Linear Combination Of Matrices: Is B A Combination Of A1 & A2?

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Hey guys! Let's dive into a super important concept in linear algebra: linear combinations of matrices. Specifically, we're going to tackle the question of how to determine if a matrix B is a linear combination of two other matrices, A1 and A2. This is a fundamental idea that pops up in various areas of physics and engineering, so understanding it well is crucial. We'll break down the procedure step-by-step and show you how to verify your results. Get ready to put on your math hats!

Understanding Linear Combinations

Before we jump into the nitty-gritty of matrices, let's make sure we're all on the same page about what a linear combination actually is. In simple terms, a linear combination of vectors (or in our case, matrices) is what you get when you multiply each vector by a scalar (a number) and then add them all together. Think of it like mixing ingredients in a recipe – each ingredient (vector) is used in a certain amount (scalar) to create the final dish (linear combination).

Mathematically, if we have matrices A1 and A2, a linear combination of them can be written as:

B = c1 * A1 + c2 * A2

Where c1 and c2 are scalars. Our goal is to figure out if we can find values for c1 and c2 that make this equation true, meaning that B can indeed be expressed as a linear combination of A1 and A2.

Step-by-Step Procedure to Determine Linear Combinations

Okay, let's get down to business. Here's the procedure we'll follow to determine if matrix B is a linear combination of matrices A1 and A2. This process might seem a bit involved at first, but once you've done a few examples, it'll become second nature. Trust me, you got this!

Step 1: Set up the Equation

The very first thing we need to do is set up the equation that represents the linear combination. As we discussed earlier, this looks like:

B = c1 * A1 + c2 * A2

Write this equation out explicitly, plugging in the actual matrices B, A1, and A2. This is where the rubber meets the road, guys. Making sure you have the correct matrices in the right spots is paramount for avoiding headaches later on.

For example, let's say we have the following matrices:

A1 = [[1, 2], [3, 4]]

A2 = [[5, 6], [7, 8]]

B = [[9, 10], [11, 12]]

Our equation would then become:

[[9, 10], [11, 12]] = c1 * [[1, 2], [3, 4]] + c2 * [[5, 6], [7, 8]]

Step 2: Create a System of Linear Equations

Now comes the clever part. We're going to use the matrix equation we just set up to create a system of linear equations. Remember that for two matrices to be equal, their corresponding entries must be equal. This gives us a set of equations, one for each entry in the matrices.

Looking at our example, we'll distribute the scalars c1 and c2 and then equate the corresponding entries:

[[9, 10], [11, 12]] = [[c1, 2c1], [3c1, 4c1]] + [[5c2, 6c2], [7c2, 8c2]]

[[9, 10], [11, 12]] = [[c1 + 5c2, 2c1 + 6c2], [3c1 + 7c2, 4c1 + 8c2]]

This gives us the following system of equations:

  1. c1 + 5c2 = 9
  2. 2c1 + 6c2 = 10
  3. 3c1 + 7c2 = 11
  4. 4c1 + 8c2 = 12

Step 3: Solve the System of Equations

Alright, we've got our system of linear equations. Now it's time to solve for the scalars c1 and c2. There are several methods we can use to do this, such as substitution, elimination, or using matrices (Gaussian elimination or row reduction). The method you choose is really a matter of personal preference and what you find easiest.

Let's use the elimination method for our example. We can multiply the first equation by -2 and add it to the second equation to eliminate c1:

-2*(c1 + 5c2) = -2 * 9 -> -2c1 - 10c2 = -18

(-2c1 - 10c2) + (2c1 + 6c2) = -18 + 10

-4c2 = -8

c2 = 2

Now that we have c2, we can substitute it back into the first equation to solve for c1:

c1 + 5 * 2 = 9

c1 + 10 = 9

c1 = -1

So, we've found that c1 = -1 and c2 = 2.

Step 4: Verify the Solution

Okay, we've got potential values for c1 and c2. But before we declare victory, it's crucial to verify that these values actually work for all the equations in our system. This step is the most important to ensure that B is indeed a linear combination of A1 and A2. Think of it as the final boss level – you've gotta beat it to win the game!

Let's plug c1 = -1 and c2 = 2 back into our original equations:

  1. -1 + 5 * 2 = 9 -> 9 = 9 (Correct!)
  2. 2 * -1 + 6 * 2 = 10 -> 10 = 10 (Correct!)
  3. 3 * -1 + 7 * 2 = 11 -> 11 = 11 (Correct!)
  4. 4 * -1 + 8 * 2 = 12 -> 12 = 12 (Correct!)

Since our values for c1 and c2 satisfy all four equations, we can confidently say that B is a linear combination of A1 and A2.

What if the System Has No Solution?

Now, here's a crucial point: what happens if, when we try to solve the system of equations, we find that there's no solution? This means that there are no values for c1 and c2 that can satisfy all the equations simultaneously. If this happens, it means that B is not a linear combination of A1 and A2. It's a bit like trying to bake a cake without the right ingredients – you just can't do it.

This is a really important takeaway, guys. The ability to determine when a linear combination doesn't exist is just as important as knowing when it does exist. It helps us understand the relationships between matrices and the spaces they span.

Example Scenario and Verification

Let's solidify this with another quick example. Suppose we have the following matrices:

A1 = [[1, 0], [0, 1]]

A2 = [[0, 1], [1, 0]]

B = [[2, 3], [4, 5]]

Following our procedure:

Step 1: Set up the Equation

[[2, 3], [4, 5]] = c1 * [[1, 0], [0, 1]] + c2 * [[0, 1], [1, 0]]

Step 2: Create a System of Linear Equations

[[2, 3], [4, 5]] = [[c1, c2], [c2, c1]]

This gives us:

  1. c1 = 2
  2. c2 = 3
  3. c2 = 4
  4. c1 = 5

Step 3: Solve the System of Equations

Notice something? We have conflicting values for c1 and c2. From equations 1 and 4, we get c1 = 2 and c1 = 5, which is impossible. Similarly, equations 2 and 3 give us c2 = 3 and c2 = 4, another contradiction.

Step 4: Verify the Solution

Since we have conflicting values and there's no consistent solution for c1 and c2, we can conclude that B is not a linear combination of A1 and A2 in this case.

Importance in Physics

Now, you might be wondering,