Find each product, if possible.

Question

[\begin{matrix} \sqrt{3} & 1 \\ 2 \sqrt{5} & 3 \sqrt{2} \end{matrix}] [\begin{matrix} \sqrt{3} & - \sqrt{6} \\ 4 \sqrt{3} & 0 \end{matrix}]

Accepted Answer

Hello, everyone. We are asked to perform multiplication of the two matrices. Given below, we are given 22 by two matrices. The first one has a first row of radical in the second row of three radical 34 radical seven. The second matrix has a first row of radical two negative radical 10. In the second row of five radical 30, we are given four answer choices all of which are two by two matrices with slightly varying signs and values. Recall that when we multiply matrices, we want to first make sure that the dimensions match as these are both two by two matrices. We can assure that the columns of the first matrix match the rows of the second matrix. So we are able to multiply this and we know that our solution matrix will be a two by two based on the dimensions of our original. So starting off the first element in my solution matrix, the one in the top left corner will come from adding the products of the first row of the first matrix and the first column of the second matrix. So we will have the first element of the first row. So radical two multiplied by the first element of the first column, which is also radical two. We're gonna add that with the second element of the first row, which is one multiplied by the second element of the first column, which is five radical three. Next, we need to multiply the first row of the first matrix by the second column of the first matrix or sorry of the second matrix. So we will have radical two multiplied by negative radical 10. And then add to that one multiplied by zero. Next, going to the second row in the first matrix, I'm gonna multiply the first element from the second row. So three radical three multiplied by the first element of the first column in the second matrix. So radical two, and we're gonna add that to the second element of the second row. So four radical seven multiplied by the second element in the first column of the second matrix. So five radical three and then for our bottom right hand element, we're gonna do again the second row of the first matrix. And now we're gonna multiply it by the second column of the second matrix. So we'll have three radical three multiplied by negative radical 10. We're gonna add to that for radical seven multiplied by zero. All right. Now that we have all of the numbers, let's see what we can simplify. So starting in the top left corner radical two multiplied by radical two would be radical four, which the square root of four is two. So I'm gonna write that as two plus and then one multiplied by five radical three is five radical three. So my first element there is two plus five radical three continuing across the row, I have radical two multiplied by negative radical which when I combine those I will get negative radical and one multiplied by zero is zero. So I don't have to change that one. So that just stays negative radical 20. The first element in the second row, I have three radical three multiplied by radical two. So that will give me three radical six plus I then have four radical seven multiplied by five radical three. I'm gonna multiply the coefficients. So I get 20 and then multiply what's under the radical. So 21. So that element is three radical six plus 20 radical 21 and then our final element. So the bottom right corner three radical three multiplied by negative radical 10 gives us negative three radical and four radical seven multiplied by zero is zero. So we don't have to worry about that. So here is our solution matrix. There's nothing else here that simplifying really would help us with. So we're gonna leave it as this. It is a two by two matrix. The top row is the elements two plus five radical three and negative radical 20. The second row is three radical six plus radical 21 negative three radical 30. And this matches answer choice D have a nice day.

Find each product, if possible.
$[\begin{matrix} \sqrt{3} & 1 \\ 2 \sqrt{5} & 3 \sqrt{2} \end{matrix}] [\begin{matrix} \sqrt{3} & - \sqrt{6} \\ 4 \sqrt{3} & 0 \end{matrix}]$

Key Concepts

Polynomial Multiplication

Distributive Property

Combining Like Terms

Watch next