Music During a marching band’s half-time show, the band...

Music During a marching band’s half-time show, the band members generally line up in such a way that a common shape is recognized by the fans. For example, as illustrated in the figure, a band might form a letter T, where an x represents a member of the band. As the music is played, the band will either create a new shape or rotate the original shape. In doing this, each member of the band will need to move from one point on the field to another. For larger bands, keeping track of who goes where can be a daunting task. However, it is possible to use matrix inverses to make the process a bit easier. The entire process is calculated by knowing how three band members, all of whom cannot be in a straight line, will move from the current position to a new position. For example, in the figure, we can see that there are band members at $(50,0),(50,15),$ and $(45,20) .$ We will assume that these three band members move to $(40,10),(55,10),$ and $(60,15),$ respectively. Source: The College Mathematics Journal. a. Find the inverse of $B=\left[\begin{array}{ccc}{50} & {50} & {45} \\ {0} & {15} & {20} \\ {1} & {1} & {1}\end{array}\right]$ b. Find $A=\left[\begin{array}{ccc}{40} & {55} & {60} \\ {10} & {10} & {15} \\ {1} & {1} & {1}\end{array}\right] B^{-1}$ c. Use the result of part b to find the new position of the other band members. What is the shape of the new position? (Hint: Multiply the matrix $A$ by a $3 \times 1$ column vector with the first two components equal to the original position of each band member and the third component equal to $1 .$ The new position of the band member is in the first two components of the product.)

Music During a marching band’s half-time show, the band members generally line up in such a way that a common shape is recognized by the fans. For example, as illustrated in the figure, a band might form a letter T, where an x represents a member of the band. As the music is played, the band will either create a new shape or rotate the original shape. In doing this, each member of the band will need to move from one point on the field to another. For larger bands, keeping track of who goes where can be a daunting task. However, it is possible to use matrix inverses to make the process a bit easier. The entire process is calculated by knowing how three band members, all of whom cannot be in a straight line, will move from the current position to a new position. For example, in the figure, we can see that there are band members at $(50,0),(50,15),$ and $(45,20) .$ We will assume that these three band members move to $(40,10),(55,10),$ and $(60,15),$ respectively. Source: The College Mathematics Journal.
a. Find the inverse of $B=\left[\begin{array}{ccc}{50} & {50} & {45} \\ {0} & {15} & {20} \\ {1} & {1} & {1}\end{array}\right]$
b. Find $A=\left[\begin{array}{ccc}{40} & {55} & {60} \\ {10} & {10} & {15} \\ {1} & {1} & {1}\end{array}\right] B^{-1}$
c. Use the result of part b to find the new position of the other band members. What is the shape of the new position? (Hint: Multiply the matrix $A$ by a $3 \times 1$ column vector with the first two components equal to the original position of each band member and the third component equal to $1 .$ The new position of the band member is in the first two components of the product.)

Key Concepts

Matrix Inversion

Matrix inversion is the process of finding a matrix that, when multiplied with the original matrix, yields the identity matrix. This operation is fundamental in solving systems of linear equations, enabling one to 'undo' a linear transformation. In problems involving coordinate transformation, inverting a matrix allows for mapping between old and new coordinate systems.

Affine Transformation

An affine transformation is a type of function that preserves points, straight lines, and planes. It consists of a linear transformation followed by a translation. This property ensures that relationships such as parallelism are maintained, which is important when calculating the new positions of objects based on a known transformation. Affine transformations are commonly used in applications involving geometric manipulations.

Homogeneous Coordinates

Homogeneous coordinates are an augmented coordinate system that introduces an extra dimension, which enables the representation of translations as linear operations along with rotations and scalings. In this system, points are typically expressed with an additional coordinate, often set to 1, making it possible to represent affine transformations using matrix multiplication. This method is crucial in computer graphics and geometric computations.

Transformation Matrix

A transformation matrix is a matrix that, when multiplied by a coordinate vector, transforms the vector from one coordinate space to another. In the context of mapping positions, a transformation matrix is computed from known corresponding points, facilitating the conversion of every point in the original configuration into its new position. Determining this matrix involves solving equations derived from these fixed correspondences.

Point Correspondence

Point correspondence refers to the process of matching points in one coordinate system to their images in another coordinate system. By using at least three non-collinear points in the source and target positions, one can uniquely determine an affine transformation. This concept is essential in many applications, including computer vision and pattern recognition, as it underlies the method for computing the transformation matrix.

Transcript

00:01 Hey, for this problem, we are going to be examining the positions of people in a marching band.

00:07 We're going to do this in a few steps.

00:10 First, we have a matrix that they've given that they're calling b.

00:15 And as you can see, if you look at how b is set up, the third row is just ones.

00:20 There's one person in each place.

00:22 But the top and the middle rows, first and second rows, show the positions of three different band members.

00:29 If you kind of think of the football field as a grid, you've got one band member at 50, 0, 1 at 5015, and one at 4520, 3 non -colinear band members.

00:41 And we've been asked to start with, we're going to find the inverse of b.

00:44 So if this is b that we're showing here, i've copied b on the screen, to find the inverse, we're going to remove that second bracket, and we're going to augment it.

00:56 We're on the right -hand side.

00:57 We're going to copy the identity matrix.

01:01 And we're going to use row manipulation to make the left -hand side look like the identity matrix, and when that's done, the right -hand side will be the inverse.

01:11 So in order to do this, the first step, we're going to leave the first row exactly the way it is.

01:21 And my goal is, since the identity matrix has an element in the first row -first column, we'll leave that first row -first column just as it is, and then we'll make everything else in that first column equal to zero.

01:34 Well, one thing that makes our job a little easier, there's already a zero in the second row.

01:39 So i can leave those numbers alone.

01:42 I don't have to do any manipulation on those.

01:45 But what about the third row? i do have a one here, and i need to change it to a zero.

01:50 So my row manipulation will be, i'm going to take the opposite of the first row, plus 50 times the third row.

01:58 And those will be my new third row entries all the way across.

02:02 And when i do that, that's going to give me a new third row of 0 .05, negative 1 ,050.

02:13 Hey, first row is done, or the first column.

02:17 Now, the second column, there is an element in the second row, second column position of the identity matrix.

02:27 So i'm going to leave the second row alone, copy it just the way it is.

02:32 And i want to have zeros everywhere else in the second column.

02:38 Well, again, we got kind of lucky.

02:40 The third row already has a zero there.

02:42 So no manipulation needed.

02:44 We can just copy it.

02:46 But what about the first row? i have a 50 here that i need to get rid of.

02:51 So what i'm going to end up doing is i'm going to take negative 10 times the second row.

02:58 And i'm going to add three times the first row.

03:02 Doing that gives me a new top row.

03:05 And my new top row is going to be 150, 0, negative 65, 3, negative 10, 0.

03:15 Hey, one more iteration should have all of our zeros in the proper place.

03:21 There is an element in the third row, third column position of the identity matrix.

03:27 So we're going to leave the third row, just the way it is, and deal with that third column.

03:37 Our goal is to make every element in the third column, except for that third row third column spot to equal zero.

03:44 So what do i need to do? well, first of all, let's look at the first row.

03:48 I have a negative 65.

03:50 So i'm going to take 13 times the third row and add it to the first row.

03:55 Doing that gives me a new top row.

03:59 150, 0, 0, negative 10, negative 10, 615.

04:08 Hey, now my second row, i have a 20 that i need to get rid of.

04:12 So i will take the opposite of four times the third row, and i'll add that to the second row.

04:18 That gives me a new second row of 0, 15, 0, 4 -1, negative 200.

04:27 Okay, we are just about done.

04:29 I'm going to just scroll a little, give us a little bit of space.

04:33 It almost looks right.

04:34 All the zeros are in the right spot.

04:36 I just need ones on that diagonal.

04:39 So i'm going to divide everything in the top row by 150, everything in the second row by 15, and everything in the third row by 5.

04:49 And that will give me my new inverse matrix.

04:54 So the left -hand side is now the identity matrix.

04:59 And the right -hand side is negative 1 -15th, negative -1 -15ths, 13 -thirds, second row is 415th, 1 .15th, negative 40 thirds.

05:18 And the bottom row is negative 1 5th, 0, and 10.

05:23 So this is the inverse of matrix b.

05:28 Great.

05:29 So we're almost ready to find our new band positions.

05:32 But first, we need to compare this b.

05:36 We have to multiply it with our grid a.

05:39 And a is going to have the positions to which they will be moving.

05:44 So this is going to be a multiplication.

05:49 I've got a times the inverse of b.

05:52 So first, before we look at where a comes from, let me just recopy this inverse matrix.

05:59 I need to be able to see that top row...

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