Evaluate each determinant in Exercises 49–52. 4 2 8 - 7- 2 0 4 15...

Question

Evaluate each determinant in Exercises 49–52. 4 2 8 - 7- 2 0 4 15 0 0 54 0 0 - 1

Matrix for Exercise 49 in college algebra, chapter on systems of equations.

Accepted Answer

hey everyone here we are asked to solve for the determinant in the following matrix. So we're given a four by four matrix which we want to reduce into smaller matrices. So to do this we first need to identify the row or column with the most zeroes and in this case column two has the most zeros. So to begin solving we want to expand and upon column two into smaller three by three matrices. And so to do this we select the first value in the column which is one. And to determine its sign, we multiply this one by the quantity of negative one to the power of its row which is one plus it's column which is to and now we multiply this value by a three by three matrix which we identify by crossing out the row which this value lies in and also crossing out its column. So now these remaining values here are the values for the three by three matrix. So the matrix is now negative and 702 and 60 negative three. And now we are done with the first value in this column. So we can move on to the second value which is just a zero. So we add zero and multiply it by the quantity of negative one to the power of its role which is two plus it's column which is also so two. And then multiply it by a three by three matrix. So again crossing out its role and the column reveals the values for the matrix. So for the first row we have 36 negative nine. Then we have 702 and finally negative three. And now we can move on to the next value which is also a zero. So we add zero and multiply it by the quantity of negative one to the power of the row which is three plus its column, which is two times a three by three matrix. Which is identified by crossing out the row and crossing out the column, revealing the values. So the first row is 36 negative nine. The second row is negative 463 and the final row is 60 negative three. And now we can move on to the final value which is also a zero. So we add zero and multiply it by the quantity of negative one to the power of its row which is four plus column which is two times its mate which is revealed by crossing out its row and crossing out its column. So this matrix has the values of 36 negative nine. For the first row then negative 463 and finally 702. So now with this equation here, this expansion we have so far we can reduce it because we have 30 terms which are just zero. So they can be ignored. Which leaves us with one times the quantity of negative one to the power of three which gives us the quantity of negative one times its matrix here which is a three by three matrix of negative 463702 and 60 negative three. So now we can continue the process we just went under because we want to continue to reduce this three by three matrix into a smaller two by two matrix. So to do this we once again identify the row or column with the most zeroes which in this case is again column two. So now we can expand upon this column. So again we take the first value which is six. And to determine the sign we multiply it by the quantity of negative one to the power of its role which is one plus its column which is two times a now. Two by two matrix which is identified by crossing out its role and crossing out the column. So we have a two by two matrix of 72 and six negative three. So now we can move on to the next terms in the column. So we move on to this zero and so we have plus zero times the quantity of negative one to the power of its row which is two plus its column which is also two times the matrix which we find by crossing out the row and crossing out the column. So we have negative +43 and six negative three. Now we can move on to the final term in the column which is yet another zero. So we have plus zero times the quantity of negative one to the power of its roll which is three plus it's column which is two times its matrix which is found by crossing out the row and column revealing the matrix negative 43 and seven to. So now we can reduce this equation. We have this expansion here because we have 20 terms which we can cancel out here. So we have six times the quantity of negative one to the power of three. So we really have the quantity of negative six times this. Two by two matrix here of 72 and six negative three. So now we can further expand this last two by two matrix because we can recall two by two. Matrix A B C. D. Is equal to the quantity A. D minus B. C. So with this we see that this two by two. Matrix is actually seven times negative three which is negative 21 minus B. C. Which is two times six which is 12. So this determinant here is negative 33. So now we have found three values. We have negative 1 -6 and -33 to find the final determinant. We must multiply all of these values. So we have negative one times negative six times negative 33 which gives us a final determinant of negative 198 which is our answer here for choice. A thanks for watching. I hope you found this video helpful, and I'll see you again next time.

Evaluate each determinant in Exercises 49–52. 4 2 8 - 7- 2 0 4 15 0 0 54 0 0 - 1

Key Concepts

Determinants

Matrix Operations

Systems of Equations

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