Q. 7 - Algebra 2

Question

Find the value of each determinant.

 $\left|\begin{array}{ccc}\mathbf{3}& \mathbf{4}& \mathbf{4}\\ \mathbf{2}& \mathbf{1}& \mathbf{5}\\ \mathbf{0}& \mathbf{-}\mathbf{8}& \mathbf{6}\\ & & \end{array}\right|$

Step-by-Step Solution

Answer

The value of the given determinant is 26.

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 $\left|\begin{array}{ccc}3& 4& 4\\ 2& 1& 5\\ 0& -8& 6\\ & & \end{array}\right|$ is given by:

$\begin{array}{c}\left|\begin{array}{ccc}3& 4& 4\\ 2& 1& 5\\ 0& -8& 6\end{array}\right|=3\left|\begin{array}{cc}1& 5\\ -8& 6\end{array}\right|-4\left|\begin{array}{cc}2& 5\\ 0& 6\end{array}\right|+4\left|\begin{array}{cc}2& 1\\ 0& -8\end{array}\right|\\ =3\left(6-\left(-40\right)\right)-4\left(12-0\right)+4\left(-16-0\right)\\ =3\left(6+40\right)-4\left(12\right)+4\left(-16\right)\\ =3\left(46\right)-4\left(12\right)+4\left(-16\right)\\ =138-48-64\\ =26\end{array}$

Step 3. Description of step.

The value of the given determinant is 26.