Matrix Operations: A + B, A - B, 3A, And -8B

Hey guys! Let's dive into some matrix operations! We've got two matrices, A and B, and we're going to tackle some basic operations like addition, subtraction, and scalar multiplication. This stuff is super fundamental in linear algebra, so let's break it down step by step. Whether you're studying for an exam or just brushing up on your math skills, this guide will help you understand how to perform these operations with ease. So, grab your pencils and paper, and let’s get started!

Problem Statement

Okay, so here’s the deal. We are given two matrices:

$A=\begin{bmatrix}7 & -9 & -1 \\ 5 & 15 & -3 \\ -11 & 3 & 10\end{bmatrix}$ and $B=\begin{bmatrix}-2 & 9 & -2 \\ -10 & 0 & -13 \\ -2 & -12 & -7\end{bmatrix}$.

Our mission, should we choose to accept it (and we do!), is to find the following:

a. $A + B$ b. $A - B$ c. $3A$ d. $-8B$

Sounds like a plan? Awesome! Let’s jump into each operation one by one.

a. Finding A + B

First up, we need to find $A + B$. Matrix addition is pretty straightforward. All we have to do is add the corresponding elements in matrices A and B. This means we add the elements in the same positions in each matrix. Remember, though, that matrix addition is only possible if the matrices have the same dimensions. Luckily for us, both A and B are 3x3 matrices, so we’re good to go!

So, how do we do it? We add each element like this:

$(A + B)_{ij} = A_{ij} + B_{ij}$

Where $A_{ij}$ represents the element in the i-th row and j-th column of matrix A, and similarly for B. Let's break it down element by element:

• Top-left corner: $7 + (-2) = 5$
• First row, middle: $-9 + 9 = 0$
• First row, last: $-1 + (-2) = -3$
• Second row, first: $5 + (-10) = -5$
• Second row, middle: $15 + 0 = 15$
• Second row, last: $-3 + (-13) = -16$
• Third row, first: $-11 + (-2) = -13$
• Third row, middle: $3 + (-12) = -9$
• Bottom-right corner: $10 + (-7) = 3$

Putting it all together, we get:

$A + B = \begin{bmatrix}5 & 0 & -3 \\ -5 & 15 & -16 \\ -13 & -9 & 3\end{bmatrix}$

Ta-da! We've just nailed our first matrix operation. Let's keep the momentum going!

b. Finding A - B

Next on our list is finding $A - B$. Just like addition, matrix subtraction involves operating on corresponding elements. Instead of adding, though, we'll be subtracting the elements of matrix B from matrix A. Again, the matrices need to have the same dimensions, which they do, so we can proceed.

The formula looks pretty similar to addition:

$(A - B)_{ij} = A_{ij} - B_{ij}$

Let's go through it element by element:

• Top-left corner: $7 - (-2) = 7 + 2 = 9$
• First row, middle: $-9 - 9 = -18$
• First row, last: $-1 - (-2) = -1 + 2 = 1$
• Second row, first: $5 - (-10) = 5 + 10 = 15$
• Second row, middle: $15 - 0 = 15$
• Second row, last: $-3 - (-13) = -3 + 13 = 10$
• Third row, first: $-11 - (-2) = -11 + 2 = -9$
• Third row, middle: $3 - (-12) = 3 + 12 = 15$
• Bottom-right corner: $10 - (-7) = 10 + 7 = 17$

So, the result is:

$A - B = \begin{bmatrix}9 & -18 & 1 \\ 15 & 15 & 10 \\ -9 & 15 & 17\end{bmatrix}$

Awesome! We’re two for two. Subtraction? Check! Now, let’s move on to scalar multiplication. This is where things get a little different, but still totally manageable.

c. Finding 3A

Now we're moving onto scalar multiplication, which sounds fancy but is actually super straightforward. When we multiply a matrix by a scalar (a regular number), we're just multiplying every single element in the matrix by that number. No fuss, no muss!

So, we want to find $3A$. This means we multiply each element of matrix A by 3.

$(3A)_{ij} = 3 imes A_{ij}$

Let's get to it:

• Top-left corner: $3 imes 7 = 21$
• First row, middle: $3 imes -9 = -27$
• First row, last: $3 imes -1 = -3$
• Second row, first: $3 imes 5 = 15$
• Second row, middle: $3 imes 15 = 45$
• Second row, last: $3 imes -3 = -9$
• Third row, first: $3 imes -11 = -33$
• Third row, middle: $3 imes 3 = 9$
• Bottom-right corner: $3 imes 10 = 30$

Putting it all together:

$3A = \begin{bmatrix}21 & -27 & -3 \\ 15 & 45 & -9 \\ -33 & 9 & 30\end{bmatrix}$

Sweet! We've just multiplied a matrix by a scalar. Feels pretty good, right? Now, let's tackle the last one, which is another scalar multiplication, but this time with a negative scalar. Don’t worry; it’s the same principle, just a little twist!

d. Finding -8B

Last but not least, we need to find $-8B$. This is another scalar multiplication, but this time we’re multiplying matrix B by -8. Remember, multiplying by a negative number just means we’re changing the sign of each element as we multiply. Same process as before, but with an extra little sign check!

So, we need to multiply each element of matrix B by -8:

$(-8B)_{ij} = -8 imes B_{ij}$

Let’s break it down:

• Top-left corner: $-8 imes -2 = 16$
• First row, middle: $-8 imes 9 = -72$
• First row, last: $-8 imes -2 = 16$
• Second row, first: $-8 imes -10 = 80$
• Second row, middle: $-8 imes 0 = 0$
• Second row, last: $-8 imes -13 = 104$
• Third row, first: $-8 imes -2 = 16$
• Third row, middle: $-8 imes -12 = 96$
• Bottom-right corner: $-8 imes -7 = 56$

Putting it all together, we get:

$-8B = \begin{bmatrix}16 & -72 & 16 \\ 80 & 0 & 104 \\ 16 & 96 & 56\end{bmatrix}$

And there you have it! We’ve conquered all the operations. High five!

Conclusion

Alright, guys, we did it! We successfully found $A + B$, $A - B$, $3A$, and $-8B$. Hopefully, this step-by-step breakdown made these matrix operations crystal clear. Remember, practice makes perfect, so feel free to try out some more examples on your own. Understanding these basics is crucial for diving into more advanced linear algebra topics. Keep up the awesome work, and you’ll be a matrix master in no time! Whether you're acing your exams or just expanding your math horizons, you're on the right track. Keep exploring, keep learning, and most importantly, keep having fun with math! You got this!