## Scalar Multiplication: One is the Loneliest Scalar?

Multiply matrices by scalars to produce new matrices, for example, when all of the payoffs in a game are doubled.

We know what you're thinking. What the heck is a **scalar**? Scalar is a fancy-panties way of saying number. By number we mean any of these things:

You can multiply any of these babies through a matrix with ease. Take this matrix:

Her name is S. We need to find 3S. Here's how that looks:

3S is really just 3 multiplied through the matrix:

So

Moving on, there are more complicated examples of scalar multiplication. For example, we already know our matrix, S. This one is K:

How do we find this?

2K + S

Basically, we just plug the matrices in for their variables and go:

We multiply the 2 through the K matrix first:

Then we're ready to just add the two matrices. We remember how to do that, of course; we just add the entries in each location:

As long as your matrices are the same size for addition and subtraction purposes, it's all good.

## Practice:

Find the product: | |

We know that 2 × 2s can always be multiplied together, and we know that the outcome must always also be a 2 × 2. Let's visualize it: Looking at where the intersections are, we can see which row of the first matrix and which column of the second matrix collide to come up with the product's entries. Below, we can see that X marks the spot where row one of matrix one and column one of matrix two crash: We know those four numbers need to crunch into one; we also know we read rows left to right, like a book, and we read columns top to bottom. We multiply the first number from the first matrix's first row with the first number from the second matrix's first column (2)(4) = 8, we multiply the second number from the first matrix's row with the second number from the second matrix's column (3)(6) = 18, and we add those together: 26. Next? We rubberneck at the collision between row one of matrix one and column two of matrix two: Right as usual, King Friday. X marks the spot, and we do the same thing. We read rows left to right and we read columns top to bottom. We multiply the first number from the first matrix's first row with the first number from the second matrix's second column (2)(1) = 2, we multiply the second number from the first matrix's first row with the second number from the second matrix's second column (3)(0) = 0, we add those together: 2 + 0 = 2. We're so almost there. X marks the spot, and we do the same thing. We read the row left to right, we read the column top to bottom. This time we are doing the second row in the product, so we multiply the first number from the first matrix's second row with the first number from the second matrix's first column (1)(4) = 4, we multiply the second number from the first matrix's second row with the second number from the second matrix's first column (2)(6) = 12, we add those together: 4 + 12 = 16. We do the same thing one last time, this time with the second row in the first matrix and the second column in the second matrix. (1)(1) = 1, (2)(0) = 0, and 1 + 0 = 1:
Our answer is
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Find the product: | |

Whenever there's an operation that can be multiplied together and one of the matrices is an identity matrix, the answer is always the same as the non-identity matrix: | |

Find the product:
| |

We know they can be multiplied together because for matrices to be multiplied together the number of columns in the first matrix must be the same as the number of rows in the second. That's true here; there are three columns in the first matrix and three rows in the second matrix. We also know that the product matrix will have to be a 2 × 2 zero matrix because whenever there's an operation that is able to be multiplied together and one of the matrices is a zero matrix, the answer is always a zero matrix that has the same number of rows as the first matrix has and the same number of columns that the second matrix has. Therefore: | |

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Find CD.

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These cannot be multiplied together. For matrices to be able to be multiplied, the number of columns in the first matrix must be the same as the number of rows in the second. In this case there are 3 columns in the first and only 2 rows in the second.

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