each-equation-in-exercises-1-4-illustrates-a-property-of-determinants-state-the-property-leftbegin-4

Question

Each equation in Exercises $1-4$ illustrates a property of determinants. State the property.
$$
\left|\begin{array}{rrr}{1} & {3} & {-4} \ {2} & {0} & {-3} \ {3} & {-5} & {2}\end{array}ight|=\left|\begin{array}{rrr}{1} & {3} & {-4} \ {0} & {-6} & {5} \ {3} & {-5} & {2}\end{array}ight|
$$

James Chok · Accepted Answer

Okay, so we want to know how the property that makes these two determinants. So to do this, we first have to know how you go from the left matrix to the right matrix. It is the right operation that we did is right to you. Minus two times we rode one, and we put that into your tooth. So we use the property that if a multiple off one rode off a is added to another road to produce a matrix B, then determines be is it determines off, eh?

Ian Shi · Answer

All right, So in this question, we&#x27;re talking about how doing elementary operations to in Matrix will change the terminate of sad matrix. So let&#x27;s let matrix b the attainment from a major say, by a one single elementary row operation. I will let for a convention. Be on. May I be the I throw of each matrix and in this case, we&#x27;re just focusing on the row multiplication by a constant. So if we have, let&#x27;s say the I equals a constant C times a I. What this implies is that the determinant of be just equals, see the same constant times, the determinant of a and that&#x27;s it.

James Chok · Answer

Okay, So in this question, you want to find out why that the determinant off this matrix is equal to the determinants off this matrix. So first thing that we have to do is to find the road operations that we went on the left hand side too. Right hand side the reparations. That dude was right three minus three times in a row one and we put that into road three. So now the reason why these two determines our equal because we use property that the multiple off one road added to another road produces the same determinants.

Christopher Stanley · Answer

on this problem. We won&#x27;t evaluate this large determined now to do this. You interested to take this piece by piece In the top left, three times five is native, 15 zero times for a zero negative. 15 minus zero is negative. 15. So the top left goes to negative 15 here in the top. Right? Negative. One times negative. One is one zero time 00 one minus zero Is one here in the bottom, right? Six times seven is 42. Now you 25 times for his native 20 40 to minus 90 of 20th 62 and then in the bottom, right. Five times four is 20 one times negative. Three is native three 20 minus negative. Three is 23. And so we end up with that determinant there. Now to evaluate it. 23 times negative 15 is 345 negative. 345. One time, 60 to 62 and then we take negative 345 minus 62. And that gives us negative 407. So this determinant should have a value of negative. 407

Lauren Shelton · Answer

Okay, So first, let&#x27;s find the determinant of em, which is just 12 minus 10 which will be to then let&#x27;s find the determinative. And two times five is 10. Minus seven is three. And now let&#x27;s multiply the matrices end times. And so this will end up becoming three times to plus two times one and then three times to excuse me three times seven plus two times five five times, two plus four times one and then doing +54 and +57 So five times seven plus four times five. So we&#x27;re running the dot product of each of these. So this will end up being, um, six plus two, which is eight. 21 plus 10 is 31 um, 10 plus four is 14 and then 35 plus 20 is 55. And now let&#x27;s find the determinate of M. And so this will be eight times 55 minus 14 times 31 then we know that eight times 35 is 4 40 minus 14 times 31. It will be 4 34 which will end up being six. Okay, So what are we trying to prove here? We can also see that the determinative of M times the determinant of end, which is three, which is two times three is six, which is the determinant of M times.

Lauren Shelton · Answer

So first I want to find the determinant of em, which is going to be for keeps for his 12 minus victims to its 10 which is two. I want to find him in verse. I know it&#x27;s one over the determinant times I switch a indeed four and three and that I make B and C negative. So if I distribute this to each term, four divided by two is too negative. One native, 2.5 and then 15 Okay, so now let&#x27;s find the determinant of em in verse. This will be two times 1.5 is three minus negative, 2.5 times one is positive. 2.5 is your 0.5, which is the same as one half. And we also know that the determine of em is to this is the same as one over the determinant of em.

Problem

Find the determinants in Exercises 5–10 by row re…

Problem 4 Easy Difficulty

Each equation in Exercises $1-4$ illustrates a property of determinants. State the property.
$$
\left|\begin{array}{rrr}{1} & {3} & {-4} \\ {2} & {0} & {-3} \\ {3} & {-5} & {2}\end{array}\right|=\left|\begin{array}{rrr}{1} & {3} & {-4} \\ {0} & {-6} & {5} \\ {3} & {-5} & {2}\end{array}\right|
$$

Answer

Adding any linear combination of rows into any row doesn't change the determinant.

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

Linear Algebra and Its Applications

Grace H.

Alayna H.

Maria G.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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Adding any linear combination of rows into any row doesn't change the determinant.

Linear Algebra and Its Applications

Grace H.

Alayna H.

Maria G.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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

Grace H.

Alayna H.

Maria G.

Michael J.

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

Linear Algebra and Its Applications