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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 .$$$A=\left[\begin{array}{rr}{-1} & {2} \\ {5} & {4} \\ {2} & {-3}\end{array}\right], \quad B=\left[\begin{array}{rr}{3} & {-2} \\ {-2} & {1}\end{array}\right]$$

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

Algebra

Chapter 2

Matrix Algebra

Section 1

Matrix Operations

Introduction to Matrices

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In mathematics, the absolu…

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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'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'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're multiplying the column Vector. I believe that's going to be too negative. One negative to one, so we'll get negative. Negative one times negative to positive, too. Plus two times one. So two plus two, we'll have five times negative, too. So negative. 10 plus four times. One. That'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'm just going to grab the matrices. Post them down here. Mhm. Okay. What? So you'll have negative one times three negative three last two times Negative, too. Then we'll have five times three plus four times negative to we'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'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's going to give us. Well, negative. Three plus two times Negative. Two. That's negative. Three minus four, Then five times negative to that is negative. 10. Eso. We'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.

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