Science:Math Exam Resources/Courses/MATH221/April 2010/Question 04

MATH221 April 2010

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Question 04
Compute the matrix product ${\begin{pmatrix}3&1&2\\2&0&0\end{pmatrix}}{\begin{pmatrix}3&1&0\\3&0&2\\5&0&0\end{pmatrix}}{\begin{pmatrix}2&0\\-1&0\\0&3\end{pmatrix}}$ .

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. In order to multiply matrices, we must first ensure that the number of columns in the first matrix matches the number of rows in the second matrix. This is because we multiply the number in the first column in the first matrix by the number in the first row in the second matrix, then the number in the second column and second row, and so on (i.e. multiply the first row by the first column). The number in the first row and column of the matrix product of these two matrices will be the sum of the numbers multiplied column by row. The first row will then be multiplied by the second column, and this will make up the number in the second column, first row in the matrix product. This pattern will continue until there are no more columns in the second matrix. Then, we move on to the second row in the first matrix and do the same thing, with the sum of the products now filling out the second row in the product matrix. We will continue this pattern until there are no more rows in the first matrix. Since we have three matrices, we will start by multiplying the first two, then multiplying this matrix product by the third matrix to get our answer. Note that the number of columns in matrix one is equal to the number of rows in matrix two, which is three. Multiplying matrix one by two, we get: ${\begin{pmatrix}3&1&2\\2&0&0\end{pmatrix}}{\begin{pmatrix}3&1&0\\3&0&2\\5&0&0\end{pmatrix}}={\begin{pmatrix}22&3&2\\6&2&0\end{pmatrix}}$ Now, multiply this matrix by matrix three. Again note that the number of columns in the new matrix is equal to the number of rows in matrix three. ${\begin{pmatrix}22&3&2\\6&2&0\end{pmatrix}}{\begin{pmatrix}2&0\\-1&0\\0&3\end{pmatrix}}={\begin{pmatrix}41&6\\10&0\end{pmatrix}}$ This result is the solution we are looking for.