in-exercises-5-and-6-compute-the-product-a-b-in-two-ways-a-by-the-definition-where-a-mathbfb_1-and-a

Question

In Exercises 5 and $6,$ compute the product $A B$ in two ways: $(a)$ by the definition, where $A \mathbf{b}_{1}$ and $A \mathbf{b}_{2}$ are computed separately, and (b) by the row-column rule for computing $A B .$
$$
A=\left[\begin{array}{rr}{-1} & {2} \ {5} & {4} \ {2} & {-3}\end{array}ight], \quad B=\left[\begin{array}{rr}{3} & {-2} \ {-2} & {1}\end{array}ight]
$$

Lucas Finney · Accepted Answer

alright for this problem, we are given the two matrices A and B. We want to show how to calculate their product a b, two different ways, the proper way where we calculate a times Be one be to in this case, which is going to give us our resulting matrix being a B one a b two and also the sort of more heuristic row column way. So a starting off with the more formally proper way a times B one going to be well, we have copy. Paste that down there. Copy. Paste this down here any times be one is going to give us. Okay, So our upper diagonal element. Oh, now I see why I was confusing myself for a second there. Yes, Times be one as a vector. I&#x27;m a doofus. Okay, so you have negative one times three is going to give us negative three there plus two times negative one. So that&#x27;s minus four Then We have five times three gives us 15 plus four times negative too. So we have 15 minus eight of two times three. So we have six less negative three times negative too. So that is going to be six plus six. So that is going to give us overall negative. Seven 15 minus eight is going to be positive. Seven and 66 is going to be 12. Then have a be to no one moment. Okay, Continuing on to calculate a Times B two, we have our same a matrix and we&#x27;re multiplying the column Vector. I believe that&#x27;s going to be too negative. One negative to one, so we&#x27;ll get negative. Negative one times negative to positive, too. Plus two times one. So two plus two, we&#x27;ll have five times negative, too. So negative. 10 plus four times. One. That&#x27;s four, two times negative, too. Being negative for plus negative. Three times one negative three. So we get for negative six and negative seven. Okay, so that tells us that a B is going to equal. It was not a big enough bracket right there. A B is going to equal negative seven 7 12 for negative six Negative seven Now doing it the alternate way long here, doing it the alternate way. Doing it the alternate way we would have just multiplying our matrices together is in a row column method. I&#x27;m just going to grab the matrices. Post them down here. Mhm. Okay. What? So you&#x27;ll have negative one times three negative three last two times Negative, too. Then we&#x27;ll have five times three plus four times negative to we&#x27;ll have 15 minus eight. Should be starting to look familiar. Two times three gives us six plus negative six times negative, too. So it&#x27;s six, and we have next up. We have, uh, negative one times negative, too. So positive, too. Plus two. We have positive five times negative, too. Ah, plus four. We have two times negative to negative. Four plus negative. Three times one so negative for minus three. So that&#x27;s going to give us. Well, negative. Three plus two times Negative. Two. That&#x27;s negative. Three minus four, Then five times negative to that is negative. 10. Eso. We&#x27;re going to get negative. Seven than 15 minus seven is positive. 76 plus six is 12. Two plus two is four. Um, negative. 10 plus four is paused or not Positive. Negative six and then negative for minus three is going to give us negative seven. Thus, we arrive at the same conclusion, using the two different methods of calculating the Matrix product.

Mike Verwer · Answer

Okay. This question. Ask us to compute the products of A and B using the definition using column vectors where you do a B one and then a beats you separately. And then again using the Row column rule for multiplying matrices. So we&#x27;ll do the first method first, and we will write out. A B one is equal to four minus two minus 30 Excuse me. Zero, 35 and be one is three column Vector one, too. And this multiplication is the three by one matrix four minus four, which zero minus three plus zero is minus three and three times one plus five times two is, um, 13 and 82 is very similar. Four minus two minus 303 and the column Vector to is three minus one. And the resulting column vector here is four times three plus minus, two times minus one. So that&#x27;s 12 plus two is 14 minus three plus times three is minus nine plus zero is my SNI in three times three is nine. Minus five is four, and so the result, then a B is the combination of these two column vectors. Zero Maya&#x27;s 3 13 and 14 minus nine and four. Okay, we&#x27;ll start a new page, and we will do the row column. It multiplication A is for minus two minus three zero three five. We&#x27;ll fix this. Pardon me? That zero and B. It&#x27;s the two by two matrix 13 two, minus one. And we used the row column method. So that&#x27;s four times one plus minus, two times minus twos without zero. Four times three plus minus, two times minus one is 12 plus two. It&#x27;s 14 and then minus three times one is minus three, and it&#x27;s your times two and then minus three times three is minus nine. That&#x27;s my nine again and three times two three times. One story is three plus 10 is 13 in three times three is nine plus minus five is for, and so you can see we have the same result as in the other page. And so we&#x27;re done showing that these two methods are equivalent

Joe Lesueur · Answer

first step in this problem is to solve for the Matrix eight B, I noticed that Matrix A is a two by three matrix and Matrix B is a three by two matrix so multiplying these together will result in a two by two matrix. Do you multiply these two matrices Together? We will first take the first comma B and multiply by the first road off a three times one is three, two times negative. Four is eight and three times A is gonna be three a. Next people take the second column of Matrix B and multiply by the first row of Matrix A for our next value in the two by two matrix. The result be times one is B plus two times a is to a plus three times B is three beats. Next we will multiply the again first calm and measured to be by the second row in matrix a day three times to six negative four times five is negative 20 and seven times a is seven. A next move Multiply the second column in Matrix B by the second row in matrix, a two times be is to be plus five times A It&#x27;s five a plus seven times b is seven beating and this is our matrix. Now, this matrix could use slight simplification before you move forward. So combining like terms, we have the matrix A B, eagle 23 a minus five seven a minus 14 to a plus four b and finally five A plus nine b. The next step in this problem is to solve for a and B lower case and be in order for eight B to equal the identity matrix and a two by two grid. To do this, we know that this matrix is going to look like the following but the diagonal being ones and the outsides being zeros, which is too too identity matrix. To do this, we can set the value three a minus five equal to one. Since we know that this most equal this for this to be an identity matrix doing this, we find that a most equal to not to solve for B, we can plug in this value instead of equal toe one as well. So five and now we know a is too, so we can put two in for a and add mine be which we know should equal want. And this gives us that B equals negative one. And if you plug in those values into these equations, if you plug in the A and B values into this matrix, you will end up getting the identity matrix of 1001 The next stop is part in this problem is to solve for the Matrix B A with the values of lower case and be in mind. So be a remember multiplication is important. We were multiple matrix B, but adding in the values of Lower case and be by matrix. A Sorry, please disregard the line drawn through a problem. And keep in mind that this multiplication is a three by two matrix multiplied by a two by three matrix. The resultant will be a three by three matrix following the same steps as we did before with multiplication. We&#x27;ll start with the first column and major today and multiplied by the first row in Matrix B. And then we will move on to Cinco Matrix B times. The first row, followed by the third column, multiplied by the first row. And then we will continue all three of these for the next rose short In this video, the final answer of Matrix B A. After doing all these multiplication is correctly is the phone.

Michael Jacobsen · Answer

In this example, we have a matrix versus a beck vector product that we would like to calculate. The very first thing that we&#x27;re going to do is analyze whether this is defined or not, and that&#x27;s determining. Determined by first looking at the size of the Matrix, say it has three rose by two columns and looking at the size of the vector. Our vector has two rows, so it&#x27;s of size to buy one now, because of these inner dimensions here are equal that tells us this is well defined. Now go to the outer dimensions, which are three in one, and it tells us that the size of the product will be all together. Three rows and one column. Okay, let&#x27;s begin multiplying. One way to organize a Matrix Times a vector product is to take the matrix and multiply as followed effectively. What we&#x27;re forming is a leaner combination where the scale er&#x27;s come from the vector. Here we place a two and a negative three, and the vectors come from the original matrix. So play six. Negative 47 five, negative 36 and then we begin to multiply as normal. So first I&#x27;m going to scaler. Multiply. We have 12 negative 8 14 plus the vector. I&#x27;m multiplying in negative three. So we have negative 15 nine negative 18 and this result in the vector Negative three. Negative one and negative four altogether. Let&#x27;s try this in a slightly different way. The way we&#x27;re molded. Multiplying this time is motivated by the need of saving a little bit of time from writing this. Where we&#x27;re going to do is build the same size of result. It&#x27;s three by one, and we&#x27;re going to go in each entry and just do some quick multiplication. So we&#x27;re going to go first from roll one take six times two minus three times five. We&#x27;re taking this entry times this century and at the result, this entry times a century. So a visualized the row, multiplying the columns, corresponding entries and adding the results so have 12 from six times two minus 15 from fight by negative three, and that gives us negative three altogether. Next go down to the second row will have negative four by two, plus negative three times negative three, which is nine. And that gives us all together a positive one. Hope, which means this was a pause of one as well. Next go down to this row, we take seven times two plus six times negative three, and the result is 14 minus 18 which is a negative for So this is the way a lot of people end up visualizing the Matrix Times, a vector product, but it&#x27;s also important to be able to write it out in this way as well.

Ankit Gupta · Answer

we are given with the metrics equals two three negative one four negative fight and to six and we held B equals two six Zito seven. Negative one. So we can also that a is a metrics off three cross to at the same time be is a Matics off dimensions to cross two hands. The number of columns in A are equal to a number of rules and be therefore a B. It&#x27;s possible so we can give a B equals two three negative one four Negative Faith 26 multiplied with 607 negative one, which is a little. Now we multiply Roy&#x27;s first through Forest Column. That gives us three times six plus negative one times seven. Then First true second color, which gives us three times zero close negative one times negative one. Similarly others we have full times six Close negative five times seven two times six plus six times seven Here the hell four times a zero plus a negative five times negative one and we have two times zero plus six times negative one, and hence our metrics. A B is given by 18 minus seven that is 11 24 minus 35. That is negative. 11 in 12 plus 42 that goes 54 here. The hell negative one day, Miss Liberty one is one five and negative six. That is over product A B.

Ankit Gupta · Answer

in discussion. We are given with the mattresses a equals to fight six, which is ah to cross one metrics and the metrics B equals two negative three negative one negative five and negative nine, which is a one cross four metrics as number of columns in a number of rules and be our same hence product. Maybe it&#x27;s possible which will be off the order to cross full The product may be can be given by multiplying 56 of it the metrics be there is negative three negative one negative five and negative nine. And hence of the gate, the man takes a B off. The order to close for that is to lose. We will get So first, though, is often by multiplying first. Ruoff and Magic see with corresponding column soft metrics be so first row. First column gives us negative 15 1st True Second column gives those negative five first through third column gives us negative 25 1st rule Fourth column gives us negative 45. Similarly, second row First column gives this negative 18 2nd row Second column gives us negative six second or third column is negative. 36 2nd rule 2nd 4th column is negative. 54. That is our answer

Problem

In Exercises 5 and $6,$ compute the product $A B$…

Problem 5 Medium Difficulty

Answer

$\left[\begin{array}{cc}{-7} & {4} \\ {7} & {-6} \\ {12} & {-7}\end{array}\right]$

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

Linear Algebra and Its Applications

Alayna H.

Caleb E.

Maria G.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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$\left[\begin{array}{cc}{-7} & {4} \\ {7} & {-6} \\ {12} & {-7}\end{array}\right]$

Linear Algebra and Its Applications

Alayna H.

Caleb E.

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

Alayna H.

Caleb E.

Maria G.

Michael J.

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

Linear Algebra and Its Applications