Science:Math Exam Resources/Courses/MATH221/April 2013/Question 06

MATH221 April 2013

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[hide] Question 06
Find the determinant of $\left[{\begin{array}{ccccc}1&2&2&-1&3\\2&6&3&-3&7\\-3&-4&-3&2&-8\\-2&-6&-2&2&-7\\1&-2&7&3&3\end{array}}\right]$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

[show] Hint 1
Spend 5 minutes wondering at the unfairness of the world in having to deal with a 5x5 matrix. Once done, start row and column reducing.

[show] Hint 2
Remember that row operation affect the determinant as follows: Adding scalar multiples of other rows does not change the determinant. Multiplying a row by a scalar multiple changes the determinant by a factor of the scalar. Swapping rows introduces a negative sign in the determinant

[show] Hint 3
The determinant of an upper (or lower) diagonal matrix is the product of its diagonal entries. Recall that an upper diagonal matrix has all zeroes below the diagonal entries.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work.
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[show] Solution
The first thing we will do is simplify the matrix by row and column operations. We start by replacing row 4 with row 2 plus row 4 (this does not change the determinant) giving: $\left[{\begin{array}{ccccc}1&2&2&-1&3\\2&6&3&-3&7\\-3&-4&-3&2&-8\\0&0&1&-1&0\\1&-2&7&3&3\end{array}}\right]$ This step helps to simplify the row reductions. Next, using row 1 as a pivot, we replace row 2 with row 2 minus 2 times row 1, row 3 with row 3 plus three times row 1 and row 5 with row 1 minus row 1 giving $\left[{\begin{array}{ccccc}1&2&2&-1&3\\0&2&-1&-1&1\\0&2&3&-1&1\\0&0&1&-1&0\\0&-4&5&4&0\end{array}}\right]$ Note that none of these changes affected the determinant either. Next, we replace row 3 with row 3 minus row 2 and we replace row 5 with row 5 plus two times row 2 to get $\left[{\begin{array}{ccccc}1&2&2&-1&3\\0&2&-1&-1&1\\0&0&4&0&0\\0&0&1&-1&0\\0&0&3&2&2\end{array}}\right]$ Again none of these changes affect the determinant. Next, using row 4 as a pivot, we replace row 3 with row 3 minus four times row 4 and replace row 5 with row 5 minus 3 times row 4 to get $\left[{\begin{array}{ccccc}1&2&2&-1&3\\0&2&-1&-1&1\\0&0&0&4&0\\0&0&1&-1&0\\0&0&0&5&2\end{array}}\right]$ Again this doesn't change the determinant. Next, use row 3 as a pivot and replace row 5 with row 5 minus five quarters of row 3 to get $\left[{\begin{array}{ccccc}1&2&2&-1&3\\0&2&-1&-1&1\\0&0&0&4&0\\0&0&1&-1&0\\0&0&0&0&2\end{array}}\right]$ None of these row operations have thus far changed the determinant. Now, we flip rows 3 and 4 which will introduce an extra negative sign in our determinant. So the quantity we want is $-\det(M)$ where M is defined by: $M=\left[{\begin{array}{ccccc}1&2&2&-1&3\\0&2&-1&-1&1\\0&0&1&-1&0\\0&0&0&4&0\\0&0&0&0&2\end{array}}\right]$ This matrix is diagonal and so the determinant of this matrix is just the product of the diagonal entries giving 16. Hence our answer is $-\det(M)=-16$ as required.