Q6. - Algebra 2

Question

Find the value of each determinant.

6.

$\left|\begin{array}{ccc}1& 3& -2\\ 7& 0& 4\\ -3& 5& -1\end{array}\right|$

Step-by-Step Solution

Answer

The value of the given determinant is $\mathbf{-}\mathbf{105}$.

Step 1 ­- Description of step.

The value of a third-order determinant by expansion along the first row can be found out by the following method:

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

Step 2 ­- Find the value of the determinant.

The value of the determinant $\left|\begin{array}{ccc}1& 3& -2\\ 7& 0& 4\\ -3& 5& -1\end{array}\right|$is given by:

$\begin{array}{c}\left|\begin{array}{ccc}1& 3& -2\\ 7& 0& 4\\ -3& 5& -1\end{array}\right|=1\left|\begin{array}{cc}0& 4\\ 5& -1\end{array}\right|-3\left|\begin{array}{cc}7& 4\\ -3& -1\end{array}\right|+\left(-2\right)\left|\begin{array}{cc}7& 0\\ -3& 5\end{array}\right|\\ =1\left(0-\left(20\right)\right)-3\left(-7-\left(-12\right)\right)-2\left(35+0\right)\\ =1\left(-20\right)-3\left(-7+12\right)-2\left(35\right)\\ =-20-3\left(5\right)-70\\ =-20-15-70\\ =-105\end{array}$

Step 3 ­- Description of step.

The value of the given determinant is $-105$.