In Exercises 31–36, use the alternative method for evaluating ...

Question

In Exercises 31–36, use the alternative method for evaluating third-order determinants on here to evaluate each determinant. $\begin{vmatrix}-3 & 4 & -5 \5 & -2 & 0 \8 & -1 & 3\end{vmatrix}$

Accepted Answer

hey everyone here we are asked to use diagonals to evaluate the following third order determinant. So to begin solving this problem, we want to copy down these three given columns and repeat the 1st and 2nd column. So for our first column we have two negative 43. For our second column we have 10 negative six. And for our third column we have negative 879. And now repeating the first two columns. Our fourth column is two negative 43 and our fifth column is 10 negative six. And now we can start drawing in our diagonals. For so for the first three columns we have diagonals going from left to right. And for the last three columns we need diagonals going from right to left. So starting with with our left to right diagonals. We want to begin at this first column here so we go from 2 to 0 to nine. We then go from 1 to 7 to three and then we go from negative eight to negative four to negative six. And now starting at the right most column we will draw diagonals going from right to left. So we have 12 negative 4 to 9. We then have 2 to 7 to negative six. And lastly we have negative 8 to 0 to three. And now we just want to take the product of the terms within each diagonal. So starting with this diagonal here, that ends at the first column here we have negative eight times zero times three which is just zero. We then move on and see that we have two times seven times negative six which is negative 84. We then have one times negative four times nine which is negative 36. And now moving on to the diagnose, going from left to right. We have two times zero times nine which is just zero. We then have one times seven times three which is 21. We then have negative eight times negative four times negative six which gives us negative 192. And now to begin solving for the determinant we want to add the products with diagonals going from left to right and we want to subtract the products with diagonals going from right to left. So beginning with the left to right diagonals we have zero plus 21 plus negative 192. And now for the products with right to left diagonals we have minus zero minus negative 84 minus negative 36. And now simplifying this expression here we have this 192 plus plus 36. And now just solving this equation we see that our determinant is negative 51 which is answer choice C Thanks for watching. I hope you found this video helpful and I'll see you again next time

In Exercises 31–36, use the alternative method for evaluating third-order determinants on here to evaluate each determinant. $\begin{vmatrix}\)-3 & 4 & -5 \\5 & -2 & 0 \\8 & -1 & 3\(\end{vmatrix}$

Key Concepts

Third-Order Determinants

Alternative Method for Evaluating Determinants

Properties of Determinants in Systems of Equations

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