Solve Matrix Equation A + 5B = C: A Step-by-Step Guide

Hey guys! Ever stumbled upon a matrix equation that looks like a cryptic code? Well, today, we're cracking one wide open! We're diving into the fascinating world of matrices, specifically focusing on the equation A + 5B = C. This might seem intimidating at first glance, but trust me, with a bit of understanding, it's totally solvable. We'll break down each component, explore the underlying principles, and walk through the solution step-by-step. So, buckle up and get ready to unravel the mysteries of matrix algebra!

Setting the Stage: Understanding the Matrices

Before we jump into solving the equation, let's get acquainted with our key players: the matrices A, B, and C. These aren't just random collections of numbers; they're carefully organized arrays that follow specific rules. Each matrix has rows and columns, and the numbers within them hold particular significance. In our case, we have:

• Matrix A: A = ${\begin{bmatrix} 3 & 5 \\ a & a \end{bmatrix}}$
• Matrix B: B = ${\begin{bmatrix} 4 & b \\ a & 9 \end{bmatrix}}$
• Matrix C: C = ${\begin{bmatrix} 2b & 29 \\ 14 & 45 \end{bmatrix}}$

Notice that each matrix is a 2x2 matrix, meaning it has two rows and two columns. The elements within the matrices are a mix of constants (like 3, 5, 4, 9, 29, and 45) and variables (represented by 'a' and 'b'). Our mission, should we choose to accept it, is to find the values of these variables that satisfy the equation A + 5B = C. This involves understanding matrix addition and scalar multiplication, which we'll explore in the next section.

Think of matrices as organized tables of data. They're used in all sorts of fields, from computer graphics and image processing to economics and physics. The beauty of matrices lies in their ability to represent complex relationships in a compact and manageable form. The values 'a' and 'b' within our matrices represent unknown quantities, and finding them will give us a complete picture of the relationships between these matrices. So, let's dive deeper into the operations we need to perform to solve for these unknowns!

Matrix Addition: Combining Like Elements

The first operation we need to understand is matrix addition. It's a pretty intuitive process: you simply add the corresponding elements of the matrices together. However, there's a crucial rule: you can only add matrices that have the same dimensions. In other words, they must have the same number of rows and columns. Luckily, our matrices A, B, and C are all 2x2 matrices, so we're good to go!

For example, if we have two matrices:

${P = \begin{bmatrix} p_{11} & p_{12} \\ p_{21} & p_{22} \end{bmatrix}}$ and ${Q = \begin{bmatrix} q_{11} & q_{12} \\ q_{21} & q_{22} \end{bmatrix}}$

Then their sum, P + Q, is calculated as:

${P + Q = \begin{bmatrix} p_{11} + q_{11} & p_{12} + q_{12} \\ p_{21} + q_{21} & p_{22} + q_{22} \end{bmatrix}}$

See? It's like adding apples to apples and oranges to oranges. You add the element in the first row and first column of P to the element in the first row and first column of Q, and so on. This element-wise addition is the key to matrix addition. Now, let's apply this to our specific matrices in the equation A + 5B = C.

Scalar Multiplication: Scaling Up the Matrix

The second operation we need to master is scalar multiplication. This involves multiplying a matrix by a scalar, which is just a regular number (like 5 in our equation). Scalar multiplication is even simpler than matrix addition. You just multiply every element in the matrix by the scalar. It's like scaling the entire matrix up or down by a certain factor.

If we have a matrix:

${R = \begin{bmatrix} r_{11} & r_{12} \\ r_{21} & r_{22} \end{bmatrix}}$

And a scalar 'k', then the scalar multiplication kR is calculated as:

${kR = \begin{bmatrix} kr_{11} & kr_{12} \\ kr_{21} & kr_{22} \end{bmatrix}}$

So, every element in the matrix gets multiplied by 'k'. In our equation A + 5B = C, we need to perform scalar multiplication on matrix B by the scalar 5. This means we'll multiply each element in matrix B by 5 before we add it to matrix A. Understanding scalar multiplication is crucial because it allows us to manipulate the magnitude of the values within the matrix, which is a fundamental tool in matrix algebra.

Cracking the Code: Solving the Equation A + 5B = C

Alright, guys, we've got the basics down! Now, let's put our knowledge into action and solve the equation A + 5B = C. This is where the fun really begins! We'll break down the equation step-by-step, performing the necessary operations and ultimately finding the values of 'a' and 'b'.

Step 1: Scalar Multiplication of Matrix B

The first thing we need to do is multiply matrix B by the scalar 5. Remember, this means multiplying each element in B by 5:

${5B = 5 \begin{bmatrix} 4 & b \\ a & 9 \end{bmatrix} = \begin{bmatrix} 20 & 5b \\ 5a & 45 \end{bmatrix}}$

So, we've successfully scaled up matrix B by a factor of 5. Now we have a new matrix, 5B, which we can use in the next step.

Step 2: Matrix Addition: A + 5B

Next, we need to add matrix A to the scaled matrix 5B. Remember, we add corresponding elements together:

${A + 5B = \begin{bmatrix} 3 & 5 \\ a & a \end{bmatrix} + \begin{bmatrix} 20 & 5b \\ 5a & 45 \end{bmatrix} = \begin{bmatrix} 3+20 & 5+5b \\ a+5a & a+45 \end{bmatrix} = \begin{bmatrix} 23 & 5+5b \\ 6a & a+45 \end{bmatrix}}$

We've now performed the matrix addition, resulting in a new matrix that represents the left-hand side of our equation. This matrix is equal to matrix C, which is our next focus.

Step 3: Equating A + 5B to C

Now comes the crucial step: equating the result of A + 5B to matrix C. This is where we'll set up equations to solve for our unknowns, 'a' and 'b'. We have:

${\begin{bmatrix} 23 & 5+5b \\ 6a & a+45 \end{bmatrix} = \begin{bmatrix} 2b & 29 \\ 14 & 45 \end{bmatrix}}$

For two matrices to be equal, their corresponding elements must be equal. This gives us a set of equations:

• Equation 1: 23 = 2b
• Equation 2: 5 + 5b = 29
• Equation 3: 6a = 14
• Equation 4: a + 45 = 45

We now have a system of equations that we can solve to find the values of 'a' and 'b'. Let's tackle them one by one!

Step 4: Solving for 'a' and 'b'

Let's start with the easiest equations to solve. From Equation 3, 6a = 14, we can solve for 'a':

${a = \frac{14}{6} = \frac{7}{3}}$

So, we've found the value of 'a'! Now, let's move on to solving for 'b'. We can use either Equation 1 or Equation 2. Let's use Equation 1, 23 = 2b:

${b = \frac{23}{2}}$

And there you have it! We've found the values of both 'a' and 'b'. We can double-check our solution by plugging these values back into the original equation and making sure everything balances out. This step-by-step approach, breaking down the problem into smaller, manageable parts, is the key to solving complex matrix equations. So, let's recap our journey and see the final result!

The Grand Finale: The Solution Unveiled

Guys, we did it! We successfully navigated the world of matrices, performed the necessary operations, and solved for the unknowns in the equation A + 5B = C. It's like cracking a secret code, right? We started by understanding the basics of matrix addition and scalar multiplication, then applied those principles to our specific equation. By equating corresponding elements, we created a system of equations that we could solve for 'a' and 'b'.

Our solution is:

• a = 7/3
• b = 23/2

This means that when we substitute these values back into the original matrices A, B, and C, the equation A + 5B = C holds true. This is a powerful demonstration of how matrix algebra can be used to represent and solve complex relationships between sets of numbers.

So, what's the takeaway from all this? Well, matrix equations might seem intimidating at first, but with a solid understanding of the fundamental operations and a systematic approach, they become much more manageable. And remember, the principles we've learned today can be applied to a wide range of problems in mathematics, science, and engineering. Keep practicing, keep exploring, and you'll become a matrix master in no time! Awesome work, everyone!