Matrix Multiplication: $(AB)C$ Vs. $A(BC)$ Explained

Hey guys! Let's dive into the fascinating world of matrix multiplication. Specifically, we're going to explore the concept of finding $(AB)C$ and $A(BC)$. This is a fundamental concept in linear algebra, and understanding it is crucial for anyone working with matrices. We will work through the math step-by-step, making it super clear and easy to follow. Get ready to flex those math muscles! We'll use the matrices $A$, $B$, and $C$ provided in the prompt to work our way through this. The cool thing about matrix multiplication is that it's associative – meaning the order in which you group the matrices when multiplying doesn't change the final result. However, we will demonstrate the step-by-step approach to make sure you fully understand the process. Trust me, it's easier than it looks, and we'll break it down into manageable chunks. Let's get started!

Step 1: Defining the Matrices

First things first, let's clearly state the matrices we'll be working with. This ensures everyone is on the same page. We've got:

• $A = \begin{bmatrix} 5 & 9 & 7 \\ 0 & 3 & -2 \end{bmatrix}$
• $B = \begin{bmatrix} 5 & 9 \\ 0 & 3 \\ 0 & 2 \end{bmatrix}$
• $C = \begin{bmatrix} 3 & 4 \\ -1 & 0 \end{bmatrix}$

Notice the dimensions of these matrices. Matrix $A$ is a 2x3 matrix (2 rows and 3 columns), matrix $B$ is a 3x2 matrix (3 rows and 2 columns), and matrix $C$ is a 2x2 matrix (2 rows and 2 columns). This is important because the dimensions must be compatible for multiplication to be possible. We'll explore these compatibilities as we go, but for now, just keep them in mind. This problem highlights the associative property of matrix multiplication which states that the way you group matrices when multiplying them doesn't change the result, but we'll manually go through each one to prove that concept. Think of it like this: If you're multiplying numbers, it doesn't matter if you do (2 * 3) * 4 or 2 * (3 * 4); you get the same answer. Matrix multiplication follows the same principle, but with a few more steps involved. Let's start with computing $(AB)C$.

Step 2: Calculating $(AB)$

Now, let's calculate the product of matrices $A$ and $B$, which we'll denote as $(AB)$. Remember, when multiplying matrices, the number of columns in the first matrix must equal the number of rows in the second matrix. Here, matrix $A$ has 3 columns, and matrix $B$ has 3 rows, so we're good to go!

The matrix multiplication process goes like this: Each element in the resulting matrix is found by taking the dot product of a row from the first matrix and a column from the second matrix. Let's break it down:

$AB = \begin{bmatrix} 5 & 9 & 7 \\ 0 & 3 & -2 \end{bmatrix} \begin{bmatrix} 5 & 9 \\ 0 & 3 \\ 0 & 2 \end{bmatrix}$

• The element in the first row and first column of $AB$ is: $(5 * 5) + (9 * 0) + (7 * 0) = 25 + 0 + 0 = 25$
• The element in the first row and second column of $AB$ is: $(5 * 9) + (9 * 3) + (7 * 2) = 45 + 27 + 14 = 86$
• The element in the second row and first column of $AB$ is: $(0 * 5) + (3 * 0) + (-2 * 0) = 0 + 0 + 0 = 0$
• The element in the second row and second column of $AB$ is: $(0 * 9) + (3 * 3) + (-2 * 2) = 0 + 9 - 4 = 5$

Therefore,

$AB = \begin{bmatrix} 25 & 86 \\ 0 & 5 \end{bmatrix}$

Notice that the resulting matrix $AB$ is a 2x2 matrix, matching the number of rows of $A$ and the number of columns of $B$.

Step 3: Calculating $(AB)C$

Now we'll multiply the result we got in step 2 (which is $AB$) by matrix $C$. So we have to multiply a 2x2 matrix by another 2x2 matrix. Excellent! The number of columns in $AB$ (which is 2) matches the number of rows in $C$ (which is also 2), so we can proceed with the multiplication:

$(AB)C = \begin{bmatrix} 25 & 86 \\ 0 & 5 \end{bmatrix} \begin{bmatrix} 3 & 4 \\ -1 & 0 \end{bmatrix}$

Let's calculate the elements of the resulting matrix:

• The element in the first row and first column is: $(25 * 3) + (86 * -1) = 75 - 86 = -11$
• The element in the first row and second column is: $(25 * 4) + (86 * 0) = 100 + 0 = 100$
• The element in the second row and first column is: $(0 * 3) + (5 * -1) = 0 - 5 = -5$
• The element in the second row and second column is: $(0 * 4) + (5 * 0) = 0 + 0 = 0$

Thus,

$(AB)C = \begin{bmatrix} -11 & 100 \\ -5 & 0 \end{bmatrix}$

This is our final result for $(AB)C$.

Step 4: Calculating $BC$

Now, let's calculate $BC$. This will be an important step for finding $A(BC)$. We will be multiplying matrix $B$ (a 3x2 matrix) and matrix $C$ (a 2x2 matrix). This means the number of columns in $B$ matches the number of rows in $C$, so we can multiply.

$BC = \begin{bmatrix} 5 & 9 \\ 0 & 3 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} 3 & 4 \\ -1 & 0 \end{bmatrix}$

Let's go through the steps:

• The element in the first row and first column is: $(5 * 3) + (9 * -1) = 15 - 9 = 6$
• The element in the first row and second column is: $(5 * 4) + (9 * 0) = 20 + 0 = 20$
• The element in the second row and first column is: $(0 * 3) + (3 * -1) = 0 - 3 = -3$
• The element in the second row and second column is: $(0 * 4) + (3 * 0) = 0 + 0 = 0$
• The element in the third row and first column is: $(0 * 3) + (2 * -1) = 0 - 2 = -2$
• The element in the third row and second column is: $(0 * 4) + (2 * 0) = 0 + 0 = 0$

Thus,

$BC = \begin{bmatrix} 6 & 20 \\ -3 & 0 \\ -2 & 0 \end{bmatrix}$

Notice that the resulting matrix $BC$ is a 3x2 matrix, matching the number of rows of $B$ and the number of columns of $C$.

Step 5: Calculating $A(BC)$

Finally, we'll calculate $A(BC)$. Now we need to multiply matrix $A$ (a 2x3 matrix) by the result we just computed, $BC$ (a 3x2 matrix). Because the number of columns of A (3) matches the number of rows of BC (3), the multiplication is valid.

$A(BC) = \begin{bmatrix} 5 & 9 & 7 \\ 0 & 3 & -2 \end{bmatrix} \begin{bmatrix} 6 & 20 \\ -3 & 0 \\ -2 & 0 \end{bmatrix}$

Let's get the individual elements:

• The element in the first row and first column is: $(5 * 6) + (9 * -3) + (7 * -2) = 30 - 27 - 14 = -11$
• The element in the first row and second column is: $(5 * 20) + (9 * 0) + (7 * 0) = 100 + 0 + 0 = 100$
• The element in the second row and first column is: $(0 * 6) + (3 * -3) + (-2 * -2) = 0 - 9 + 4 = -5$
• The element in the second row and second column is: $(0 * 20) + (3 * 0) + (-2 * 0) = 0 + 0 + 0 = 0$

Therefore,

$A(BC) = \begin{bmatrix} -11 & 100 \\ -5 & 0 \end{bmatrix}$

Step 6: Comparing the Results and Conclusion

Congratulations, we've done it! We have successfully calculated both $(AB)C$ and $A(BC)$. Notice something interesting? The final matrices are:

• $(AB)C = \begin{bmatrix} -11 & 100 \\ -5 & 0 \end{bmatrix}$
• $A(BC) = \begin{bmatrix} -11 & 100 \\ -5 & 0 \end{bmatrix}$

They're identical! This demonstrates the associative property of matrix multiplication. The order in which we grouped the matrices didn't change the final answer. This is a crucial concept to understand because it simplifies calculations and lets you rearrange your matrix multiplications for easier computation. This is especially helpful in more complicated matrix calculations. Keep practicing, and you'll become a matrix multiplication master in no time! Matrix multiplication is used in many different fields, including computer graphics, data science, and physics, so it is a key skill to develop! Hope this helped, and happy calculating!