& -4\\ -2& 9& 6\\ 0& 0& 0\end{array}\right|=0\left(18-\left(-36\right)\right)-0\left(42+8\right)+0\left(63-\left(-6\right)\right)\\ =0\end{array}$