Evaluate each determinant. See Example 3.

Question

Accepted Answer

Hello, everyone. We've been asked to find the value of the determinant below, we are given a three by three matrix where the first row reads six, negative 11 14, the second row reads and the third row reads negative 950. Our answer choices are a negative B 303 C negative 303 D 807. Before we start to solve this, I want to set up what our format is. So in a three by three matrix, I'm filling in generic values. So the top row is ABC second row would be D E F and the third row G H I, this way we know which place in the matrix we are referring to, to find the value of the determinant, we would do the value of a multiplied by the determinant of the minor matrix which is a two by two composed of the rows and columns that do not include A. So in this case, the first row would be E F, the second row would be H I minus the value of B multiplied by the determinant again of the two by two matrix that will be created by the rows and columns that do not include B. So this would be D F and G I and then plus the value of C multiplied by the determinant of the two by two matrix created by the rows and columns that do not include C so D E for the first row and G H for the second row. Now I'm gonna fill in these spotss with the values from our matrix. So A would be six multiplied by the determinant of the two by two matrix. That would be E F. So first row is 25, the second row would be 50 minus the value of B which is negative 11, which I'm putting in parentheses just to separate my minus sign from my negative. And that'll be multiplied by the determinant of the two by two matrix that has values D F. So 35 in the first row and G I so negative 90 in the second row plus the value of C which is multiplied by the determinant of the two by two matrix composed of D E or 32 in the top row and G H or negative 95 in the second row. The next thing to recall is that the determinant of A two by two matrix is found by the difference of the diagonal products starting at the top left corner. So what this would look like is we would have six, still multiplied by the parentheses. And in the parentheses are the values from finding my determinant. So two multiplied by zero, which is zero minus five, multiplied by five, which is 25 closed parentheses. Next I have minus and negative 11. So I'm going to rewrite this as plus 11 multiplied by the parentheses. And I will find this determinant by doing three multiplied by zero, which is zero minus five multiplied by negative nine, which is negative 45. I put the negative 45 in a set of parentheses to separate it from my minus sign. Then plus 14 multiplied by parentheses. And in my parentheses, the determinant would be three multiplied by five which is 15 minus two multiplied by negative nine, which is negative 18. Again, I put the negative 18 in parentheses to separate the minus sign from the negative sign. Next, I'm gonna simplify what's in my parentheses. So I have six multiplied by zero, minus 25 is negative 25 closed parentheses plus 11 multiplied by, we have zero minus negative 45 which is like zero plus 45. So 45 stays in the parentheses plus 14 multiplied by 15 minus negative 18 which is like adding and this will be a 33. Next, I need to find the values when I do the products and then we'll add it all together. So six multiplied by negative 25 is negative 150 11 multiplied by 45 is plus 14. Multiplied by 33 is from there. Combining those all together, it looks like we get the value and 807 should be our determinant which is answer choice. D have a nice day.

Evaluate each determinant. See Example 3.

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Key Concepts

Determinants

Matrix Operations

Cofactor Expansion