the-expansion-of-a-3-times-3-determinant-can-be-remembered-by-the-following-device-write-a-second-2

Question

The expansion of a $3 	imes 3$ determinant can be remembered by the following device. Write a second copy of the first two columns to the right of the matrix, and compute the determinant by multiplying entries on six diagonals:
Graph cannot copy
Add the downward diagonal products and subtract the upward products. Use this method to compute the determinants in Exercises $15-18 .$ Warning: This trick does not generalize in any reasonable way to $4 	imes 4$ or larger matrices.
$$
\left|\begin{array}{rrr}{0} & {3} & {1} \ {4} & {-5} & {0} \ {3} & {4} & {1}\end{array}ight|
$$

James Chok · Accepted Answer

in this example. We want to find the determines all the three by three matrix using this strip right here. So the first step to perform industry is we have to copy the 1st 2 columns. Yeah, to copy them over. We have zero for three, 35 and four. Then we&#x27;ll multiply down on these, diagnose and multiply up on these. So is there times five times? One gives you zero plus three times during times free, which gives you the right plus one times four times for which gives you 16 and who, minus by three times five times of one, which is 50 minus by four times. There are times right, which is doing minus by one times four times street, which should give you 12. This is 16 minus 15 miles to which is one months, 12 and six equal to my system.

James Chok · Answer

okay. In this example, we want to find the determines off this matrix using this trick right here. So the first thing we have to copy the 1st 2 rows First columns onto the right side. So have 123 33 and three. The multiplying the diagonals. And these diagonals we get one times three times to, which is six plus three times one time, three get in line. Plus four times. Two times three, which is eight times three, which is six times 4 24 minus three times, three times, for which is nine times for which should give you 36. Why does three times? One times one, which is three minus two times two times three, which is 12. So this is equal to 39 minus by 36 months. Three is 39 minus 12. So this is my soul.

Harry Evans · Answer

this question. We want to run the determinant office three by three matrix just to working with the first column. So we will start with the minor for the one of free and into two. So by doing this, we have eliminating the brain. Bacanovic Minor. We have the determinant off one minus for 36 on Mindless because we work in is Joice on the matrix, US minus, plus minus, plus minus on plus minus. Plus, there&#x27;s minus free and then the limit. Never in the current containing the phrase search minus free to three and six on, then with the two that&#x27;s a plus two. So plus two, limiting their own carbon containing the to touch minus three to on Andi brightness for and then I cochlear. The determinant of image makes metrics. We have one time. Six Just six plus 12 minus free and then minus. That&#x27;s minus 18 minus six. Andi hunters dropped this subsidence. No way! Um, enough. Hey, that&#x27;s 12 minus two. And then just simply work in itself is just 18 minus. And then that&#x27;s minus 24. Ah, for us too. 10. What&#x27;s your secret? 18 Minus 72 US 20. Fitch is equal to minus very for

Sriram Soundarrajan · Answer

everyone today we&#x27;re going to start problem about five heavy needs to find the determent off the three by three matrix which has given us to three minus tree four zero three, 615 So if you expand, why do they get They will be like minus three in the 20 minus 18 second element is zero into some value minus morning. Six press developed. So it comes to be like minus three in your tooth my last one in the 18. So it will be like my last 24.

Kayleah Tsai · Answer

so he I&#x27;m going to show you another way to evaluate a three by three determinant. So the first day we&#x27;re going to do is we&#x27;re going to copy the 1st 2 columns and write them on the side here. So we have three, 21213 Then we&#x27;re going to draw, diagnose and each diagnose going to have three numbers in it. So what? We have three times one times one that we have to native tuned. One negative. 12 and three. They were going have ones that are going the other way. So we have 2 to 1 three negative. 23 native 11 So we have a bunch of, um, hatches everywhere, right? So I would have basically so they&#x27;re not confusing. Okay, so now what we have to do if we have to find the product of the numbers on each read, diagnose and something together, and then we&#x27;re going to do the product of all the numbers on the Blue Dagnall and some them together, they were going to subtract the products on the red on the blue from the red. So let&#x27;s multiply on the red first. So we have three times one times one plus to negative two times one plus Native, one times, two times three. So we&#x27;re going to end up with three plus negative for plus negative six. So we get negative one negative seven. So negative seven is all the bits. Now let&#x27;s go do on the blues. So the blues we have negative one, huh? Negative One times. One times one plus the really negative. Two and three three times Native two times three plus two times, two times one plus two times, two times one. So we get negative one plus as negative six. Um, negative 18 plus four. So we get negative. 19 plus four or negative 15. Now we&#x27;re going to subtract the some of the blues from the some of the red. So we&#x27;re going to get negative seven minus negative 15 which, because they give seven plus 15 we&#x27;re simply becomes eight. So the value of our determinate is eight

James Chok · Answer

Okay, so we want Thio find a determinant off this matrix using expansion across the first road and then we&#x27;ll find it by the second caller. So the first thing that we have to do is to record to remember what the signs up. So if you want to expand over the first road, it can&#x27;t be positive, negative and positive. So we have zero times by the determinants off when we removed the road and color off 3031 and then it&#x27;s negative. Wonderful. Minus full tons of determinants off 502 and one. And finally, it&#x27;s a plus sign under one time, by the determinant off five minus 32 and three. So this is zero multiplied by voxant determinants that this is zero. So you can stop that in this advice for time by the determinant, which is by time by one minus two times zero plus one times five times three minus two times like native three. This is four miles four time bye bye plus five pounds Freeze 15 15 plus six which is equal to 20 What? Minus 20 plus 21 which is simply one Now if you want todo an expansion on the second column. We have to remember that this is negative. This is positive. And this is negative. So the easiest way to remember is this is 11 The position of this block 11 and one plus one is two even. So, some positive. This is one to which is through one plus two years. Three, This is going to be negative. And this is 22 which is four. So it will be a positive. This is 32 which is five. And so we need to. So you will have minus four. Tether the determinants off 50 to 1 plus by minus three kinds. Like the determinant off your one Jew one in minus three times is your one five and zero. This is minus four times by five times one plus two times zero minus three times zero times one minus. Two times one minus three time. Zero times there, minus five times one. And this is minus four times by five. Minus three times minus two and minus three times minus five, which is minus 20 plus six plus 15 which is equal to one

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The expansion of a $3 \times 3$ determinant can b…

Problem 16 Easy Difficulty

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David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications

Heather Z.

Alayna H.

Kayleah T.

Kristen K.

Absolute Value - Example 1

Absolute Value - Example 2

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Linear Algebra and Its Applications

Heather Z.

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Absolute Value - Example 1

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David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Heather Z.

Alayna H.

Kayleah T.

Kristen K.

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications