Video: Constructing a Matrix given a General Equation for Its Elements

Find the matrix 𝐴 = (𝑎_(𝑥𝑦)), with an order of 3 × 3, whose elements are given by the formula 𝑎_(𝑥𝑦) = 5𝑥 + 4𝑦.

04:09

Video Transcript

Find the matrix 𝐴 equals 𝑎 𝑥𝑦, with an order of three by three, whose elements are given by the formula 𝑎 𝑥𝑦 equals five 𝑥 plus four 𝑦.

So the first thing we know from the question is that we’re gonna be making a three-by-three matrix. And we’ve got a formula to help us do this. Now, the way that we find out what our matrix is going to be is by dealing with it a row at a time. So what we’re gonna do is we’re gonna start with the first row.

So to work out our first row, what we do is we fix 𝑥 to be equal to one. And then we vary 𝑦 from one to three. So when we do that, we can say that 𝑎 one one which is gonna be our first term in the top-left corner, well, this is gonna be equal to five multiplied by one plus four multiplied by one. And that’s cause we have five 𝑥. And we said that 𝑥 was gonna be equal to one, so five multiplied by one, and then we have four 𝑦. And we started with 𝑦 being equal to one. And when we do that, we get a value of nine because we have five plus four which is nine. So that’s gonna be our first term in the first row.

So now, to find the second term in the first row, what we’re gonna do this time is vary our 𝑦. So our 𝑦 is now gonna be equal to two. So we’re gonna get the second term. It’s gonna be equal to five multiplied by one plus four multiplied by two which is gonna give us five plus eight which is gonna give 13. So I’ve now added that to our matrix.

So then to find the last term in the first row, what we’re going to do is we’re gonna substitute in 𝑦 equals three. So we get five multiplied by one plus four multiplied by three which should give us five plus 12 which will give us our final term in the first row which is 17. So, therefore, we completed our first row. So great, now let’s move on to the second row.

So now, for the second row, what we’re gonna do is fix 𝑥 being equal to two. And then we’re gonna vary 𝑦 again this time from one to three. So then to find the first term in the second row, what we’re gonna do is five multiplied by two plus four multiplied by one which is gonna give us 10 plus four which is 14. So the first term in the second row is 14. So the only change again for the second term in the second row is change in the 𝑦 to be equal to two. So we get five multiplied by two plus four multiplied by two. So this is 10 plus eight which is gonna give us 18.

Then for the last term in the second row, once again, we’re gonna substitute in 𝑦 is equal to three which means we get five multiplied by two plus four multiplied by three which gives us 10 plus 12. So we get 22. So that’s the last term in the second row which is 22. So we’ve finished that row. And now, we can move on to the final row of our matrix.

Now, for the final row of the matrix, what we’re gonna do is fix 𝑥 being equal to three and then, once again, vary 𝑦 from one to three. So for the first term, we’re gonna have five multiplied by three plus four multiplied by one which is gonna give us 19 cause it’s 15 plus four. Then for the second term, once again, we just change our 𝑦, this time to be equal to two. So we get five multiplied by three plus four multiplied by two which is 23 because it’s 15 plus eight.

Then for the final term in our matrix, we’re gonna substitute in 𝑦 is equal to three. So we get five multiplied by three plus four multiplied by three which is gonna give us 15 plus 12 which is gonna give us 27. So, therefore, we’ve finished our matrix. So, therefore, we can say that the matrix with an order of three by three whose elements are given by the formula, 𝑎𝑥𝑦 equals five 𝑥 plus four 𝑦, is nine, 13, 17, 14, 18, 22, 19, 23, 27.

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