In Exercises 9 - 16, find the following matrices: d. - 3A + 2B2 -...

Question

In Exercises 9 - 16, find the following matrices: d. - 3A + 2B2 - 10 - 2 6 10 - 2A = 14 12 10 B = 0 - 12 - 44 - 2 2 - 5 2 - 2

Matrices A and B for exercise 15 in college algebra, chapter 7 on systems of equations.

Accepted Answer

hey everyone in this problem we are asked to perform the following matrix operation and the operation is to a minus five. B. K. Were given A. And B which are three by three matrices, K. A. Has first row, three negative two negative three. Second row 7 to 2. And third row five negative 10. And B has first row 12 negative eight. Second row zero negative six negative three. And third row negative 21 and five. Okay, so let's just go ahead and write out what we're trying to find. Two A minus five B. Let's write this in terms of our matrices. So we get two times the matrix A. 375. In the first column negative 22 negative one in the second column negative 3 to 0 in the third column. Okay minus five times the matrix B. 10 negative two in the first column to negative 61 in the second column and negative eight negative 35 in the third column. Alright, so we see we have some multiplication here. And we also have the subtraction. Similarly to regular multiplication, division addition. We want to do the multiplication first. Okay And then we'll do the subtraction. So let's go ahead and do the multiplication. Now when we're multiplying matrices by a constant, the constant is going to come into that matrix and multiply to every single element in it. Okay, so two times matrix A. We're gonna end up with two times three. Two times seven. Two terms five. Okay, two times negative two, two times two. two times negative one. two times negative three. Two times two And two times 0. Okay, so we've just taken that too. We've multiplied it to every single element. Okay, We have our subtraction and then we're gonna do the same for the second. Matrix. We have five times the matrix B. That constant five is gonna come in and multiply to every single element. So we get five times one. Five times zero. five times negative. two, five times two, five times negative. 6, 5 times one. five times negative. eight. five times negative. 3. 5 times five. Alright, so now we have two matrices. Let's go ahead. And just simplify within those matrices. Do those multiplication. Okay. And this is just regular multiplication. Two times three is 6, two times 7, 14. 2 times 5. 10. two times negative. 2 -4. Four negative two -6, 4 and zero. And then we're subtracting Or other matrix which is gonna be -10. 10 -35 negative negative 15 and 25. All right. Now we're doing subtraction. When we're subtracting two matrices, who want to make sure that they have the same dimension? In this case we have three by three matrices. Both of them have the same dimension. So we're okay to go ahead and do the subtraction. Ok. And when we subtract two matrices we do it element wise. Okay, So we take an element from the first matrix And we subtract the corresponding element from the second matrix. Okay, so let's go ahead and see what that looks like. Alright, so let's choose first row. First column Okay. In the first matrix we have six. In the second matrix we have five. Okay, so in the first row, first column of our resulting matrix we're gonna get six. Our operation is subtract so six minus five. We can do the same thing for the second row. First column so we have 14 in the first matrix. In the second matrix we have zero. Okay, so in the same position, second row, first column in our resulting matrix we're gonna get 14 0. Alright, doing the same for the rest of the elements. The bottom left, we have 10 minus negative 10. Okay, column to roll one. We have negative four minus 10. Row two column 24 minus negative 30. And row three column two negative two minus five K. In In the last column, row one negative six minus negative 44 minus negative 15 and zero minus 25. So we're just subtracting each corresponding element. All right, let's move down so we can finish this off. Okay? And we are gonna get and again we are finding to a -5. B and the result is going to be six minus 51. 14 minus 0, 14. 10 minus negative 10 is going to be 20 negative four minus 10 is negative 14. 4 minus negative 30 is, 34 negative two minus five is negative seven negative six minus negative 40. Gonna give us 34 4 minus negative is 19 and finally zero minus 25 is gonna give us 25 sorry negative 25. Okay. And so the result of two a minus five B. Is a three by three matrix. It has the first row one negative 14 30 for the second row 14 34 19 and the third row 20 negative seven and negative 25. And that's going to correspond with answer D. Thanks everyone for watching. I hope this video helped see you in the next one.

In Exercises 9 - 16, find the following matrices: d. - 3A + 2B2 - 10 - 2 6 10 - 2A = 14 12 10 B = 0 - 12 - 44 - 2 2 - 5 2 - 2

Key Concepts

Matrix Addition and Scalar Multiplication

Matrix Dimensions

Order of Operations in Matrix Calculations

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