Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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Matrix Operations - Problem 2

Carl Horowitz
Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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Multiplying a matrice by a scalar. So a scalar is basically the same thing as what we are used to talking about as a coefficient on a normal variable. So what a scalar is basically this number out in front so what we are looking at right now is 4C. So C is a matrix represented by this guy over here, so really what we have is 4 times 3 1 -6 4 make the matrice a little bit bigger 2 and 0.

So what the scalar does it just goes to every element in this matrix okay so this 4 gets distributed into everything. We just multiply every term by 4, so our new matrix is 4 times 3 12 4 -24 16 8 and 0. Just multiply that scalar by every single term in the matrix that we on your term.

Okay one other way we can use scalar multiples is in combinations with addition and subtraction. So let’s take a look at one of the problem where I asked you to find 2A minus 3B. So what 2A is saying is multiplying everything in 2, everything in the A by 2, what 3B is saying is take everything in B and multiply by 3 now we're just going to take the difference.

Okay when we are taking the difference we need to be aware that we have the same dimensions because we can’t combine, we can’t add or subtract matrices if their dimensions are different. Rows by columns this is 2 by 3 rows by columns 2 by 3 as well so our dimensions are the same so let’s go ahead and calculate this.

So 2A is basically taking everything in A multiplying that by 2. I’m going to save writing out 2 times this think we can do it in our head so we have 2 times every element there 0 2 8, 8 -4 14. Took 2 multiplied by everything in there. Okay, taking B multiplying everything by 3 9 -6 12, 0 3 6.

Now it is dealing with the subtraction of two matrices just combine like elements we can add same dimension, our dimensions are going to stay in the same of our end product and we're left with a 2 by 3 as well. I often draw this little underlines in just so I sort of can keep track of where everything is going to go and we just subtract. So 0 minus 9 -9, 2 minus -6 minus and minus turns into addition 8, 8 minus 12 -4, 8 minus 0 8 -4 minus 3 plus the negative turns into -7, 14 minus 6 is 8.

So whenever you are using the scaling multiple make sure it goes to every single element in the matrix addition and subtraction rules still hold as they would before.

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