Q12. - Algebra 2

Question

Evaluate each determinant using diagonals.

 

12. $\left|\begin{array}{ccc}1& 6& 4\\ -2& 3& 1\\ 1& 6& 4\end{array}\right|$

Step-by-Step Solution

Answer

The determinant of a matrix $\left|\begin{array}{ccc}1& 6& 4\\ -2& 3& 1\\ 1& 6& 4\end{array}\right|$ using diagonals is 0. 

Step 1 - Rewrite the columns to the right side

In method of calculating determinant by using diagonals first two columns are rewritten outside to the right side of the determinant.

 

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

Step 2 - Find the products

The products are calculated diagonally in two ways – bottom products and top products. For bottom product, draw diagonals from each element of the top row of the determinant downward to the right. Find the product of the elements of each respective diagonal.

For top product, draw diagonals from each element of the bottom row of the determinant upward to the right. Find the product of the elements of each respective diagonal.

Step 3 - Add and subtract the products

In order to find the determinant, add the bottom products and subtract the top products.

 

$12+6-48-12-6+48=0$

 

Therefore, $\left|\begin{array}{ccc}1& 6& 4\\ -2& 3& 1\\ 1& 6& 4\end{array}\right|=0$.