Science:Math Exam Resources/Courses/MATH221/December 2007/Question 05 (a)

MATH221 December 2007

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Question 05 (a)
Find the determinant of the matrix $B={\begin{pmatrix}1&1&1&1\\1&2&3&4\\1&3&7&10\\1&4&10&t\end{pmatrix}}.$

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

Checking a solution serves two purposes: helping you if, after having used the hint, 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. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Recall that adding multiples of one row to another does not change the determinant. So to make our lives easier when calculating the determinant of B, we will simplify by row reduction as much as possible. Subtracting the first row from the second, third, and fourth, results in the matrix ${\begin{pmatrix}1&1&1&1\\0&1&2&3\\0&2&6&9\\0&3&9&t-1\end{pmatrix}}$ whose determinant is the same as B. Now subtracting twice the second row from the third, and three times the second row from the fourth, gives ${\begin{pmatrix}1&1&1&1\\0&1&2&3\\0&0&2&3\\0&0&3&t-10\end{pmatrix}}$ whose determinant is still equal to that of B. A final step is to subtract 3/2 times the third row from the fourth to have ${\begin{pmatrix}1&1&1&1\\0&1&2&3\\0&0&2&3\\0&0&0&t-10-9/2\end{pmatrix}}.$ This matrix is upper-triangular, so we know how to calculate its determinant: it is simply the product of the diagonal entries. We therefore conclude that $\det(B)=1\cdot 1\cdot 2\cdot (t-10-9/2)=2t-29.$

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